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Telangana TSBIE TS Inter 2nd Year Chemistry Study Material 1st Lesson Solid State Textbook Questions and Answers.

TS Inter 2nd Year Chemistry Study Material 1st Lesson Solid State

Very Short Answer Questions (2 Marks)

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
An amorphous solid is a substance whose constituent particles do not possess a regular orderly arrangement. Ex : Glass, plastics, rubber, starch etc.

Question 2.
What makes a glass different from a solid such as quartz?
Answer:
Glass is an amorphous form of silica and do not posses long range regular structure. It do not posses sharp melting point.

Quartz is a crystalline form of silica and have long range regular structure. It has sharp melting point. If the molten quartz Is cooled rapidly it becomes glass.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
a) Tetra phosphorous deoxide
b) Graphite
c) Brass
d) Ammonium phosphate (NH₄)₃ PO₄
e) SIC
f) Rb
g) I₂
h) LiBr
i) P₄
j) Si
k) plastic.
Answer:

Question 4.
What is meant by the term coordination number?
Answer:
The number of nearest neighbours of a particle in a packing is called coordination number, eg : in hexagonal close packing coordination number is 12.

Question 5.
What is the coordination number of atoms in cubic close – pack structure ?
Answer:
In cubic close – packed structure coordi-nation number is 12.

Question 6.
What is the coordination number of atoms in a body – centered cubic structure ?
Answer:
In a body centred cubic structure coordi-nation number is 8.

Question 7.
Stability of a crystal is reflected in the magnitude of Its melting point. Comment.
Answer:
If the interparticle forces are stronger more energy is required to separate them. The crystals in which jilter particle forces are stronger are stable. More stable crystals have higher melting points than less stable crystals. Hence we can say that stability of a crystal is reflected in magnitude of its melting point.

Question 8.
How are the intermolecular forces among the molecules affect the melting point ?
Answer:
If the intermolecular forces among the molecules are stronger, more energy is required to separate them. Such solids require more amount of energy to melt. Hence they have higher melting points. If the intermolecular forces are weak, less energy is sufficient to melt them. Such solids have low melting points.

Question 9.
How do you distinguish between hexagonal close packing and cubic close packing structures ?
Answer:
In hexagonal close – packing every third layer is similar to the first layer. Thus there is ABABAB ………. arrangement. In cubic close packing every fourth layer is similar to first, there is ABC ABC…. arrangement.

Question 10.
How do you distinguish between crystal lattice and unit cell ?
Answer:
A crystal lattice may be defined as a regular three dimensional arrangement of identical points in space.
A unit cell may be defined as a three dimensional group of lattice points that generates the whole lattice by translation or stacking. It is the smallest repetitive portion of the crystal lattice.

Question 11.
How many lattice points are there in one unit cell of face – centered cubic lattice ?
Answer:
A face centred cubic unit cell has lattice points at the corners of the cube and at face centres. So it contain 8 + 6 = 14 lattice points.

Question 12.
How many lattice points are there in one unit cell of face centered tetragonal lattice ?
Answer:
A face centred tetragonal unit cell has 4 atoms per unit cell.

Question 13.
How many lattice points are there in one unit cell of body centered cubic lattice ?
Answer:
A body centred cubic unit cell has lattice points at the corners of the cube and at the body centre. So the total number of lattice points in one unit cell of a body centred cubic lattice is 9.

Question 14.
What is a semiconductor ?
Answer:
Semiconductors are solids with conductivi-ties in the intermediate range from 10^-6 to 10⁴ ohm^-1 m^-1. Their conductivity increases with increase in temperature.

Question 15.
What is Schottky defect ? (AP & TS ’15)
Answer:
The Schottky defect is simply a vacancy defect in ionic solids. This is due to missing of equal number of positive and negative ions. It is more common in ionic compounds

with high coordination number and in the ionic scJlids which contain positive and negative ions of equal size. This defect decreases density of the substance, eg. NaCl, KCl, CsCl, AgBr.

Question 16.
What is Frenkel defect ? (AP & TS ’15) (Mar. 2018 . TS)
Answer:
Frenkel defect arises when the smaller ion of ionic solids is dislocated from its normal site to an interstitial site. Frenkel defects

are common in ionic solids having low coordination number and large difference in the sizes of positive and negative ions. The density of the substance do not change, eg. ZnS, AgCl, AgBr and Agl.

Question 17.
What is interstitial defect ?
Answer:
If the constituent particles of a solid occupies the interstitial sites instead of lattice points, it is called interstitial defect. It is observed in non – ionic solids.

Question 18.
What are F – centres ?
Answer:
A compound may have excess metal ion if a negative ion is absent from its lattice site, leaving a hole which is occupied by electron to maintain electrical neutrality. The holes occupied by electrons are called F – centres.

Question 19.
Explain ferromagnetism with suitable example.
Answer:
The substances which are strongly attracted by external magnetic field are called ferromagnetic substances. They contain unpaired electrons, eg. iron, cobalt, nickel.

Question 20.
Explain paramagnetism with suitable example.
Answer:
Substances which are weakly attracted by a magnetic field are called paramagnetic substances. This is due to the presence of one or more unpaired electrons. Ex : O₂, Cu⁺², Fe⁺³, Cr⁺³ etc.

Question 21.
Explain Ferromagnetism with suitable example.
Answer:
When the magnetic moments of the domains in a substance are aligned in parallel and anti parallel directions in unequal numbers, then they are said to exhibit Fern magnetism. They are weakly attracted by magnetic field.
Eg: Fe₃O₄, MgFe₂O₄, NiFe₂O₄ etc.

Question 22.
Explain Antiferro magnetism with suitable example.
Answer:
When the magnetic moments of the domain in a substance are aligned in opposite directions and cancel out each others moment, then they are said to exhibit Anti ferromagnetism.

Question 23.
Why x-ray are needed to probe the crystal structures ?
Answer:
According to a fundamental principle of optics, the wavelength of light used to observe an object must be no greater than twice the length of the object itself. Since atoms have diameters of around 2 × 10^-10 m and the visible light detected by our eyes has wavelength of 4 – 7 × 10^-7 m, it is impossible to see atoms using even the tinest optical microscope. Hence, to see atoms we must use light wdth a wave length of approximately 10^-10 m, which is in the x – ray region of electromagnetic spectrum.

Short Answer Questions (4 Marks)

Question 24.
Explain similarities and differences between metallic and ionic crystals.
Answer:
Similarities:

1. In metallic and Ionic solids the constituent particles are in regular arrangement. So both have a regular shape.
2. In both solids the bond is non – directional.
3. Both solids are hard.

Differences:

1. In metallic crystals the constituent particles are positive ions immersed in sea of mobile electrons but in ionic solids the constituent particles are positive and negative ions.
2. In metallic crystals binding force is metallic bond whereas in ionic crystals is ionic bond.
3. Metallic crystals are conductors of electricity while ionic crystals are insulators in solid state and act as conductors in molten state and aqueous solutions.
4. Metallic crystals are malleable and ductile while ionic crystals are brittle.

Question 25.
Explain why ionic solids are hard and brittle.
Answer:
Ionic solids are hard because in ionic solids, ions are held together by strong electrostatic attractive forces. If sufficient force is applied on ionic crystals, the ions with similar charges come close due to displacement. Thus the crystal breaks due to repulsions between similar charges.
Hence Ionic solids are brittle.

Question 26.
Calculate the efficiency of packing In case of a metal of simple cubic crystal.
Answer:
Packing efficiency in simple cubic lattice:
In a simple cubic lattice the atoms are located only on the corners of the cube. The particles touch each other along the edge.
Thus, the edge length or šide of the cube
‘a’, and the radius of each particle, r are related as a = 2r

Simple cubic unit cell. The spheres are in contact with each other along the edge of the cube.
The volume of the cubic unit cell = a³
= (2r)³ = 8r³
Since a simple cubic unit cell contains only 1 atom.
The volume of the occupied space = \(\frac{4}{3} \pi r^3\)
Packing efficiency

Question 27.
Calculate the efficiency of packing in case of a metal of body – centred cubic crystal.
Answer:
From Fig it is clear that the atom at the centre will be in touch with the other two atoms diagonally arranged.
In ΔEFD
b² = a² + a² = 2a²
b = \(\sqrt{2} a\)
Now in ΔAFD
c² = a² + b² = a² + 2a² = 3a²
c = \(\sqrt{3} a\)

(sphere along the body diagonal are shown with solid boundaries).
The length of the body diagonal c is equal to 4r.

where r is the radius of the sphere (atom), as all the three špheres along the diagonal touch each other.
Therefore, \(\sqrt{3} a\) = 4r = a = \(\frac{4 r}{\sqrt{3}}\)
Also we can write, r = \(\frac{\sqrt{3}}{4} a\)
In this type of structure, total number of atoms is 2 and their volume is 2 × \(\left(\frac{4}{3}\right) \pi r^3\)
Volume of the cube, a³ will be equal to
\(\left(\frac{4}{\sqrt{3}} \mathrm{r}\right)^3\) or a³ = \(\left(\frac{4}{\sqrt{3}} \mathrm{r}\right)^3\)
Therefore
Packing efficiency

Question 28.
Calculate the efficiency of packing in case of face – centered cubic crystal.
Answer:
Both types of close packing (hep and ccp) are equally efficient. Let us calculate the efficiency of packing in ccp structure. In Fig. let the unit cell edge length be a’ and face diagonal AC = b.
In Δ ABC
AC² = b² = BC² + AB² = a² + a² = 2a² (or)
b = \(\sqrt{2}\)
If r is the radius of the sphere, we find b = 4r = \(\sqrt{2 a}\) (or) a = \(\frac{4 r}{\sqrt{2}}\) = \(2 \sqrt{2 r}\)
(we can also write, r = \(\frac{a}{2 \sqrt{2}}\))

Cubic close packing other sides are not provided with spheres for sake of clarity

We know, that each unit cell in ccp structure, has effectively 4 spheres. Total volume of four spheres is equal to 4 × (4/3) πr³ and volume of the cube is a³ or \((2 \sqrt{2} r)^3\).
Therefore,
Packing efficiency
Volume occupied by four spheres

The same packing efficiency occurs in hep arrangement also.

Question 29.
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body – centre. What is the formula of the compound ? What are the coordination numbers of P and Q ?
Answer:
The atom at corner makes \(\frac{1}{8}\) contribution while atom at body centre makes 1 contri-bution to the unit cell.
No. of atoms of Q per unit cell
= 8 (at corners) × \(\frac{1}{8}\) = 1
No. of atoms of P per unit cell
= 1 (at body centre) × 1 = 1
∴ Formula of the compound is PQ.
The atom at the body centre would be in contact with all the atoms at the corners. Hence the coordination number of P would be 8. Similarly coordination number of Q is also 8.

Question 30.
If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’, derive relation between r and R.
Answer:
The octahedral void can be represented as follows.

Though an octahedral void is surrounded by six spheres only four are shown in the figure. The spheres present above and below the void are omitted.
Let the length of one side of the square ABCD is a cm.
In right angled ΔABC,
The diagonal AC
= \(\sqrt{\mathrm{AB}^2+\mathrm{BC}^2}\) = \(\sqrt{a^2+a^2}\) = \(\sqrt{2} a\)
But AC = 2R + 2r ∴ 2R + 2r = \(\sqrt{2} a\)
But a = 2R ∴ 2R + 2r = \(\sqrt{2} .2 \mathrm{R}\)
Dividing by 2R we get 1 + \(\frac{r}{R}\) = \(\sqrt{2}\)
\(\frac{r}{R}\) = \(\sqrt{2}\) – 1 = 1.414 – 1 = 0.414

Question 31.
Describe the two main types of semiconductors and contrast their conduction mechanism. (Mar. ’19, ’18. AP)
Answer:
Semiconductors are o two types. They are
1) Intrinsic semiconductors
2) Extrinsic semiconductors.

1) Intrinsic semiconductors: In this type of semiconductors, electrical conductivity increases with rise in temperature, since more electrons can jump to conduction band. Eg :Silicon and Germanium.

2) Extrinsic semiconductors: In this type of semiconductors, conductivity is increased by adding a appropriate amount of suitable impurity. This process is called Doping. Doping can be done with an impurity which is electron rich or electron deficient. These are of two types.

a) n – type semiconductors : The semi-conductors in which the increase in conductivity is due to the negatively charged electron are called n-type semiconductors.
Eg : Silicon and germanium of group 14 doped with a group 15 elements like P or As form n – type semiconductors.

b) p – type semiconductors : The semi-conductors in which the increase in conductivity is due to the positively charged holes are called p – type semi-conductors.
Ex : Silicon or germanium can also be doped with group B elements like B, Al or Ga from p-type semiconductors.

Question 32.
Classify each of the following as either a p – type or a n – type semiconductor.
1) Ge doped with In
2) Si doped with B.
Answer:
1) Ge is an element of group -14. Its outer electronic configuration is 4s² 4p². When it is doped with In group – 13 element having 5s² 5p¹ configuration it can form three covalent bonds with germanium. The fourth bond of germanium contains only one electron and hence it is an electron deficient bond or a hole is created. Conductivity is due to these holes. Therefore it is a p – type semiconductor.

2) Silicon is an element of group -14 with an outer electron configuration 2s² 2p². When it is doped with Boron, an element of group -13 with 3s² 3p¹ configuration, it can form three covalent bonds with silicon. The fourth bond of silicon contain only one electron and hence it is an electron deficient bond or a hole is created. Conductivity is due to these holes. Therefore it is a p – type semi-conductor.

Question 33.
Analysis shows that nickel oxide has the formula Ni_0.98O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions ?
Answer:
From the given formula it is clear that for every 100 oxide ions there are only 98 nickel ions. Suppose, out of 98 nickel ions x exist as Ni²⁺ and the remaining (98 – x) exist as Ni³⁺
Total positive charge on 98 nickel ions = x × 2 + (98 – x) 3
Total negative charge on 100 oxide ions
= 100 × 2 = 200
Due to electrical neutrality,
x × 2 + (98 – x)3 = 200
2x + 294 – 3x = 200
x = 94
∴ Fraction of nickel as Ni²⁺ = \(\frac{94}{98}\) × 100
= 96%
Fraction of nickel as Ni³⁺ = (100 – 96)% = 4%

Question 34.
Gold (atomic radius = 0.144nin) crystallizes in a face centred unit cell. What is the length of a side of the unit cell?
Answer:
Radius of gold atom r = 0.144 nm
In face centred unit cell, lace diagonal
= 4r = \(\sqrt{2} a\)
∴ a = \(\frac{4 \mathrm{r}}{\sqrt{2}}\) = \(2 \sqrt{2}\) × r = 2 × 1.414 × 0.144mn
= 0.407 nm
Edge length of unit cell, a = 0.407 nm

Question 35.
In terms of band theory, what is the difference between a conductor and an insulator?
Answer:
In a metal the outer orbitals of very large number of atoms overlap to form a very large number of molecular orbitals that are delocalised over the metal. As a result a large number of energy levels are crowded together into bands. The highest occupied energy band is called the valence band while the lowest unoccupied energy band is called conduction band. The energy difference separating the valence band and conduction band is called band gap or energy gap.

In conductors the valence band is only partially filled. So electrons may easily excited from lower energy level to higher energy level by supplying a very small amount of energy. When a voltage is applied to a metal crystal, electrons are excited to the unoccupied orbitals in the same band and move towards the positive terminal. Thus, a material with partly filled energy band is a conductor.

In case of insulators, the highest occupied band is completely filled. The energy gap between the fully filled valence band and the vacant conduction band is very large and it is not possible to excite the electrons to the conduction band. So they become insulators.

Question 36.
In terms of band theory, what is the’ difference between a conductor and a semiconductor?
Answer:
In a metal the outer orbitals of very large number of atoms overlap to form a very large number of molecular orbitals that are delocalised over the metal. As a result a large number of energy levels are crowded together into bauds. The highest occupied energy band is called the valence hand while the lowest unoccupied energy band is called conduction band. The energy difference separating the valence band and conduction band is called band gap or energy gap.

In conductors the valence band is only partially filled. So electrons may easily excited from lower energy level to higher energy level by supplying a very small amount of energy. When a voltage is applied to a metal crystal, electrons are excited to the unoccupied orbitals in the same band and move towards the positive terminal. Thus, a material with partly filled energy band is a conductor.

In the case of semiconductors, the forbidden band i.e., the energy gap bet-ween valence band and conduction band is little. The thermal energy available at room temperature is enough to excite some electrons from the highest occupied band to the next permitted energy band. So conductivity occurs but their conductivity is in between the conductors and insulators.

Question 37.
If NaCl is doped with 1 × 10^-3mol percent of SrCl₂, what is the concentration of cation vacancies?
Answer:
Every Sr²⁺ ion causes one cation vacancy (because two Na⁺ ions are replaced by one Sr²⁺).
Therefore, introduction of 10^-3 moles of SrCl₂ per 100 moles of NaCl would introduce 10^-3 mole cation vacancies in 100 moles of NaCl.
No. of vacancies per mole of NaCl
= \(\frac{10^{-3}}{100}\) × 6.02 × 10²³
= 6.02 × 10¹⁸ vacancies mol^-1.

Question 38.
Derive Bragg’s equation. (TS Mar. 19) (AP 17, 16, 15; TS 16, 15; IPE 14)
Answer:
When a beam of X – rays strikes a crystal, the constituent particles of the crystal scatter or deflect some of the X – rays from their original path. The scattered X – rays can be detected on a photographic plate. From the diffraction pattern, the distance between constituent particles in a crystal can be studied.

In the above figure X – rays with wave-length λ strike a crystal face at an angle θ and then reflect at the same angle. The rays that strike an atom in the second layer are diffracted at the same angle θ, but because the second layer of atoms away from the X – ray source, the distance that the X – rays has to travel is more to reach the second layer. The path difference is indicated by AB in the figure. AB is equal to the distance between atomic layers d(=zB) times the sine of the angle θ.

sin θ = \(\frac{\mathrm{AB}}{\mathrm{d}}\) ; so AB = d sin θ
The X – rays have to travel a total extra distance AB + BC.
AB + BC = 2d sin θ (∴ AB = BC)
The different X – rays striking the two layers of atoms are in – phase initially and also they will be in – phase after reflection only the extra distance AB + BC is equal to a whole number of wavelengths nX where n is an integer (1, 2, 3, 4….)
∴ AB + BC = 2d sin θ = nλ
This is known as Bragg’s equation.
By knowing the value of φ, λ the value of θ can be measured, and the value of n is a small integer, usually 1. Thus the distance d between layers of atoms in a crystal can be calculated.

Long Answer Questions (8 Marks)

Question 39.
How do you determine the atomic mass of an unknown metal if you know its density and dimension of its unit cell ? Explain.
Answer:
From the dimensions of the unit cell, it is possible to calculate the volume of the unit cell. From the knowledge of the density of unit cell we can calculate the mass of atoms in the unit cell or vice versa.
If we know the edge length of a cubic crystal of an element or compound, we can easily calculate its density as described below.

Consider a unit cell of edge ‘a’ picometer (pm)
The length of the edge of the cell = a pm
= a × 10^-12m
= a × 10^-10 cm
Volume of unit cell = (a × 10^-10 cm)³
= a³ × 10^-30 cm³

Mass of unit cell = Number of atoms in unit cell × mass of each atom = Z × m
Where Z = number of atoms in unit cell and m = mass of each atom.

Mass of unit cell = Z × \(\frac{M}{N_o}\)
Substituting in (1) we get
∴ Density of unit cell = \(\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{a}^3 \times 10^{-30} \times \mathrm{N}_o}\) gm cm^-3

If ‘a’ is in pm, then density will be in g cm^-3. Density of unit cell is the same as the density of the substance. li the density of the metal and dimensions of unit cell are known the atomic mass can be calculated
by using the above equation.

Question 40.
Silver crystallizes In FCC lattice. If edge of the cell is 4.07 × 10^-8cm and density is 10.5 gcm^-3. Calculate the atomic mass of silver.
Answer:
The edge length of unit cell, .
a = 4.077 × 10^-8cm.
Number of atoms in FCC unit cell Z = 4
Density d = 10.5 g cm^-3
Atomic mass M =?
Avogadro constant N₀ = 6.023 × 10²³
We know that density
d = \(\frac{\mathrm{ZM}}{\mathrm{a}^{-3} \mathrm{~N}_{\mathrm{o}}}\) or M = \(\frac{d a^3 \mathrm{~N}_{\mathrm{o}}}{\mathrm{Z}}\)
M = \(\frac{10.5 \times\left(4.077 \times 10^{-8}\right)^3 \times 6.023 \times 10^{23}}{4}\)
= 107.8
∴ Atomic mass of silver = 107.8U

Question 41.
Niobium crystallizes In body – centred cubic structure. If density is 8.55 g cm^-3. Calculate atomic radius of niobium using its atomic mass 93U.
A. Density = 8.55 g cm³
Let length of tie edge = a cm
Volume of unit cell = a³ cm³

Number of atoms per unit cell = 2 (bcc)
Mass of unit cell = \(\frac{2 \times 93}{6.023 \times 10^{23}}\)
Volume of unit cell =

= \(\frac{2 \times 93}{6.023 \times 10^{23} \times 8.55}\) = 36.12 × 10^-24 cm³
∴ a³ = 36.12 × 10^-24 cm³
Edge length a = (36.12 × 10^-4)^1/3
= 3.306 × 10^-8 cm
= 3.306 × 10^-10 m
Now radius, in body centred cubic r = \(\frac{\sqrt{3}}{4} a\)
= \(\frac{\sqrt{3} \times 3.306 \times 10^{-10}}{4}\)
= 1.431 × 10^-10 m
= 0.143 nm. (or) 143 pm.

Question 42.
Copper crystallizes Into a fcc lattice with edge length 3.61 × 10^-8cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^-3.
Answer:
Volume of the unit cell = (361 × 10^-10)³ cm³
= 47.4 × 10^-24 cm³
Molar mass of copper = 63.5 g mol^-1
In a face centred cubic unit cell, there are four atoms per unit cell.
Mass of unit cell = \(\frac{4 \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1}}{6.02 \times 10^{23} \mathrm{~mol}^{-1}}\)

Thus the calculated value of density is in agreement with the measured value of density.

Question 43.
Ferric oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.
Answer:
In hexagonal close packed arrangement of oxide ions, each oxide ion has one octahedral hole associated with it.
For each O^2- ion there is one octahedral void in the structure and two of three are occupied by Fe³⁺ ion.
∴ Number of Fe³⁺ ions present per oxide ion = \(\frac{2}{3}\)
Thus the formula of ferric oxide should be Fe_2/3 O₁ or Fe₂ O₃.

Question 44.
Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit cell ?
(ii) How many unit cells are there in 1.00 cm3 of aluminium ?
Answer:
i) In cubic close packed structure, the face diagonal of the unit cell is equal to four times the atomic radius.
Face diagonal = 4 × r = 4 × 125 pm = 500 pm
But face diagonal = \(\sqrt{2}\) × edge length

= 354 pm.

ii) Volume of one unit cell = a³
= (3.54 × 10^-8 cm)³
No of unit cells in 1.00 cm³
= \(\frac{100}{\left(3.54 \times 10^{-8}\right)^3}\) = 2.26 × 10²²

Question 45.
How do you obtain the diffraction pattern for a crystalline substance ?
Answer:
The diffraction pattern for a crystalline sub-stance can be obtained by Debye – Scherrer method.

A monochromatic X – ray beam from a source ‘S’ is made incident on a crystal powder sample filled in a thin walled glass capillary tube ‘C’. As a result of diffraction from all sets of lattice planes, diffraction maxima arise which are recorded with a detector placed on the circumference of a circle centred on the powder specimen. The essential features of a diffractometer are shown in figure. A powder specimen C is supported on a table H, which can be rotated about an axis 0 perpendicular to the plane of the drawing. The X – ray source S is also normal to the plane of the drawing and therefore parallel to the diffractometer axis O. X – rays diverge from the source S and are diffracted by the specimen to form con-vergent diffracted beam which comes to a focus at the slit F and then enters ionisation chamber or photographic plate G.

The ioni-sation chamber contains methyl bromide vapours. The energy of the X – rays causes ionisation so that there is a flow of current. The current is measured with electrometer E. The extent of ionisation in the vapours is shown by the electrometer reading. As the intensity of the diffracted X – rays increase, the degree of ionisation also increases. For a set of parallel planes making an angle ‘θ’ with the incident beam of X – rays that satisfy Bragg’s equation, the intensity of the diffracted beam is measured. In Debye – Scherrer powder method the angle 2θ for the diffracted X – rays corresponding to each and every lattice plane is measured along with the intensities of diffracted X – rays with the help of the electrometer. A graph is drawn between the diffracted angle (2θ) and intensity of X – rays. This gives the diffraction pattern for a crystal line substance.

Intext Questions – Answers

Question 1.
Why are solids rigid ?
Answer:
In solids

1. the intermolecular distances are short,
2. intermolecular forces are strong,
3. the constituent particles (atoms, molecules or ions) have fixed positions and can only oscillate about their mean positions. So solids are rigid and incompressible.

Question 2.
Why do solids have a definite volume ?
Answer:
In solids the strong intermolecular forces bring the particles (atoms, molecules or ions) so close that they cling to one another and occupy fixed positions. So solids have definite volume and shape.

Question 3.
Classify the following as amorphous or crystalline solids : Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Answer:
Amorphous solids : Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass.
Crystalline solids : Naphthalene, benzoic acid, potassium nitrate, copper.

Question 4.
Why is glass considered a super cooled liquid ?
Answer:
Glass is an amorphous solid. The structure of amorphous solids is similar to that of liquids. So glass is considered as a super cooled liquid.

Question 5.
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of the solid. Would It show cleavage property?
Answer:
In amorphous solids the particles do not have long range order and their arrangement is irregular along all directions. So the value of any physical property would be same along any direction. This property is known as isotropy.

Question 6.
Classify the following solids into different categories based on the nature of inter- molecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer:

1. Molecular solids: Benzene, urea, ammonia, water and argon. Among these benzene and argon contain dispersion or London forces. Urea, ammonia and water contain hydrogen bonding.
2. Ionic solids : Potassium sulphate, zinc sulphide.
3. Metallic solids: Tin, rubidium.
4. Covalent or network solids : Graphite and silicon carbide.

Question 7.
Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it ?
Answer:
In covalent or network solids the adjacent atoms are bonded through covalent bonds throughout the crystal. Covalent bonds are strong and directional in nature, therefore atoms are held very strongly at their positions. Such solids are very hard and brittle. They have extremely high melting points. Hence solid A is covalent solid.

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
Ionic solids do not conduct electricity in solid state because ions are held by strong electrostatic forces and are not free to move. However, in molten state the ions become free to move. Hence ionic substances conduct electricity in molten state.

Question 9.
What type of solids are electrical conductors, malleable and ductile ?
Answer:
Metallic solids are electrical conductors, malleable and ductile.

Metals are orderly collection of positive ions surrounded by and held together by a sea of free electrons which are mobile and spread throughout the crystal. These free moving electrons are responsible for high electrical and thermal conductivity of metals.

Question 10.
Give the significance of a ‘lattice point’.
Answer:
If the three dimensional arrangement of constituent particles in a crystal is repre-sented diagramatically, in which each particle is depicted as a point, the arrangement is called crystal lattice and the points are called lattice points. Thus a lattice point represents a particle (atom, molecule or ion) in a crystal.

Question 11.
Name the parameters that characterise a unit cell.
Answer:
A unit cell is characterised by

1. its dimensions along the three edges a, b and c. These edges may or may not be mutually perpendicular.
2. Angles between the edges α (between b and c), β (between a and c) and γ (between a and b).
   Thus a unit cell is characterised by six parameters, a, b, c, α, β and γ.

Question 12.
Distinguish between
i) Hexagonal and monoclinic unit cells.
ii) Face centred and end – centred unit cells.
Answer:
(i) In hexagonal unit cell a = b ≠ c and α = β = 90°, γ = 120° but in monoclinic unit cell a ≠ b ≠ c and α = γ = 90°, β ≠ 120°.
(ii) In face centred unit cell a ≠ b ≠ c and α = β = γ = 90° but in end – centred unit cell a ≠ b ≠ c and α = β = γ = 90°.

Question 13.
Explain how much portion of an atom located at
i) corner and
ii) body centre of a cubic unit cell is part of its neighbouring unit cell.
Answer:
i) Each atom at a corner is shared between eight adjacent unit cells. Therefore only \(\frac{1}{8}\)th of an atom belongs to a particular unit cell.
ii) The atom at the body centre wholly belongs to the unit cell in which it is present.

Question 14.
What is the two dimensional coordination number of a molecule in square close – packed layer?
Answer:
In a square close packed layer, the two dimensional coordination number of a molecule is 4.

Question 15.
A compound forms hexagonal close – packed structure. What is the total number of voids in 0.5 mol of it ? How many of these are tetrahedral voids ?
Answer:
Let the number of close packed spheres be ‘N’ then
The number of octahedral voids = N
The number of tetrahedral voids = 2N
∴ Total number of voids = 3N
Since 1 mol of a substance contain N_o (Avogadro number) particles (atoms or molecules or ions). The total number of voids in one mol is 3N_o. In 0.5 mol the total number of voids is 1 .5N_o. Among these the number of tetrahedral voids is 1.0N_o.

Question 16.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3 rd of tetrahedral voids. What is the formula of the compound?
Answer:
In cubic close packing structure with each atom of N there would be two tetrahedral voids. Because only 1/3 rd of tetrahedral voids are occupied by M there would be 2/3 M for each N. Thus formula of the compound is M_2/3 N or M₂N₃.

Question 17.
Which of the following lattices has the highest packing efficiency
i) simple cubic
ii) body centred cubic and
iii) hexagonal close – packed lattice ?
Answer:
The efficiency of
i) simple cubic is 52.4%
ii) body centred cubic is 68% and
iii) hexagonal close – packed lattice is 74%. So hexagonal close – packed lattice has highest packing efficiency.

Question 18.
An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 10^-3 kg m^-3, what is the nature of the cubic unit cell ?
Answer:
Density d = 2.7 × 10³ kg m^-3
Molar mass M = 2.1 × 10^-2 kg mol^-1
Edge length, a = 405 pm or 405 × 10^-12 m
Density d = \(\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}\)
or Z = \(\frac{\mathrm{d} \times \mathrm{a}^3 \times \mathrm{N}_{\mathrm{A}}}{\mathrm{M}}\)
= \(\frac{2.7 \times 10^3 \times\left(405 \times 10^{-12}\right)^3 \times 6.022 \times 10^{23}}{2.7 \times 10^{-2}}\) = 4
Since the number of atoms per unit cell is 4 the cubic unit cell must be face centred cube (fcc) (or) ccp. „

Question 19.
What type of defect can arise when a solid is heated ? Which physical property is affected by it and in what way ?
Answer:
On heating a solid vacancy defects may arise. Due to the vacancies the density of the solid would be effected. The density of the solid would decrease due to vacancy defects.

Question 20.
What type of stoichiometric defect is shown by
i) ZnS
ii) AgBr ?
Answer:
i) ZnS shows Frenkel defect.
ii) AgBr shows Frenkel defect as well as Schottky defect.

Question 21.
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Answer:
If a cation of higher valence is introduced into an ionic solid as impurity, it occupies some site that occupied by host cations. In order to maintain electrical neutrality some host ions are missing from their sites leaving behind vacancies. For example, when Sr²⁺ ions are added to NaCl as impurity, each Sr²⁺ ion replaces two Na⁺ ions. Sr²⁺ ion occupies the site of one Na⁺ ion and the other site remains vacant. Thus, the number of vacancies equal to the number of Sr²⁺ ions added are produced.

Question 22.
Ionic solids, which have anionic vacan-cies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Answer:
The colour is due to the excitation of electrons in F- centres by absorbing energy from visible light falling on the crystals, e.g. the anion vacancies in NaCl impart yellow colour to the compound.

Question 23.
A group 14 element is to be converted into n – type semiconductor by doping it with a suitable impurity. To which group should this impurity belong ?
Answer:
To convert the group 14 element into n -type semiconductor it should be doped with an impurity containing extra electron required for bonding. The impurity element should belong to group -15. It may be phosphorous or arsenic. Silicon doped with group – 15 element behaves as a n – type semiconductor.

Question 24.
What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic ? Justify your answer.
Answer:
Ferromagnetic substances would make better permanent magnets because in ferromagnetic substances all the domains get oriented in one direction. When such a substance is placed in a magnetic field. A strong magnetic effect is produced. These substances remain magnetised even when magnetic field is removed.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 16th Lesson Communication Systems Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 16th Lesson Communication Systems

Very Short Answer Type Questions

Question 1.
What are the basic blocks of a communication system? [TS June ’15]
Answer:
The basic blocks of a communication system are

1. Transmitter
2. Communication channel
3. Receiver.

Question 2.
What is “world wide web” (www)? [IMP]
Answer:
It is an encyclopedia of knowledge accessible to everyone round the clock throughout year with the help of computer connectivity.

Question 3.
Mention the frequency range of speech signals.
Answer:
Frequency range of speech signals is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation? (IMP)[AP May ’16; June ’15]
Answer:
In the frequency range from a few MHz upto about 30 MHz, long distance communication can be achieved by the ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave pro-pagation and it is used by short wave broadcast services.

Question 5.
Mention the various parts of the ionosphere? [TS May ’16]
Answer:
Parts of ionosphere are

1. D – Layer at a height of 65 to 75 km from ground.
2. E – Layer at a height of nearly 100 km.
3. F₁ – Layer at a height of 170 to 190 km.
4. F₂ – Layer at a height of 250 to 400 km.
   Note : F₁ and F₂ will merge during night.

Question 6.
Define modulation. Why is it necessary? [AP Mar. 18, 17, 16, 14; May 17, 14; TS Mar. 19, 16, 15, May 18, 17]
Answer:
Modulation :
The process of combining audio frequency signal with high frequency signal is called modulation. Modulation is necessary for the following reasons.

1. to reduce the size of antenna.
2. to increase the effective power radiated by antenna.
3. to avoid mixing up of signals from different transmitters.

Question 7.
Mention the basic methods of modulation [AP Mar. 19. 16; TS Mar.’ 18, ’17, ’15]
Answer:
The basic methods of modulation are :
a) Amplitude modulation
b) Frequency modulation and
c) Phase modulation.

Question 8.
Which type of communication is employed in mobile phones? [AP May ’18, Mar. ’15]
Answer:
Space wave communication is employed in mobile phones.

Short Answer Questions

Question 1.
Draw the block diagram of a generalized communication system and explain it briefly.
Answer:
The block diagram of a generalized communication system are as shown. It consists of the following three parts.

1. Transmitter
2. Communication channel and
3. Receiver.

In this system, the transmitter and receiver are located at two different places and the channel is the physical medium that connects them depending upon the type of communication system. A channel may be in the form of wires (or) cables connecting the transmitter and the receiver, or it may be wireless.

Using the transmitter, the original message is converted into a form so that it can be transmitted over the communication channel. Hence it is called as a-message signal.

The uses of these three parts are :
1) Transmitter :
Purpose of transmitter is to convert the message signal produced by source into a form suitable for transmission through channel.

2) Channel :
Channel is used for transmission of signals. It may consist a medium like coaxial cable, optical fibre, (or) even space is used as channel in wireless communication where E.M waves are used.

3) Receiver :
Receiver will convert the signals received through channel into a recognisable form of the original message signal.

Question 2.
What is a ground wave? When is it used for communication?
Answer:
Ground wave :
In A.M. broadcasting ground based vertical metallic towers called antennas are used for transmission. For such antennas ground has strong influence on transmission. Near transmitting antennas energy of E.M wave is high. They have strong influence on propagation of signals along the ground surface. A strong E.M. wave near antenna will induce current on surface of ground over which it passes. So E.M waves slides along surface of ground.

For ground wave propagation attenuation on surface of earth is high. The magnitude of attenuation is proportional to frequency. Due to very high energy absorption of ground.

The ground wave propagation is limited to few kilometers from transmitting antenna.

Question 3.
What are sky waves? Explain sky wave propagation, briefly.
Answer:
Sky waves :
The electromagnetic waves which suffers reflection of ionosphere and reaches earth are called sky waves.

Sky wave propagation :
Electromagnetic waves with in the range of 30 to 40 MHz are reflected by ionosphere. The degree of ioni-sation varies with height above ground. Due to ionospheric reflection long range com-munication is possible with sky waves. Because these reflected waves reach earth at a longer distance. Short wave broadcast for long distance is possible with sky waves. Generally E.M waves with in the frequency range of 3 to 30 MHz are highly suitable for sky wave propagation. E.M waves of higher frequencies i.e., frequency > 30 MHz will penetrate ionosphere. Hence sky wave propagation is not possible with high frequency electro magnetic waves.

Question 4.
What is space wave communication? Explain.
Answer:
Space wave :
A space wave will travel from transmitting antenna to receiver in the form of straight lines. Space waves will obey line of sight properties.

Space wave propagation :
This type of propagation of E.M waves is called space wave propagation.

Space wave propagation is mainly used in television transmission. Due to curvature of earth range of space wave propagation is limited to certain distance only. Range depends on height of the antenna. Let height of transmitting antenna is ‘h_T’ and radius of earth is ‘R’ the distance upto which signals will reach with LOS (hne-of-sight) properties is d_r = \(\sqrt{2Rh_T}\). If receiving antenna is at a height h_R then also distance d_r upto which signals can be reached with LOS properties will increase. In this case d_r = \(\sqrt{2Rh_T}+\sqrt{2Rh_R}\).

Question 5.
What do you understand by modulation? Explain the need for modulation.
Answer:
Modulation :
The process of super imposing a low frequency signal onto a high frequency carrier wave is known as modulation.

Modulation is necessary for long range transmission of signals.

Modulation is a very essential part of communication. Transmission of low frequency of voice signals (frequency range 20 Hz to 20,000 Hz) to longer distances is not possible because energy associated with low frequency signals is less.

So our voice signals after converting into electrical signals must be superposed on to a high frequency signal called carrier wave. Generally frequency of carrier wave is high. This will help for efficient transmis¬sion and reception. Modulation is necessary for the following reasons,

1. To reduce the size of antenna for E.M. waves to be transmitted as antenna is necessary.
2. To increase effective power radiation by transmitter.
3. To avoid mixing up of signals from different transmitters.

To obtain the above advantages we are using modulation techniques where a low frequency signal is superposed on a high requency carrier wave.

Question 6.
What should be the size of the antenna or aerial? How the power radiated is related to length of the antenna and wavelength?
Answer:
Antenna and its size :
It is a must for every transmitter to have an antenna. Without antenna, it is not possible to radiate electrical energy output of transmitter into space in the form of electromagnetic radiation.

Antenna size must be comparable to wavelength of the signal to be transmitted.

To transmit all information contained in a signal without any loss in quality the minimum size of antenna must be at least \(\frac{\lambda}{4}\).

For good transmission output power of transmitter must be high.

Relation between length of antenna ‘l’ and wavelength ‘ λ’ and power radiated is
Power radiated ∝ (\(\frac{1}{\lambda}\))²

Question 7.
Explain amplitude modulation.
Answer:
Amplitude modulation (A.M.) :
In amplitude modulation (A.M.) the amplitude of carrier wave is varied in accordance with the voice signal superimposing on it.

Theory :
Let time varying carrier wave is c(t) = A_c sin ω_ct

Time varying modulating wave is m(t) = A_m sin ω_mt

when these two waves are superposed modulated wave is given by
c_m(t) = (A_c + A_m sin ωt) sin ω_ct

Because modulating wave changes only amplitude of carrier wave but not its frequencies

is the modulated wave. Where (ω_c – ω_m) and (ω_c + ω_m) are lower and upper side frequencies. Production of amplitude modulated wave.

Question 8.
How can an amplitude modulated wave be generated?
Answer:
The block diagram to produce amplitude modulated wave is as shown. Here modulating signal A_m sin ω_mt is mixed with a carrier signal A_c sin ω_ct in a square law device.

The output of square law device y(t) = B x (t) + Cx²t

Where B and C are constants.
∴ y(t) = BA_m sin ω_mt + BA_c sin ω_ct + C (A_m sin ω_mt + A_c sin ω_ct)² …………… (1)

By using trigonometric relations (1)
sin²A = (1 – cos 2A) / 2

and (2) sin A sin B = \(\frac{1}{2}\) [cos (A – B) – cos (A + B)] equation (1) can be written as
y(t) = BA_msin ω_mt + BA_c sin ω_ct
+ C\(\frac{\mathrm{A}_{\mathrm{m}}^2}{2}\) + A²_c – \(\frac{c\mathrm{A}_{\mathrm{m}}^2}{2}\) cos 2ω_mt – cos 2ω_mt
\(\frac{C\mathrm{A}_{\mathrm{c}}^2}{2}\) cosω_ct + cA_mA_ccos(ω_c – ω_m
CA_mA_c cos(ω_c

This signal is passed through a b filter which rejects the D.C compc sinusoidal frequencies ω_m , 2ω_m a, ailows the frequencies ω_c, ω_c ω_m + ω_m at the out put side. This me signal is transmitted through ante

In this way amplitude modulat is produced.

Question 9.
How can an amplitude modulated wave be detected?
Answer:
The block diagram of amplitude modulated wave detector is as shown.

The receiving antenna will collect the amplitude modulated waves from space. These are passed through an amplifier to increase strength of signals.

These signals are passed through intermediate frequency stage (I.F. stage) to frequency of carrier.

For recovery of original massage signal m(t) with angular frequency ω_m the modulated signal is passed detector which consists (1) a rectifier where lower half (- ve half) p modulated signals is eliminated. (2) Later it was sent through envelope detector to recive the original signal.

This signal is passed through a amplifier and finally signal with sufficient strength is going to a microphone to convert electrical signals into audio signals.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits

Very Short Answer Type Questions

Question 1.
What is an n-type semiconductor? What are the majority and minority charge carriers in it?
Answer:
When intrinsic Germanium / Silicon crystal is doped with a pentavalent impurity then “n-type semiconductor” will be formed.

The majority charge carriers are electrons and minority charge carriers are holes.

Question 2.
What are intrinsic and extrinsic semiconductors? [AP May ’18, Mar, ’15; June ’15]
Answer:
Intrinsic semiconductor :
Ultra high pure semiconductor are called “intrinsic semi-conductor”.

Extrinsic semiconductor :
The doped semi-conductor is called extrinsic semiconductor.

Question 3.
What is a p-type semiconductor? What are the majority and minority charge carriers in it? [TS May ’18, Mar. ’17; AP Mar. ’17]
Answer:
When third group impurities like boron, aluminium, galium, indium etc., are added to intrinsic semiconductor then it is called “p-type semiconductor.”

In p-type semiconductor majority, charge carriers are holes and minority charge carriers are electrons.

Question 4.
Give examples of “photosensitive substances”. Why are they called so? [AP May ’16]
Answer:

1. Metals like zinc, cadmium and magnesium will respond to ultraviolet rays.
2. Alkalimetals such as sodium, potassium, caesium and rubidium will respond to visible light.

These substances are called photo sensitive surfaces because they will emit electrons when light falls on them.

Question 5.
What is p-n junction diode? Define depletion layer. [TS Mar. ’19 May ’16]
Answer:
p-n semiconductor diode :
When a semiconductor material such as silicon or germanium crystal is doped in such a way that one side of it becomes a p-type and the other side becomes n-type then a p-n semi-conductor diode is formed.

Depletion layer :
The formation of a narrow region on either side of the junction which becomes free from mobile charge carriers is called “depletion layer”.

Question 6.
How is a battery connected to a Junction diode in i) forward and ii) reverse bias?
Answer:
In Forward Bias :
In a p-n junction diode if p-side is connected to positive terminal of a battery and n – side to negative terminal it is called “forward biased”.

In Reverse Bias :
If p-side is connected to negative terminal of the battery and n-side to positive terminal of the cell it is called “reverse biased”.

Question 7.
What is die maximum percentage of rectification in half wave and full wave rectifiers?
Answer:
The maximum percentage of rectification in a half wave rectifier is 40.6% and in a full wave rectifier maximum percentage of rectification is 81.2%.

Question 8.
What is zener voltage (V_Z) and how will a zener diode be connected in circuits generally?
Answer:
When a zener diode is reverse biased at a particular voltage the current increases suddenly. The voltage at which the current increases is called “breakdown voltage” or “zener voltage”, so zener is always connected in “reverse bias”.

Question 9.
Write the expressions for the efficiency of a half wave rectifier and a full wave rectifier.
Answer:
Efficiency of half wave rectifier

Efficiency of full wave rectifier

Question 10.
What happens to the width of the depletion layer in a p – n junction diode when it is i) forward biased and ii) reverse biased? [AP Mar. ’19]
Answe:
When a p – n diode is forward biased thickness of depletion layer decreases and in reverse bias condition the thickness of depletion layer increases.

Question 11.
Draw the circuit symbols for p – n – p and n – p – n transistors. [TS & AP Mar. ’18, May ’17; AP May ’16, ’14; Mar. ’14; TS Mar. ’16]
Answer:

Question 12.
Draw the symbol of NOT gate and explain its operation. Give its truth table. [TS June ’15]
Answer:
NOT gate :
It has one input terminal and one output terminal. The output of NOT gate is the opposite of input i.e., if input is ‘0’ then output is ‘1’. If input is ‘1’ then output is ‘O’.

Implementation of NOT gate using a transistor :
NOT gate can be implemented with transistor. If A = 0 the emitter base junction is open and there is no current through the transistor. The current through the resistor. R_L = 0 and Q becomes equal to a potential of 5V i.e., Q when A = 1 then Q = 1 in the emitter base junction. So large current flows and Q approximately 0 volt i.e., Q =0. Thus output is same as that of a NOT gate.

Truth tables of NOT gate :
The truth tables of NOT gate interms of low and high (0 and 1) are as given below.

Question 13.
Define amplifier and amplification factor.
Answer:
Amplifier :
Raising the strength of a weak signal is known as amplification and the device used for this purpose is called amplifier.

Amplification factor :
It is the ratio between output voltage to the input,
Voltage, (A) = \(\frac{V_0}{V_i}\)

Question 14.
In which bias, can a zener diode be used as voltage regulator? [AP Mar. 16; TS June 15]
Answer:
In reverse bias, zener diode can be used as voltage regulator.

Question 15.
Which gates are called universal gates? [TS Mar. ’15]
Answer:
NAND and NOR gates are known as the basic building blocks of logic gates or universal gates.

Question 16.
Write the truth tables of NAND gate. How does it differ from AND gate?
Answer:

The output of NAND gate is opposite to output of AND gate.

Short Answer Questions

Question 1.
What are n-type and p-type semiconductors? How is a semiconductor junction formed?
Answer:
n-type semiconductors :
When pentavalent impurities such as phosphorous (P), arsenic (As), antimony (Sb) are added to intrinsic semiconductors then they are called n-type semiconductors.

p-type semiconductors :
When trivalent. impurities such as Boron (B), Aluminium (AI), Galium (Ga), Indium (In) etc. are added to intrinsic semiconductor then it is called p-type semiconductor.

p-n junction :
A p-n junction is formed by adding a small quantity of pentayalent impurities in a highly controlled manner to a p-type silicon/germanium wafer.

• During the formation of p-n junction diffusion and drift of charge carriers takes place.
• In a p-n junction concentration of holes is high at p – side and concentration of electrons is high at n-side.

Due to the concentration gradient between p-type and n-type region holes diffuse to n- region and electrons diffuse to p-region.

Due to diffusion of changes a chargeless region is formed near junction called depleted layer.

Question 2.
Discuss the behaviour of p-n junction. How does a potential barrier develop at the junction?
Answer:
p-n junction :
A p-n junction is formed by adding a small quantity of pentavalent impurities in a highly controlled manner to a p-type silicon/germanium wafer.

• During the formation of p-n junction diffusion and drift of charge carriers takes place.
• In a p-n junction concentration of holes is high at p – side and concentration of electrons is high at n-side.

Due to the concentration gradient between p-type and n-type region holes diffuse to n- region and electrons diffuse to p-reglon. This leads to diffusion current.

Due to diffusion of electron an ionised donor is developed at n-region and due to diffusion of holes to n- region an ionised acceptor. These ions are immobile. So some – ve charge is developed in p – region and positive charge is developed in n-region. This space charge prevents further motion of electrons and holes near junction.

Depletion layer :
Both the negative and positive space charge regions near junc-tion are called depletion region.

Question 3.
Draw and explain the current-voltage (I-V) characteristic curves of a junction diode in forward and reverse bias.
Answer:
When a graph is plotted between junction potential ‘V’ and junction current T of a p-n junction then it is called V-I characteristics.

In forward bias p-side of p-n junction is connected to + ve terminal and n – side is connected to – ve terminal of external voltage ‘V’. The external voltage ‘V’ is gradually increased and junction current T is measured.

Initially junction potential is slowly increased in steps of 0.1 V. Until junction potential reaches a minimum value called threshold potential junction current is almost zero. ;

Once applied potential crosses threshold potential then junction current increases exponentially with applied voltage. On the reverse bias side nearly a steady current of few micro amperes was observed with applied voltage. When reverse bias potential reaches a high valued suddenly the diode is thrown into conduction. This is called breakdown potential.

Question 4.
Describe how a semiconductor diode is used as a half wave rectifier. [TS May. 18. Mar. 16; AP Mar. 16, 14]
Answer:
A junction diode allows current through it in forward bias only. In a half wave rectifier an a.c. source, p-n junction and load resistance (R_L) are connected in series as shown.

For ‘+ve’ half cycle p-n junction is for-ward biased so current flows through diode and we will get output current across load resistance.

For ‘+’ ve half cycle the p-n junction is reverse biased so current does not flow through p-n junction. So we are not able to get current through load resistance.

In half wave rectifier the out put voltage changes sinusoidally. But still it is flowing in only one direction through load resistance (R_L) so input a.c. voltage is rectified.

Question 5.
What is rectification? Explain the work¬ing of a full wave rectifier. [AP Mar. ’18, ’15; May ’17, ’14; TS Mar. ’19. ’15, May ’17]
Answer:
Rectification :
The process of converting alternating current (a.c) into direct current (d.c) is called rectification. Instruments used for rectification is called rectifier.

In a full wave rectifier, two p-n diodes are connected at the output side of a center-tapped transformer through a load resistance as shown.

The centre tap will divide the output a.c. wave exactly into two equal halves say + ve half cycle and – ve half cycle.

Let diode D₁ is connected to ‘+ve’ half cycle then it is forward biased. So current flows through D₁ at the same time ‘-ve’ half cycle is applied to diode D₂ so it is reverse biased and current does not flow through it. Hence we will get output through diode D₁.

As the applied a.c. wave is progressing we will get ‘+ve’ half cycle to diode D₂ at that time diode D₂ is forward biased so current flows through D₂. But now ‘-ve’ half cycle is applied to diode D₁. So it is in reverse bias hence current does not flow through D₁.

In full wave rectifier diodes D₁ and D₂ will conduct current alternately. Even though out put current oscillates between a minimum and maximum value it always passes through same direction through load resistance (R_L). So input a.c. is rectified.

Question 6.
Distinguish between half-wave and full-wave rectifiers. [AP Mar. ’19. ’17; May. ’16; TS Mar. ’18, May, ’18]
Answer:

Question 7.
Distinguish between zener breakdown and avalanche breakdown.
Answer:
Avalanche breakdown:

1. In avalanche breakdown, the thermally generated electrons and holes acquire sufficient energy from the applied potential to produce new carries by removing valence electrons from their bonds.
2. These new carries, in turn produce additional carriers again through the process of disrupting bonds.
3. This cumulative process is referred to as avalanche multiplication. It results in the large flow of current, and the diode finds itself in avalanche breakdown.
4. This occurs in lightly doped diodes at high reverse bias voltages.

Zener breakdown:

1. If a diode is heavily doped, direct rupture of covalent bonds takes place because of strong electric field at the junction.
2. As a result of heavy doping of p and n regions, the depletion region width becomes very small and an applied voltage causes an electric field of 10⁷ V/m of the junction making condition suitable for zener breakdown. This occurs in heavily doped diodes at low reverse bias voltages.

Question 8.
Explain hole conduction in intrinsic semiconductors.
Answer:
Intrinsic semiconductors :
Semiconductors with ultra high pure state are called “intrinsic semiconductors”.

In pure germanium (Ge) or silicon(Si) crystal every germanium or Silicon atom forms four covalent bonds with neighbouring Ge/Si crystal.

At very low temperatures intrinsic semiconductors are insulators. When temperature increases electrons absorbs more thermal energy and it may become a free electron and that atom will become positive.

In intrinsic semiconductors number of free electrons (n_e) is equal to number of holes (n_h)

Due to thermal energy some of the electrons escapes from the bonds and an empty spaces left behind in the valence band. This vacancy in the valence band is called a “hole”.

Due to applied electric field the holes drift in opposite direction to the electrons with lesser speed and behave like positive charge carriers and current is produced due to the both electrons and holes.

Current contribution by electrons (l_e) and holes (I_h) is same.
∴ In an intrinsic semiconductor n_e = n_h = n_i.
Total current I = I_e + I_h

Question 9.
What is a photodiode? Explain its working with a circuit diagram and draw its I-V characteristics.
Answer:
Photodiode :
A photodiode consists of a p-n junction diode with a transparent window to allow light to fall on to the junction.

When light photons falls on this p-n junction it will produce electron – hole pair. These pairs are separated by the applied potential ‘V’ before they recombine. The magnitude of photo current depends on intensity of incident light.

i.e., current, i ∝ intensity of light I. So a photodiode can convert light intensity variations into current variations. This property is used to detect optical signals.

The I-V characteristics of photodiode can easily studied with reverse bias potential on it. The junction voltage ‘V’ and current ‘I’ characteristics are as shown in figure.

Question 10.
Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
Answer:
Light emitting diode (LED) :
It is a highly doped p-n junction. Under forward biased condition it emits spontaneous radiation in visible region. This p-n junction is coated with a transparent cover so that emitted light can come out.

Working :
When LED is forward biased holes are driven to n-region and electrons are driven to p-region due to electric potential of battery. As a result charge concentration near junction region increases. The electrons and holes while recombining with them they will release the recombination energy in the form of light photons.

Generally LED breakdown voltages are very low such as 5V. For the fabrication of visible LED the energy band gap of 1.8 eV to 3 eV is necessary. The minimum energy gap must be 1.8 eV. It LED materials are selected with in this range then we can get visible light in the wavelength range of 0.7 pm to 0.4 pm or in wavelength range of 7000Å to 4000Å.

Generally LEDs are biased to emit light with maximum efficiency. Intensity of light emitted depends on strength of current. Colour of light depends on energy band width.

For gallium arsenide-phosphide energy gap is ≅ 1.9 eV, it will emit red light.

For galium Arsenide Ga, As energy gap is ≅ 1.4 eV, it will emit infrared light. These LEDs are widely used in T.V remote control and in burglar alarm systems.

Advantages of LED:

1. They require low operational voltage and consumes less power.
2. They are surged and life period is very high.
3. They will respond very quickly to current changes.

Question 11.
Explain the working of a solar cell and draw its I-V characteristics.
Answer:
Solar cells :
A solar cell is also a p-n junction which generates emf when solar radiation falls on it. A p-type silicon wafer of nearly 300 pm is taken. A n-type impurity layer of nearly 0.3 µm is developed on it through diffusion process, junction surface area Is kept large. A metallic grid is deposited on n-region and the other p- side is also coated with metal. These two will provide electrical contact.

Working :
Generally semiconductors width a band width of nearly 1.5 eV or less are selected in solar cells when solar radiation falls on p-n junction electron hole pair is produced.

Separation of electrons and holes will takes place (before they recombine) due to the electric field produced across depleted region.

Electrons on reaching n – side are collected by front metallic grill. Holes reaching p – side are collected by back contact.

V.I characteristics of solar cells are drawn in fourth quadrant of the coordinate axis. It is as shown in figure.

Solar cells are widely used to tame solar energy from which electricity is produced.

These solar cell plays an important role in supplying electrical energy to satellites and in remote forest areas.

Question 12.
Explain the different transistor configurations with diagrams.
Answer:
Transistors are connected in three ways. They are :

1. Common base configuration
   Common emitter configuration
2. Common collector configuration

1) Common base configuration :
In this configuration, base is common to both input and output. Base terminal is earthed and input is given across base emitter and output is taken across base collector.

2) Common emitter configuration :
In this configuration emitter is common to both input and output. The emitter is earthed and input is given across base emitter and output is taken across collector emitter.

3) Common collector configuration :
In this configuration, collector is common to both input and output. The collector is earthed and input is given across base collector and output is taken across emitter-collector.

Question 13.
Explain how transistor can be used as a switch?
Answer:
In a transistor D.C. input voltage V_i = I_BR_B + V_BE …….. (1)

Sum of D.C potential between emitter base i.e., and product of base current and base resistance (V_BE) / I_BR_B.

D.C. output voltage V₀ = V_CC – I_CR_C ……….. (2)

As far as V_i is less than active region minimum voltage of 0.6 V (nearly), output voltage V₀ is high because Collector current (I_C) is zero.

When input voltage V_i increases collector current I_C increases and output voltage V₀ decreases because V₀ = V_i – I_CR_C. i.e., when V_i is less V₀ is high and whenV_i is high output voltage V₀ is less. By changing the input potential a transistor can be made to move between high and low states or ON and OFF states. Hence by selecting proper input voltage V_i a transistor can be used as a switch.

Question 14.
Explain how transistor can be used as an oscillator.
Answer:
In an oscillator, we will get out put without any external input. The oscillator circuit is as shown in figure.

When switch ‘S₁‘ is closed a surge current will flow in transistor and it produces output in collector circuit. Current in coil T₂ will gradually increase from minimum value ‘x’ to maximum value ‘y’ with changing time. Current in T₂ is connected to oscillator LC in collector circuit. This induces current in coil T₁ due to inductive coupling between T₁ and T₂. Coil T₁ is connected to emitter base circuit. So a part of output is given as feedback to input.

When collector current reaches maximum value rate of change in collector current is zero. So induced current in T₁ is zero i.e., input feedback is zero. So collector current begin to decrease while decreasing again emf is induced in T₁ and feedback is given to emitter base circuit. Like this the signal is self sustained
Frequency of oscillator υ = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

In this way a transistor can be used as an oscillator.

Question 15.
Define NAND and NOR gates. Give their truth tables. [TS Mar. ’17; AP June ’15]
Answer:
NOR Gate :
It is the combination of OR gate and NOT gate OR + NOT = NOR.

In this logic gate the output of OR gate is given to the input of NOT gate as shown in figure.

NAND gate :
It is the combination of AND gate and NOT gate.
AND + NOT = NAND.
In this logic gate the output of AND gate is given to the input of NOT gate.

The NAND and NOR gates are basic build¬ing blocks of logic gates; because any logic gates can be constructed by using only NAND or only NOR gates.

Question 16.
Explain the operation of a NOT gate and give its truth table. [TS June ’15]
Answer:
NOT gate :
It has one input terminal and one output terminal. When the input is low, the output is high and when the input is high, the output is low.

Implementation of NOT gate using a transistor :
NOT gate can be implemented with transistor. If A = 0 the emitter base junction is open and there is no current through the transistor. The current through the resistor. R_L = 0 and Q becomes equal to a potential of 5V i.e., Q when A = 1 then Q = 1 in the emitter base junction. So large current flows and Q approximately 0 volt i.e., Q =0. Thus output is same as that of a NOT gate.

Truth tables of NOT gate :
The truth tables of NOT gate interms of low and high (0 and 1) are as given below.

Long Answer Questions

Question 1.
What is a junction diode? Explain the for-mation of depletion region at the junction. Explain the variation of depletion region in forward and reverse biased condition.
Answer:
p-n junction :
A p-n junction is formed by adding a small quantity of pentavalent impurities in a highly controlled manner to a p-type silicon/germanium wafer.

During the formation of p-n junction diffusion and drift of charge, carriers takes place.

In a p-n junction concentration of holes is high at p – side and concentration of electrons is high at n-side. Due to the concentration gradient between p-type and n-type regions holes diffuse to n-region and electrons diffuse to p-region. This leads to diffusion current.

Due to diffusion of electron an ionised donor is developed at n-region and due to diffusion of holes to n- region an ionised acceptor. These ions are immobile. So some – ve charge is developed in p – region and positive charge is developed in n-region. This space charge prevents further motion of electrons and holes near junction.

Depletion layer :
Both the negative and positive space charge regions near junction are called depletion region.

Variation of depletion region :
In forward bias due to applied voltage electrons from n – region crosses junction layer and goes to p-region. Similarly holes from p-region crosses junction and goes to n-region. Since charge carriers are freely crossing the junction depleted region vanishes in forward bias condition.

In reverse bias condition electrons in n- region are attracted by + ve terminal of bat-tery connected to it. Similarly holes in p – region are attracted by ’-ve’ terminal con-nected to it. As a result charges will travel towards battery terminals. We can not find charge carriers near p-n junction.

So in reverse bias condition width of depleted region increases.

Question 2.
What is a rectifier? Explain the working of half wave and full wave rectifiers with diagrams.
Answer:
The process of converting alternating current (a.c) into direct current (d.c) is called “rectification”. Instruments used for rectification is called “rectifier”.

A junction diode allows current through it in forward bias only. In a half wave rectifier an a.c. source, p-n junction and load resistance (R_L) are connected in series as shown.

For ‘+ve’ half cycle p-n junction is forward biased so current flows through diode and we will get output current across load resistance.

For + ve half cycle the p-n junction is reverse biased so current does not flow through p-n junction. So we are not able to get current through load resistance.

In half wave rectifier the out put voltage changes sinusoidally. But still it is flowing in only one direction so input a.c. voltage rectified.

In a full wave rectifier two p-n diodes are connected at the output side of a center tapped transformer through a load resistance as shown.

The centre tap will divide the output a.c. wave exactly into two equal halves say + ve half cycle and – ve half cycle.

Let diode D₁ is connencted to ’+ve’ half cycle then it is forward biased. So current flows through D₁ at the same time ‘-ve’ half cycle is applied to diode D₂. So it is reverse biased and current does not flow through it. Hence we will get output through diode D₁.

As the applied a.c. wave is progressing we will get ‘+ve’ half cycle to diode D₂ at that time diode D₂ is forward biased so current flows through D₂. But now -ve’ half cycle is applied to diode D₁. So it is in reverse bias hence current does not flow through D₁.

In full wave rectifier diodes D₁ and D₂ will conduct current alternately. Even though out put current oscillates between a minimum and maximum value it always passes through same direction. So input a.c. is rectified.

Question 3.
What is a zener diode? Explain how it is used as a voltage regulator.
Answer:
Zener diode :
A zener diode is a highly doped p-n junction with sharp breakdown voltage. Generally zener is operated in “reverse bias condition”.

Note:
In forward bias condition zener diode will also act as ordinary p-n junction.

Zener diode as voltage regulator :
Principle :
Zener diode has sharp breakdown voltage in “reverse bias condition”.

In zener diode release of large scale of charge carriers at breakdown voltage is due to field emission of electrons from host atoms. For field ionisation nearly an electric field of 10⁶ V/m is necessary.

The speciality of zener diode is even though current through zener increases largely its potential (zener potential) remains almost constant.

Working :
Let a zener diode is connected to unregulated supply through a series resistance R_s. The value of zener voltage is selected such that zener voltage V_z is less than the minimum value of unregulated supply. A load resistance R_L is connected parallel to zener. Output is taken across load resistance R_L.

Case – I :
Let voltage of unregulated supply is at Its minimum value say V_{i min}. Zener potential V_z is constant. So V_{i min} – V_z is voltage drop across R_s. Current through R_s = I_min
= \(\frac{V_{i min}-V_z}{R_s}\)
Load current, I_L = \(\frac{V_z}{R_L}\)
∴ Current through zener I_{z min} = I_min – \(\frac{V_z}{R_L}\)

Case – II :
When applied voltage is maximum say, V_{i max} then voltage drop across
R_s = V_{i max} – V_z

Maximum current through R_s is
I_max = \(\frac{V_{i max}-V_z}{R_s}\)
Current through zener is also max.
I_{z max} = I_max = \(\frac{V_z}{R_L}\)

In zener diode voltage regulator voltage changes in unregulated supply are converted into current changes in series resistance R_s. These current changes are absorbed by zener.

As a result we will get a constant voltage and current at output. In this way zener diode acts as a voltage regulator.

Question 4.
Describe a transistor and explain its working.
Answer:
Transistor :
A transistor is a three layered electronic device. These layers are called emitter, base and collector.

Transistors are two types 1) p-n-p 2) n-p-n.

Emitter :
Emitter region is of moderate size. It is heavily doped. It supplies large number of majority charge carriers for the current flow through transistor.

Base :
Width of base region is very less. It is lightly doped nearly with 3 to 5% Impurity concentration of emitter.

Collector :
Size of collector region is larger than emitter. It is moderately doped (i.e., impurity concentration is less than emitter).

Biasing of transistor :
In a transistor for transistor action to takes place (i) Emitter base region must be forward biased, (ii) Base collector region must be reverse biased, (iii) Forward bias emitter base potential V_EB must be less than reverse bias collector base potential V_CB.

If a transistor is biased as above, then the transistor is said to be in active state.

Working principle :
Electrons/holes in emitter region are injected into base region due to forward bias potential in emitter base region.

In base region nearly 3 to 5% of electrons/ holes are recombined so a small amount of current (I_B) will flow in emitter base circuit. Due to less impurity concentration the remaining charges injected into base will cross base collector region.

The high reverse bias potential in base collector region will act as forward bias for the charges left unrecombined in base region.

So these charges are attracted by collector region. These charges will recombine at collector region and some current (I_C) will flow in base collector region.

In a transistor emitter current (I_E) = Base current (I_B) + Collector current (I_C)
∴ I_E = I_B + I_C

Hence transistor is a current controlled device.

Question 5.
What is amplification? Explain the working of a common emitter amplifier with necessary diagram.
Answer:
Amplifier :
An amplifier is an electronic device used to strengthen weak signals.

Transistor as an amplifier (C.E. configuration) :
A transistor can be used as an amplifier in its active region. In this region output voltage V₀ increases drastically even for a small change in input voltage’ V_i‘.

In transistor amplifier the mid point of active region is taken as operating point (also called input potential) on which varying signal voltage is superposed. This variation is magnified at output side by a factor equals to amplification factor p.

In a transistor output voltage of collector, V_OC = V_CE + I_LR_L

Input voltage V_BB = V_BE + I_BR_B when input signal voltage V_i ≠ 0 the,

V_BE + V_i + V_BE + I_BR_B + ∆I_B(R_B + r_i)

Where r_i is input resistance.

Input signal voltage
V_i = ∆I_B(R_B + r_i) = r∆I_B.

Current amplification factor β is defined as the ratio of change in collector current ∆I_C to change In base current ∆I_B when V_CC Is constant.

Voltage amplification factor A_v :
It is defined as the ratio of output signal voltage (V₀) to input signal voltage (V_i).

Question 6.
Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
Answer:
Implementation of OR gate using diodes :
Let ‘D₁‘ and ‘D₂‘ represent two diodes. A potential of 5V represent the logical value 1 and a potential of 0V represents the logical value zero.

When A = 0, B = 0 both the diodes are reverse biased and there is no current through the resistance. So, the potential at ‘Q’ is zero, i.e., Q = 0. When A = 0 or B = 0 and the other equal to a potential behaves like a closed switch. The output potential then becomes 5V. i.e., Q=l. When both A and B are 1, both the diodes are forward biased and the potential at Q’ is same as that at A and B which is 5V i.e., Q =1. The output is same as that of the OR gate.

Truth table of OR gate :
The truth table of OR gate interms of low, high; 0 and 1 are below.

Logical symbol of OR gate :
The logical function OR is represented by the symbol plus. So that the output, Q = A + B

Questuion 7.
Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors?
Answer:
AND gate :

1. It has two input terminals and one output terminal.
2. When both the inputs low or one of the input is low the output is low in an AND gate.
3. The output of the gate is high only when both the inputs are high.
4. The output of the gate is high only when both the inputs are high.
5. If the input of the gate are A and B and the outputs in Q than Q is logical function of A and B. The value of Q for different combinations of A and B is shown by means of a table called truth table.

Truth Table :
It is defined as the table that shows the values of the output of all possible combinations of the value of the input variables.

Implementation of AND gate using diodes:

1. Let D₁ and D2 represent two diodes. A potential of 5V represents the logical value 1 and potential of OV represents the logical value zero (0).
2. When A = 0, B = 0, both the diodes D₁ and D₂ are forward biased and they behave like open switches. There is no current through the resistance R’ making the potential of Q equal to 5V i.e., Q = 1. The output is same as that of an AND gate.

Doping – Conductivity of semiconductors:
When a p – type impurity is doped it will develop acceptor energy levels near valance band in Forbidden region. So Forbidden gap width decreases and electrons can easily sent to conduction band.

When n-type impurities are doped they will develop donor energy levels near conduction band in Forbidden gap. As a result Forbidden gap decreases and electrons in valance band will be sent to conduction band with very little energy (< 0.01 eV).

As a result due to doping conductivity of semiconductors will increase.

Problems

Question 1.
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier.
Answer:
Internal resistance of diode r_f = 20 Ω;
Load resistance R_L = 2kΩ = 2000 Ω
Efficiency of half wave rectifier =

\(\frac{812}{2020}\) = 0.4019
% of η = 0.4019 × 100 = 40.19% = 40.2%

Question 2.
A full wave p-n junction diode rectifier uses a load resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Answer:
Load resistance R_L = 1300 Ω;
Internal resistance of diode r_f = 9 Ω
Efficiency of full wave rectifier

% of efficiency η = 0.8064 × 100 = 80.64%

Question 3.
Calculate the current amplification factor b (beta) when change in collector current is 1 mA and change in base current is 20 mA.
Answer:
Change in collector current ∆I_C = 1mA = 1 × 10^-3 A
Change in base current ∆I_b =20µA = 20 × 10^-6 A
Current amplification

Question 4.
For a transistor amplifier, the collector load resistance R_L = 2k ohm and the input resistance R_i = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Answer:
Load resistance RL = 2 kD = 2000 Ω;
Input resistance Rj = 1 kQ = 1000 Ω
Current gain b = 50 ; Voltage gain A_v = ?

Telangana TSBIE TS Inter 2nd Year Physics Study Material 14th Lesson Nuclei Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 14th Lesson Nuclei

Very Short Answer Type Questions

Question 1.
What are isotopes and isobars?
Answer:
Isotopes :
The nuclei having the same atomic number (Z) but different mass number (A) are called isotopes.
Ex: ₈O¹⁶ , ₈O¹⁷, ₈O¹⁸.

Isobars :
The nuclei having the same mass number (A) but different atomic numbers (Z) are called isobars.
Ex: 14 ¹⁴₆C, ¹⁴₇N.

Question 2.
What are isotones and isomers?
Answer:
Isotones :
The nuclei having same neutron number (N) but different atomic number (Z) are called isotones.
Ex: ₈₀Hg¹⁹⁸, ¹⁹⁷₇₉Au.

Isomers :
Nuclei having the same atomic number (Z) and mass number (A) but different nuclear properties such as radioactive decay and magnetic moments are called isomers.
Ex: ⁸⁰₃₅Br^m, ⁸⁰₃₅Br^g. Here’m’ denotes metastable state and ‘g’ denotes ground state.

Question 3.
What is a.m.u.,? What is its equivalent energy?
Answer:
Atomic mass unit (lu) :
1/12th mass of ¹²₆C atom is taken as atomic mass unit.

Energy equivalent of 1 u = 931.5 MeV.

Question 4.
What will be the ratio of the radii of two nuclei of mass numbers A₁ and A₂?
Answer:
We know the radius of the nucleus
R = R₀ A^1/3

Question 5.
Natural radioactive nuclei are mostly nuclei of high mass number, why?
Answer:
As the atomic number increases coulombian repulsive force increases in the nucleus and hence the stability of the nucleus decreases.

That is why the nuclei after lead are unstable and they exhibit natural radioactivity.

Question 6.
Does the ratio of neutrons to protons in a nucleus increase, decrease or remain the same after the emission of an a – particle?
Answer:
α – particle means Helium nucleus (₂He⁴).

If the nucleus emits α -particle, it loses 2 protons and 2 neutrons. But in the nuclei of radio active elements number of neutrons is greater than the number of protons.

When neutrons number and protons number decreases equally, their ratio increases.

Question 7.
A nucleus contains no electrons but can emit them. How?
Answer:
Natural radioactive elements undergo β – decay. Then a neutron loses one electron and converts into proton.

This electron is ejected out with high velocity called β – ray. The proton remains inside the nucleus.

Question 8.
What are the units and dimensions of the disintegration constant?
Answer:
Since \(\frac{N}{N_0}\) = e^λt . λt has no unit and no
dimensions. So unit of λ = s^-1
Dimensional formula λ = T^-1

Question 9.
Why do all electrons emitted during β^– – decay not have the same energy?
Answer:
In negative beta decay (β^–) emission of electron is accompanied by a neutrino. These neutrinos have very small mass cempared to electron and some kinetic energy. Due to this neutrino energy of electrons liberated in β^– decay is not constant.

Question 10.
Neutrons are the best projectiles to produce nuclear reactions. Why?
Answer:
Neutron is an uncharged particle so it is not deflected by electric and magnetic fields and has high penetrating power. So the neutron required lesser energy than a positive charged particle for producing nuclear reactions. Hence neutron is the best projectile for producing nuclear reactions.

Question 11.
Neutrons cannot produce ionization. Why?
Answer:
Since neutron does not posses any charge, its ionising power is very less when compared to α and β rays.

Question 12.
What are delayed neutrons?
Answer:
The neutrons which are liberated over a period of time after the fission process has taken place are called delayed neutrons.

They play an important role in the nuclear reactor.

Question 13.
What are thermal neutrons? What is their importance?
Answer:
Slow neutrons are called thermal neutrons. Their energy is 0.025 eV. They produce nuclear fission.

Question 14.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction?
Answer:
In controlled chain reaction K = 1.
In uncontrolled chain reaction K > 1.

Question 15.
What is the role of controlling rods in a nuclear reactor?
Answer:
They can control chain reaction by absorbing neutrons in nuclear reactor.

Question 16.
Why are nuclear fusion reactions called thermo nuclear reactions?
Answer:
As nuclear fusion reactions occur at very high temperature of the order of 10⁷K, the fusion reactions are known as thermonuclear reactions.

Question 17.
Define Becquerel and Curie.
Answer:
Becquerel (Bq) :
It is a unit to measure radioactivity of a substance. If a radioactive substance undergoes one disintegration or decay per second then it is called Becquerel.

Curie :
It is a unit to measure radioactivity of a substance. If a radioactive substance undergoes 3.7 × 10¹⁰ decays per second the radioactivity of that substance is called curie.
1 Curie = 3.7 × 10¹⁰ Bq (Becquerel)

Question 18.
What is a chain reaction?
Answer:

1. Chain reaction : If a fission reaction is self-maintained due to the neutrons released in that reaction then it is called chain reaction.
2. For chain reaction to takes place (1) At- least one external neutron is necessary 2) neutron multiplication factor K ≥ 1.
3. Initial mass of uranium must be greater than critical mass.

Question 19.
What is the function of moderator in a nuclear reactor?
Answer:
The function of moderator in a nuclear reactor is, to slow down the neutrons, thus neutrons will participate actively in fission reaction. It decreases the speed of neutrons from 2 MeV to 0.025 eV in the nuclear reactor. A good moderator does not absorb the neutrons.

Question 20.
What is the energy released in the fusion of four protons to form a helium nucleus?
Answer:
Energy released due to fusion of four protons is 26.7 MeV.

Short Answer Questions

Question 1.
Why is the density of the nucleus more than that of the atom? Show that the density of nuclear matter is same for all nuclei.
Answer:
a) In an atom nearly 99.9% of mass is concentrated in a very small volume called nucleus. Volume of nucleus is 10^-12 times less than volume of atom. From above explanation density of nucleus is clearly more than density of atom.

b) Density = ρ = mass / volume. But mass of nucleus m = A.m_p.

The above equation is independent of mass number of nucleus A. So density of nuclear matter is same for all nuclei.

Question 2.
Write a short note on the discovery of neutron.
Answer:
Chadwick observed emission of neutral radiation when beryllium nuclei were bombarded with alpha particles.

This neutral radiation when passed through lighter elements like helium, CO₂ and nitrogen knock out a proton.

If the particle is a photon, then its energy must be for higher than the α – particle participating in the reaction.

The energy of neutral particle and also that of a proton released when that neutral particle passes through the lighter element gave some energy discrepancy.

To explain this energy discrepancy Chadwick proposed that the neutral particle is not photon. But a new particle with its mass equal to that of proton and he called it neutron.

Mass of neutron is m_n = 1.00866 u.

In this way Chadwick discovered Neutron.

Question 3.
What are the properties of a neutron?
Answer:
Properties of a neutron :

1. Neutron is a chargeless particle and its mass is almost equal to mass of proton.
2. A free neutron is unstable when it is out side the nucleus. Its mean life period is 1000 sec.
3. Inside nucleus neutron is stable.
4. Number of neutrons in an atom is (A – Z) where A is mass number and Z is atomic number.

Question 4.
What are nuclear forces? Write their pro-perties. .
Answer:
Nuclear forces :
Forces between nucleons present inside the nucleus are called nuclear forces. Properties of nuclear forces are

1. A nuclear force is mych stronger than the coulomb force and the gravitational force.
2. Nuclear force between two nucleons are distance dependent.
3. From potential energy graph of a pair of nucleons these forces are found to be attractive forces when separation between nucleons is 0.8 Fermi or more. These forces are found to be repulsive forces when separation between nucleons is less than 0.8 Fermi.
4. Nuclear forces are saturated forces.
5. Nuclear forces are independent on charge. So nuclear force between proton-proton, proton – neutron and neutron – neutron are equal.

Question 5.
For greater stability, a nucleus should have greater value of binding energy per nucleon. Why?
Answer:
When a graph is plotted between binding energy per nucleon (E_bn) and mass number (A) for different elements it is called
E_bn – A graPh

Salient features of the graphs :
i) The binding energy per nucleon (E_bn) is constant in mass number range of 30 to 170 these elements are found to be more stable. E_bn value is maximum at 8.75 MeV / nucleon for Iron A = 56 which is highly stable.

ii) For heavy nuclei (A > 170) Binding energy per nucleon gradually decreases with increasing mass number.
Ex : Uranium has low binding energy per nucleon of 7.6 MeV. To obtain greater stability under suitable conditions it always tries to break up into two intermediate masses.

For E_bn Vs mass number A graph for region A > 30 to A < 170 is almost same we cannot break easily in two separate nuclei.

Hence atoms with high E_bn are more stable.

Question 6.
Explain α – decay.
Answer:
Alpha decay :
In α – decay ⁴₂He nuclie is emitted from given radioactive substance. Mass number of product nucleons (called daughter nucleus) is decreased by Four units and Atomic number is decreased by two units. Equation of α – decay is

Disintegration energy (OR) Q – value :
The Q – value of a nuclear reaction is the difference between the initial mass energy and total mass-energy of decay products. For α – decay Q value is given by Q = (m_x – m_y – m_He)c².

Question 7.
Explain β – decay.
Answer:
Beta decay (β) :
In Beta decay an electron (or) a positron is liberated from given radioactive substance.

Positive beta decay (β⁺) : In this decay a positron (e⁺) and a neutrino (v) are liberated from radioactive substance.
Ex: ²²₁₁Na → ²²₁₀Ne + e⁺ + v

In + Ve beta decay a proton looses positron and converts into neutron.
⇒ p → n + e⁺ + ν

Negative beta decay (β^–) :
In this decay an electron (e^–) and an antineutrino (\(\overline{\mathrm{ν}}\))are liberated.

In – Ve beta decay a neutron loses electron and converted into proton i.e.,
n → p + e“ + \(\overline{\mathrm{ν}}\)

Question 8.
Explain γ – decay.
Answer:
Nucleus also has discrete energy levels like that of an atom. The energy difference bet-ween these energy levels is in the order of MeV. They are called ground state and exi-ted states.

When a nucleon is in an excited state it will spontaneously return ground state.

While returning to ground state the nucleons will emit energy photons whose energy is equal to difference of those two energy levels.

Since energy difference is in the order of MeV. The photons emitted are highly energetic. This highly energetic radiation is called gamma(γ) radiation.

Question 9.
Define balf-life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half-life period (T_½) : It is the time taken for the number of nuclei (N) to become half of initial nuclei (N₀) i.e., N = \(\frac{N_0}{2}\).

Decay constant (λ) :
Let N is the number of nuclie in a sample. The number of nuclei (∆N) undergoing radioactive decay during the time ‘∆t’ is given by \(\frac{\triangle N}{\triangle t}\) ∝ N or \(\frac{\triangle N}{\triangle T}\) = λN.

where λ = disintegration constant (or) decay constant.

Relation between half-life peirod T_½ and decay constant λ :
From definition of halflife period N = \(\frac{N_0}{2}\) ; dt = T_½ …………… (1)
But N (t) = N₀ e^-λt ………….. (2) i:e., Number of nuclei after a time ‘t’
From above equation
T_(½) = \(\frac{log 2}{\lambda}=\frac{0.693}{\lambda}\)

Question 10.
Define average life of a radioactive substance. Obtain the relation between decay constant and average life.
Answer:
Average life time :
In a radioactive substance, some nuclei may live for a long time and some nuclei may live for a short time. So we are using average life time τ.

Relation between decay constant and average life time:

When integrated with in the limits o to ∞
⇒ τ = \(\frac{1}{\lambda}\).

Question 11.
Deduce the relation between half-life and average life of a radioactive substance.
Answer:
1) Half-life period (T_½) :
The half-life period of a radioactive nuclide is the time taken for the number of nuclei (N) to become half of initial nuclei (N₀) i.e., N = \(\frac{N_0}{2}\).

2) Average life time :
In a radioactive substance, some nuclei may live for a long time and some nuclei may live for a short time. So we are using average life time τ.

Relation between half-life period and Average life period :
half-life period τ_½ =\(\frac{log 2}{\lambda}\) …………. (1)
But from average life period τ = \(\frac{1}{\lambda}\) ….. (2)
From equations (1) & (2)
∴ τ_½ = t log²_e

Question 12.
What is nuclear fission? Give an example to illustrate it.
Answer:
Nuclear fission:
It is a nuclear reaction in which a heavy nucleus is divided into smaller nuclei along with some energy.

Explanation:
When uranium isotope ²³⁵₉₂U is bombarded with a neutron it will break up into two intermediate mass nuclei.

The atomic number of products is not constant it may vary from 35 to 57.

The fragments produced are radioactive nuclei. They will emit β – particles to achieve stable end products.

Energy liberated in this process is in the form of kinetic energy of fragments and neutrons. Later it is transferred into surroundings in the form of heat energy.

Question 13.
What is nuclear fusion? Write the Conditions for nuclear fusion to occur.
Answer:
Fusion :
The process of combining lighter nuclei to form a heavy nucleus is called “fusion reaction” (or) “nuclear fission”.

Conditions for fusion to occur :

1. For fusion to take place the two nuclie must come close enough so that nuclear attractive short range force is able to effect them.
2. Since protons have positive charge on them coulomb repulsive forces will come into account. These forces will develop nearly a potential barrier of 400 keV.
3. Fusion will be achieved if temperature is raised to nearly 108 K. So that protons will overcome coulomb repulsive forces. At this temperature the particles have enough kinetic energy to overcome coulomb repulsive force.

Question 14.
Distinguish between nuclear fission and nuclear fusion. [TS June ’15]
Answer:

Question 15.
Explain the terms ‘chain reaction’ and ‘multiplication factor’. How is a chain reaction sustained?
Answer:
1) Chain reaction :
If a fission reaction is self maintained due to the neutrons released in that reaction, then it is called “chain reaction”.

2) Multiplication factor K :
The ratio of number of fissions produced by neutrons of present generation to the number of fissions produced in preceding generation is called “multiplication factor”.

3) Condition to sustain chain reaction :
For chain reaction to takes place

1. Atleast one external neutron is necessary.
2. Neutron multiplication factor K ≥ 1.
3. Initial mass of uranium must be greater than critical mass.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number? What is its significance?
Answer:
Mass defect :
The difference in mass of nucleus and its constituents is known as mass defect. In every nucleus the theoretical mass (Mt) is always less than practical mass (M).

Mass defect ∆m = [Zm_p + (A – Z) m_n] – M

Binding energy :
The energy released while forming a nucleus is called Binding energy E_b. Binding energy E_b = ∆mc².

When a certain number of protons and neutrons are brought together to form a nucleus then certain amount of energy E_b is released.

To divide a nucleus into its constituents, we have to supply an amount of energy equals to E_b from outside.

Binding energy per nucleon :
The ratio of Binding energy E_b to mass number ‘A’ of nucleus is cated’Binding energy per nucleon”.

Binding energy per nucleon E_bn = \(\frac{E_b}{A}\).

Binding energy per nucleon (E_bn) – Mass number (A) graphs :
When a graph is plotted between binding energy per nucleon (E_bn) and mass number (A) for different elements it is called “E_bn -A graph”.

Salient features of the graphs:

1. The binding energy per nucleon (E_bn) is almost constant within mass number range of 30 to 170.
2. For mass number A < 30 Binding energy per nucleon is less it gradually increases with increasing of mass number.
   This region suggested that when lighter nuclei are fused to form large nucleus then they release energy (Fusion process).
3. For heavy nuclei(where A > 170) Binding energy per nucleon gradually decreases with increasing mass number (A).

This region suggested that when larger nuclei are divided into small nuclei then energy is released (Fission process).

Question 2.
What is radioactivity? State the law of radioactive decay. Show that radioactive decay is exponential in nature. [TS May 18. 16; Mar. 16]
Answer:
Radioactive decay :
The spontaneous disintegration of unstable nucleus is referred as “radioactivity or radioactive decay”.

When a nucleus undergoes radioactive decay three types of radioactive decay takes place.

1. α – decay : In this process ⁴₂He nuclei are emitted.
2. β – deay : In this process electrons or positrons are emitted.
3. γ – decay : In this process high energy photons (E.M. Waves) are liberated.

Law of radioactive decay :
Let N is the number of nuclie in a sample. The number of nuclie (∆N) undergoing radioactive decay during the time ‘∆t’ is given by
\(\frac{\triangle N}{\triangle t}\) ∝ N or \(\frac{\triangle N}{\triangle t}\) = λN

Where λ is disintegration constant or decay constant.

Decay rate (R) (OR) activity :
The total decay rate of a sample is the number of nuclei disintegrating per unit time.
∴ Total decay rate R = – \(\frac{dN}{dt}\) (OR)
R = R₀ e^-λt (or) R = λN (activity)

Total decay rate is also called activity. Radioactive decay is exponential because the total decay rate R = R₀ e^-λt Where R0 is a constant.
∴ R ∝ e^-λt Here e-W is exponential function whose value decreases with time.

Hence total radioactive decay of a substance decreases exponentially.

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. [TS Mar. 17, 15, AP Mar. ’17, ’16, ’15, ’14; May 18, 16, 14; AP & TS Mar. 19, 18; May 17; AP June 15]
Answer:
Principle :
A nuclear reactor works on the principle of “sustained and controlled chain reaction”.

The labelled diagram of nuclear reactor is as shown. The important parts of nuclear reactor are

1) Core :
The core of the reactor is the site of nuclear fission. Generally it is made with graphite bricks. Core contains fuel elements in suitably fabricated form.

2) Fuel :
Generally used fuel is enriched uranium which contain ²³⁵₉₂U at high conentration than in natural uranium. Fuel is responsible for fission reaction in reactor.

3) Reflector :
Generally inner part of core is provided with some reflecting material which prevents leakage of neutrons from core.

4) Coolant :
The purpose of coolant is to remove heat energy produced in reactor, generally water is used as a coolant.

5) Moderator :
Core is filled with Heavy water. It is a good moderator. Purpose of moderator is to slow down the neutrons. Slow neutrons will effectively participate in fission reaction.

6) Control rods :
Purpose of control rods is to reduce number of neutrons inside the core. Generally control rods are made with neutron absorbing material like boron or cadmium.

Radiation shield :
The whole assembly of core is shielded with suitable material which prevents leakage of radioactive radiation from core.

Working :
Due to fission reaction energy is released in the core. It is transferred to an outside tank by means of coolant. As a result water in the tank is heated and steam is produced. This steam is used to run a turbine and power is produced. Power produced in this method is called atomic power.

Question 4.
Explain the source of stellar energy. Explain the carbon-nitrogen cycle, proton-proton cycle occurring in stars. [TS June ’15]
Answer:
The source of stellar energy is nuclear fusion reaction, for which hydrogen is the fuel.

Energy from Sun and Stars :
Sun and Stars have been radiating huge amounts of energy by nuclear fusion reactions taking place in their core, where the temperature is of the order 10⁷ K. More scientists proposed two types of cyclic processes for the source of energy in the Sun & Stars. They are :

Carbon – Nitrogen Cycle :
Carbon – Nitrogen cycle is one of the most important nuclear reactions for the production of solar energy by fusion.

The entire cycle can be summed up as,
4₁H¹ → ₂He⁴ + 2 _{+ 1}e⁰ + 3γ + 2υ + 26.70MeV

The energy released during this process is 26.70 MeV.

Proton-Proton Cycle : [TS June ’15]
At higher temperature, the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form higher nuclei of helium ³₂He. Two such helium nuclei combine with another proton releasing a total amount of energy 25.71 MeV. The nuclear fusion reactions are given below.
₁H¹ + ₁H¹ → ₁H² + ₊₁e⁰ + υ + 0.42MeV ………. (1)
e⁺ + e^– → γ + γ + 1.02MeV …………. (2)
₁H² + ₁H² → ₂He³ + γ + 5.49MeV ………. (3)
₂He³ + ₂He³ → ₂He⁴ + 2 ₁H¹ + energy 12.86 MeV ………….. (4)

For this equation to takes place first three equations must occur twice.

The nef result of the above reactions is 4 ₁H¹ → ₂He⁴ + 2 ₊₁e⁰ + 2γ + 2υ + 26.70 MeV

The energy released during this process is 26.70 MeV.

Problems

Question 1.
Show that the density of a nucleus does not depend upon its mass number (density is independent of mass).
Answer:
Volume of nucleus

But volume V ac A (mass number )
Density of nucleus matter

The above expression is independent of mass number ‘A’. Hence density of nuclear matter is independent of mass number of nucleus.

Question 2.
Compare the radii of the nuclei of mass numbers 27 and 64.
Answer:
Ratio of radii of nuclie is

Question 3.
The radius of the oxygen nucleus ¹⁶₈O is 2.8 × 10^-15m. Find the radius of lead nucleus ²⁰⁵₈₂Pb.
Answer:
Given, Radius of the oxygen nucleus R_o = 2.8 × 10^-15 m ;
Mass number of oxygen, A₀ = 16 ;
Mass number of lead = A_pb = 205 ; Radius of the lead nucleus R_pb = ?
Radius of nucleus

Question 4.
Find the binding energy of ⁵⁶₂₆Fe. Atomic mass of Fe is 55.9349 u and that of Hydrogen is 1.00783u and mass of neutrons is 1.00876 u.
Answer:
Mass of Hydrogen atom M_H = 1.00783u;
Mass of neutron m_n = 1.00867u
Mass of Iron atom M₁ = 55.9349u
Atomic number Z = 26;
Mass number A = 56;
Mass defect ∆m = [Zm_H + (A – Z)m_n – M_I]
∆m = [26 × 1.00783 + (56 – 26) 1.00867] – 55.9349 u
∆m = 0.52878 u
Binding energy = ∆mc² = 0.52878 × (3 × 10⁸)²
∴ B.E. = 0.52878 × 9 × 10¹⁶ or 0.52878 × 931.5 MeV
∵ [lu × C²] = 931.5 MeV energy
∴ B.E. =492.55
Binding energy per nucleon BE/ n = \(\frac{492.55}{56}\)
= 8:79 MeV

Question 5.
How much energy is required to separate the typical middle mass nucleus ¹²⁰₅₀Sn k its constituent nucleons? (Mass of ¹²⁰₅₀Sn = 119.902199 u, mass of proton = 1.007825 u and mass of neutron = 1.008665 u)
Answer:
Mass of proton m_p = 1.007825 amu ;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of Sn, Z = 50 ;
Mass number of Sn, A = 120 ;
Mass of nucleus of Sn atom = 119.902199 amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z) m_n – M_N]
= [ (50) (1.007825) + (120 – 50) (1.008665) – 119.902199]
= (50 × 1.007825 + 70 × 1.008665 -119.902199)
= (50.39125 + 70.60655 – 119.902199)
∆m = [120.9978 – 119.902199] = 1.095601 amu

(ii) Energy required to separate the nucleons = Binding energy of the nucleus
B.E = ∆mc² (∵ lamu × c² = 931.5MeV)
= ∆m × 931.5MeV = 1.095601 × 931.5 MeV
= 1021 MeV

Question 6.
Calculate the binding energy of an α – particle. Given that mass of proton – 1.0073u, mass of neutron = 1.0087u, and mass of α – particle = 4.0015u.
Answer:
An α – particle is nothing but helium nucleus ⁴₂He.
It contains 2 protons, 2 neutrons with a mass number A = 4.
Mass of proton m_p = 1.0073 amu ;
Mass of neutron m_n = 1.0087 amu ;
Atomic number of helium Z = 2 ;
Mass number of helium A = 4 ;
Mass of helium atom M_N = 4.0015 amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z)m_n – M_N]
= [(2) (1.0073) + (4 – 2) (1.0087) – 4.00260]
= (2 × 1.0073 + 2 × 1.0087-4.00260)
= (2.0146 + 2.0174)-4.0015 Am = [4.032 – 4.0015] = 0.0305 amu

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ lamu × c² = 931.5MeV)
= ∆m × 931.5 MeV = 0.0305 × 931.5 MeV
= 28.41 MeV

Question 7.
Find the energy required to split ¹⁶₈O nucleus into four α – particles. The mass of an α – particle is 4.002603u and that of oxygen is 15.994915u.
Answer:
The energy required to split
Q = [ Total mass of the products
– Total mass of the reactants] c²
= [Mass of four ⁴₂He – Mass of ¹⁶₈O] × c²
= [(4 × 4.002603) -15.994915] amu × c²
= [16.010412 – 15.994915] amu × c²
= (0.015497) 931.5 MeV = 14.44 MeV.

Question 8.
Calculate the binding energy per nucleon of ³⁵₁₇Cl nucleus. Given that mass of ³⁵₁₇Cl nucleus = 34.9800u, mass of proton = 1.007825u, mass of neutron = 1.008665 u and lu is equivalent to 931 MeV.
Answer:
Given, Mass of proton m_p = 1.007825 amu;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of ³⁵₁₇Cl, Z = 17 ;
Mass number of ³⁵₁₇Cl, A = 35 ;
Mass of nucleus of ³⁵₁₇Cl atom =34.98 amu.

(i) Mass defect ∆m = [Zm_p + (A – Z)m_n – M_N]
= [(17) (1.007825) + (35 – 17) (1.008665) – 34.98]
= (17 × 1.007825 + 18 × 1.008665 – 34.98)
= (17.13303 + 18.15597 -34.98)
∆m = [35.289 – 34.98] = 0.3089

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ 1amu × c² = 931.5 MeV)
B.E = ∆m × 931.5 MeV = 0.3089×931.5 MeV
= 287.83 MeV

(iii) Binding energy per nucleon of Cl = \(\frac{B.E}{A}\)
= \(\frac{287.83}{35}\) = 8.22 MeV

Question 9.
Calculate the binding energy per nucleon of ⁴⁰₂₀Ca. Given that mass of ⁴⁰₂₀Ca nucleus = 39.962589 u, mass of a proton = 1.007825 u ; mass of neutron = 1.008665 u and 1 u is equivalent to 931 MeV.
Answer:
Given, Mass of proton m_p = 1.007825 amu;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of ⁴⁰₂₀Ca Z = 20 ;
Mass number of ⁴⁰₂₀Ca A = 40 ;
Mass of nucleus of ⁴⁰₂₀Ca atom = 39.962589 amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z)m_n – M_N]
= [ (20) (1.007825) + (40 – 20) (1.008665) – 39.962589]
= (20 × 1.007825 + 20 × 1.008665 – 39.962589)
= (20.1565 + 20.1733 – 39.962589)
∆m = [ 40.3298 – 39.962589 ] = 0.3672 amu

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ lamu × c² = 931.5 MeV)
∴ B E = ∆m × 931.5MeV
= 0.3672 × 931.5 MeV = 342.06 MeV

(iii) Binding energy per nucleon of Ca = \(\frac{B.E}{A}\)
= \(\frac{342.06}{40}\) = 8.55 MeV

Question 10.
Calculate(i) maw defect, (ii) binding energy and (iii) the binding energy per nucleon of ¹²₆C nucleus. Nuclear mass of ¹²₆C 12.000000 u; maw of proton = 1.007825 u and maw of neutron = 1.008665 u.
Answer:
Given, Mass of proton m_p = 1.007825 amu;
Mass of neutron m_n = 1.008665 amu ;
Atomic number of ¹²₆C, Z = 6 ;
Mass number of ¹²₆C, A = 12;
Mass of nucleus of ¹²₆C atom = 12.00amu.

(i) Mass defect, ∆m = [Zm_p + (A – Z)m_n – M_N]
= [6(1.007825) + (12 – 6) (1.008665) – 12.00]
= (6 × 1.007825 + 6 × 1.008665 – 12.00)
= (6.04695 + 6.05199 – 12.00)
∆m = [12.09894 – 12.00] = 0.09894 amu

(ii) Binding energy of the nucleus B.E = ∆mc²
(∵ lamu × c² = 931.5 MeV)
∴ B.E = ∆m × 931.5 MeV
= 0.09895 × 931.5 MeV = 92.16 MeV

(iii) Binding energy per nucleon of carbon =
\(\frac{B.E}{A}=\frac{92.16}{12}\) = 7.86 MeV

Question 11.
The binding energies per nucleon for deuterium and helium are 1.1 MeV and 7.0 MeV respectively. What energy in joules will be liberated when 10v deuterons take part in the reaction.
Answer:
Given, Binding energies per nucleon for deuterium (\(\frac{B.E}{A}\))_D = 1.1 MeV
Mass number of deuterium A = 2
Binding energy of deuterium
B.E = (\(\frac{B.E}{A}\))_D × A = 1.1 × 2 = 2.2 MeV
Binding energies per nucleon for helium (\(\frac{B.E}{A}\))_He = 7.0 MeV
Mass number of helium A = 4
Binding energy of helium
B.E = (\(\frac{B.E}{A}\))_D × A = 7 × 4 = 28MeV
We know, ₁H² + ₁H² → ₂He⁴
Energy released = B.E. of 10⁶ deuterons
– B.E. of \(\frac{1}{2}\) × 10⁶ Helium atoms
Binding energy = 2.2 × 10⁶ – \(\frac{1}{2}\) × 10⁶ × 28
= 10⁶ (2.2 – 14) = – 11.8 × 10⁶ MeV
= -11.8 × 10⁶ × 1.6 × 10^-31J
= -18.88 × 10^-7J
(- ve sign indicates that energy is released)
∴ Energy released = 18.88 × 10^-7J

Question 12.
Bombardment of lithium with protons gives rise to the following reaction :

Find the Q – value of the reaction. The atomic masses of lithium, proton, and helium are 7.016u, 1.008u, and 4.004u respectively.
Answer:
Given, Mass of Lithium = 7.016 amu ;
Mass of Proton = 1.008 amu ;
Mass of Helium = 4.004 amu ;
1 a.m.u = 931.5 MeV energy
Q = [Total mass of the reactants – total mass of the products] c²
= [mass of Lithium + mass of Proton – ( 2 × mass of Helium)] 931.5 MeV
= [ 7.016 + 1.008 – 2 ( 4.004 ) ] 931.5 MeV
= [8.024 – 8.008] 931.5. MeV
∴ Energy Q. = 0.016 × 931.5 = 14.904 MeV

Question 13.
The half-life of radium is 1600 years. How much time does 1 g of radium take to reduce to 0.125 g. [TS May, ’18, ’16, Mar. ’16]
Answer:
1g become \(\frac{1}{2}\) g after one half-life period and it become \(\frac{1}{4}\) g after 2 half-life periods and \(\frac{1}{8}\) after 3 half-life periods.
But \(\frac{1}{8}\) g = 0.125g.
So number of half-life periods, n = 3.
Time taken = n x half-life period
= 3 × 1600 = 4800 years.

Question 14.
Plutonium decays with a half-life of 24.000 years. If plutonium is stored for 72,000 years, what fraction of it remains?
Answer:
Given, Half-life of Plutonium = 24,000 years;
The duration of the time = 72,000 years
Initial mass = x g ; Final mass = mx g

Fraction of Plutonium that remains

Question 15.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Answer:
Given, Fraction of substance decays

Question 16.
The half-life period of a radioactive sub¬stance is 20 days. What is the time taken for 7 / 8th of its original mass to disintegrate?
Answer:
Given, Half-life period = 20 days

∴ Time taken to disintegrate
= n × Half-life time = 3 × 20 = 60 days

Question 17.
How many disintegration per second will occur in one gram of ²³⁸₉₂U if its half – life against ∝ – decay is 1.42 × 10¹⁷ s?
Answer:
Given, Half-life period T = 1.42 × 10¹⁷s.;
mass of uranium m = 238

Number of disintegrations / sec = nλ

Question 18.
The half – life of a radioactive substance is 100 years. Calculate in how many years die activity will decay to 1/10th of its initial value.
Answer:
Given, Half – life period = 100 years

∴ Time taken to disintegrate
= n × Half-life time = 3 × 20 = 60 days

Question 19.
One gram of a radium is reduced by 2 milligram in 5 years by β – decay. Calculate the half-life of radium.
Answer:
Initial Mass of radium M₀ = 1 gram ;
∴ Mass reduced = 2mg
Final mass of radium M = 1 – 0.002
= 0.998 mg;
Time taken to reduce the mass t = 5 years
But number of atoms N α mass of the sample

Question 20.
The half-life of a radioactive substance is 5000 years. In how many years, its activity will decay to 0.2 times of its initial value? Give log₁₀ 5 = 0.6990.
Answer:
Half-life period T = 5000 years ;
Activity A = Nλ = 0.2 times initial value
Initial activity A₀ = N₀λ
In radioactivity N = N₀e^-λt

Question 21.
An explosion of atomic bomb releases an energy of 7.6 x 1013 J. If 200 MeV energy is released on fission of one atom calculate CD the number of uranium atoms undergoing fission. 00 the mass of uranium used in the bomb.
Answer:
i) Energy released E = 7.6 × 10¹³J
Energy released per fission = 200 MeV
= 200 × 10⁶ × 1.6 × 10^-19J
= 200 × 1.6 × 10^-13J = 3.2 × 10^-11 J

ii) Number of Uranium atoms participated

Question 22.
If one microgram of ²³⁸₉₂U is completely destroyed in an atom bomb, how much energy will be released? [AP Mar. ’19]
Answer:
Given, Mass of Uranium destroyed = 1
Micro gram = 1 × 10^-6 gr = 10^-9 Kg.
According to Einstein mass energy relation Energy released E = me² = 10^-9 × [ 3 × 10⁸]² (∵ Velocity of light c = 3 x 108 m/s)
∴ Energy released = 9 × 10^-9 × 10¹⁶ = 9 × 10⁷J

Question 23.
Calculate the energy released by fission from 2g of ²³⁸₉₂U in kWh. Given that the energy released per fission is 200 MeV.
Answer:
Given, Mass of Uranium (m) = 2g ;
Energy per fission = 200 MeV
Number of atoms in 2 grams of Uranium

Energy released per fission = 200 MeV
= 200 × 10⁶ × 1.6 × 10^-19J
= 200 × 1.6 × 10^-13 J = 3.2 × 10^-11J

Total Energy released ’Q’ = 5.1256 × 3.2 × 10^-11 = 1640.2 × 10⁸J
But, 1 kWh = 1000 × 60 × 60 = 36 × 10⁵J

Question 24.
200 MeV energy is released when one nucleus of ²³⁸U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Answer:
Given, Total power produced = 1 Mega Watt = 10⁶ Watt
Energy per fission = 200 MeV
= 200 × 1.6 × 10^-13 J = 3.2 × 10^-11 J
∴ Number of fissions per second

Question 25.
How much ²³⁸U is consumed in a day in an atomic power house operating at 400 MW, provided the whole of mass ²³⁸U is converted into energy?
Answer:
Power P = 400 MW = 400 × 10⁶ watt;
Time T = 1 day = 86,400 sec.
∴ Energy produced per day
= 400 × 10⁶ × 86400 = 3.456 × 10¹³J
But Energy E = me² × m = E/c²
Mass of uranium consumed

Telangana TSBIE TS Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 11th Lesson Electromagnetic Waves

Very Short Answer Type Questions

Question 1.
What is the average wavelength of X-rays?
Answer:
Wavelength range of X-rays = 10^-8 m to 10^-13 m i.e., 1 nm to 10^-4 nm.
∴ Average wavelength of X-rays is \(\frac{1+10^{-4}}{2}\) nm = 0.5 nanometers (nearly)

Question 2.
Give any one use of infrared rays. [TS Mar. ’19; AP May ’18. ’17]
Answer:

1. Infrared detectors are used in satellites both for military purpose and to observe growth of crops.
2. Infrared radiation is responsible to keen the atmosphere warm through green house effect.

Question 3.
If the wavelength of electromagnetic radiation is doubled, what happens to die energy of photon? [TS Mar. ’16; June ’15]
Answer:
Energy of electromagnetic photon E = hυ = \(\frac{hc}{\lambda}\)

So when wavelength λ is doubled energy of photon is reduced to half.

Question 4.
What is the principle of production of electromagnetic waves?
Answer:
Electromagnetic waves are produced by accelerating charges through conductors.

Question 5.
How are Microwaves produced? [AP Mar. ’15]
Answer:
Microwaves can be produced by special type of vacuum tubes. Namely Klystrons, magnetrons and yun diodes.

Question 6.
What is sky wave propagation? [AP June ’15]
Answer:
In the frequency range from a few MHz upto about 30 MHz, long distance communication can be achieved by the ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation and it is used by short wave broadcast services.

Question 7.
What is the ratio of speed of infrared rays and ultraviolet rays in vacuum?
Answer:
Ratio =1:1
In vacuum speed of electromagnetic waves v = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
It is applicable to whole range of electromagnetic spectrum.
∴ In vacuum velocity of UV rays = Velocity of I.R rays.

Question 8.
What is the relation between the amplitudes of the electric and magnetic fields in free space for an electromagnetic wave?
Answer:
Amplitude of Magnetic field

Relation between amplitude of Magnetic field B0 and Electric field E0 in vacuum is B₀ = \(\frac{E_0}{c}\)

Question 9.
What are the applications of microwaves? [AP Mar. 18, 17, 15; May 18, 16 , 15; June 15; TS Mar. 18, 17,15, May. 18]
Answer:

1. Microwaves are widely used in radar because of their small wavelength.
2. Microwaves are used in microwave ovens due to the frequency of microwave region is nearly equals to resonant frequency of water molecule.

Question 10.
Microwaves are used in Radars, why?
Answer:
Due to short wavelengths microwaves are suitable for the radar systems used in air craft navigation.

Question 11.
Give two uses of infrared rays. [TS Mar. ’17; May ’16; AP Mar. ’19, ’16; May ’14]
Answer:

1. Infrared detectors are used in satellites both for military purpose and to observe growth of crops.
2. Infrared radiation is responsible to keen the atmosphere warm through green house effect.

Question 12.
The charging current for a capacitor is 0.6 A. What is the displacement current across its plates?
Answer:
Charging current i = conduction current

∴ Displacement current i_d = 0.6 A across the plates.

Short Answer Questions

Question 1.
What does an electromagnetic wave consists of? On what factors does its velocity in vacuum depend?
Answer:
According to Maxwell’s theory, accelerated charges radiate electromagnetic waves.

Suppose a charge oscillates with some frequency. It produces an oscillating electric field in space. This oscillating electric field produces oscillating magnetic field. It again regenerates oscillating electric field. These electric and magnetic fields are mutually perpendicular and also perpendicular to direction of propagation. Electric field component E_x = E₀ sin (kz – ωt) Magnetic field component By = B0 sin (kz – ωt)

where k = \(\frac{2 \pi}{\lambda}\) and speed of wave v = \(\frac{\omega}{k}\)

According to Maxwell’s equations, relation between E₀ and B₀ \(\frac{E_0}{B_0}\) = c or B₀ = \(\frac{E_0}{c}\)

In vaccum velocity of electromagnetic wave c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
∴ Velocity of electromagnetic wave in vacuum depends on permeability µ₀ and permittivity ∈₀ of vacuum.

Question 2.
What is Greenhouse effect and its contribution towards the surface temperature of earth?
Answer:
Greenhouse effect :
Infrared rays plays an important role in maintaining average temperature of earth’s atmosphere.

During daytime, earth absorbs the energy of incoming Infrared & Visible radiation from Sun. As a result earth’s surface get heated. Hot earth will reradiate energy in the form of infrared radiation.

Many molecules such as Carbon dioxide (CO₂), Ammonia (NH₃), Chloro Fluoro Carbons, Methane (CH₄), etc. will absorb the long range infrared radiation because the resonant frequency of these gases matches with infrared region. So energy reradiated by earth is trapped by these molecules as a result earth’s atmosphere is heated. This is known as “greenhouse effect”.

Effect of greenhouse gases :
When concentration of greenhouse gases such as CO₂, NH₃, CH₄ etc., increases average temperature of .earth is also gradually increase. This effect is called global warming.

Long Answer Questions

Question 1.
Give the brief history of discovery of knowledge of electromagnetic waves.
Answer:
Maxwell conducted an experiment to apply the ampere’s law to study the magnetic field at a point outside the capacitor by applying a time varying current.

Consider a point ‘P’ outside the parallel plate capacitor [See Fig (a)]. Consider a loop of radius ‘r’. Its direction is perpendicular to the conductor.
From Amperes Law B(2πr) = µ₀ i (t) → (1)

Consider another different surface [See Fig (b)]. It is a pot like dielectric surface without electrical contact with capacitor plate. But it encloses the capacitor plate.

By applying ampere’s Law LHS of eq. (1) is not changed but RHS = 0. Because there is no current flowing in the plate.
∴ B(2πr) = 0 → (2)

Consider another surface like a tiffin box [See Fig (c)] apply ampere’s circuital law with same parameters LHS of eq (1) is same. But RHS is zero
i.e. B(2πr) = 0 → (3)

From eq. (1) there is a magnetic field at ‘P’ But for the figures (b) & (c) field at ‘P’ is zero. So there is some contradiction in calculating magnetic field.

Maxwell suggested that there is some electric field E = \(\frac{Q}{Ae_0}\) between the plates
and it is perpendicular to the plates. E has some value in between the plates and vanishes outside the plate.

Electric flux between the plates Φ_E = |E|A\(\frac{Q}{\epsilon_0}\)

Charge Q on plates of capacitors changes with time as a result Φ_E changes with time.

Changing electric flux will produce the current called displacement current i_d = ∈₀\(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\).

So Maxwell suggested that source of magnetic field is not only due to conduction current i_c

but due to a total current i = i_c + i_d and

So source of magnetic field is not just conduction current, it is also produced by displacement current.

Displacement current will have same physical properties just like conduction current.

Basing on this he formulated that a time varying electric field will produce a time varying magnetic field. Similarly a time varying magnetic field will produce a time varying electric field. In this way energy oscillates between perpendicular electric and magnetic fields in an electromagnetic wave.

Question 2.
State six characteristics of electromagnetic waves. What is Greenhouse effect?
Answer:
From Maxwell’s theory

1. accelerated charges radiate electromag¬netic waves.
2. Frequency of electromagnetic wave is equal to frequency of oscillator.
3. Energy associated with the propagation of wave is obtained from oscillating source.
4. From Maxwell’s equations relation between E₀ and B₀ is \(\frac{E_0}{B_0}\) = c or B₀ = \(\frac{E_0}{c}\)
5. In vacuum velocity of electromagnetic wave c = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)
6. Hertz experiments on electromagnetic waves showed that electromagnetic waves of wavelength 10 million times more than light waves could be diffracted, reflected and polarised.
   Electromagnetic waves carry energy and momentum like other waves.
7. In vacuum speed of electromagnetic waves v = \(\frac{1}{\sqrt{\mu_0 \epsilon_0}}\)

It is applicable to whole range of electromagnetic spectrum.

Greenhouse effect :
Infrared rays plays an important role in maintaining average temperature of earth’s atmosphere.

During daytime earth absorbs the energy of incoming Infrared & Visible radiation comming from Sun. As a result earth’s surface get heated. Hot earth will reradiate energy in the form of infrared radiation.

Many molecules such as CO₂ Ammonia (NH₃), Chloro Fluro Carbons, Methane CH₄ etc., will absorb the long range infrared radiation because the resonant frequency of these gases matches with infrared region. So energy reradiated by earth is trapped by these molecules as a result earth’s atmosphere is heated. This is known as “greenhouse effect”.

Intext Question and Answers

Question 1.
What physical quantity is the same for X- rays of wavelength 10^-10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 in?
Answer:
The speed of light (3 × 10⁸ m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.

Question 2.
A plane electromagnetic wave travels in ‘ vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.

Frequency of the wave, o = 30 MHz
= 30 × 10⁶ s^-1
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Wavelength of a wave is given as, λ = \(\frac{c}{v}\)
\(\frac{3\times10^8}{30\times10^6}\) = 10 m

Question 3.
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
A radio can tune to minimum frequency, υ₁ = 7.5 MHz = 7.5 × 10⁶ Hz
Maximum frequency, υ₂ = 12 MHz
= 12 × 10⁶ Hz ;
Speed of light, c = 3 × 10⁸m/s
Relation between λ and υ is λ₁ = \(\frac{c}{υ_1}\)

Thus the wavelength band of the ratio is 40 m to 25 m.

Question 4.
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 10⁹ Hz.

Question 5.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
Magnetic field B₀ = 510 nT = 510 × 10^-9 T ;
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀ = 3 × 10⁸ × 510 × 10^-9 = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

Telangana TSBIE TS Inter 2nd Year Physics Study Material 13th Lesson Atoms Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 13th Lesson Atoms

Very Short Answer Type Questions

Question 1.
What is the angular momentum of electron in the second orbit of Bohr’s model of hydrogen atom?
Answer:
Angular momentum L = \(\frac{nh}{2 \pi}\)
For 2nd orbit ⇒ n = 2.
∴ Angular momentum of 2nd orbit
L₂ = \(\frac{2h}{2 \pi}=\frac{h}{\pi}\)

Question 2.
What is the expression for fine structure constant and what is its value?
Answer:
Fine structure constant a = \(\frac{2 \pi e^2}{ch}=\frac{1}{137}\)
= 7.30 × 10^-3.

Question 3.
What is the physical meaning of ‘negative energy of an electron’?
Answer:
Energy is always +ve. There is no negative energy. But negative energy of electron means it is the force of attraction with which an electron is bounded to the nucleus.

Question 4.
Sharp lines are present in the spectrum of a gas. What does this indicate?
Answer:
When gases are excited they will absorb the exact quanta of energy equals to energy level difference of the orbits.
i.e., E = hv = E_f – E_i

So electron goes to higher orbit. As a result we will see sharp lines in the spectrum of gases corresponding to that energy.

Question 5.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Angular momentum (L) = mvr,
Dimensions = ML²T^-1
Planck’s constant h has same dimensions of L.
∵ L = \(\frac{nh}{2 \pi}\), where \(\frac{n}{\pi}\) is dimensionless.

Question 6.
What is the difference between α-particle and helium atom?
Answer:
α – particle is helium nucleus. But not helium atom.

α – particle contains 2 protons and two neutrons where as helium atom contains 2 protons, two neutrons and two electrons.

Question 7.
How is impact parameter related to angle of scattering?
Answer:
When impact parameter is minimum head on collision between α – particle and nucleus takes place and α- particle will be deviated by θ = π radians, when impact parameter is high or large the α – particle goes nearly undeviated.

Question 8.
Among alpha, beta and gamma radiations, which get affected by the electric field?
Answer:
α and β particles are affected by electric fields because both of them carries charge on them.

γ – ray is radiation so it is not affected by electric field.

Question 9.
What do you understand by the phrase ‘ground state atom’?
Answer:
The lowest state of the atom which is the lowest energy, with the electron revolving in the orbit of smallest radius is called the ground state.
For ground state hydrogen atom E = – 13.6 eV,

Question 10.
Why does the mass of the nucleus not have any significance in scattering in Rutherford’s experiment?
Answer:
In Rutherford experiment gold foil is used. For gold Z = 79 and its atomic weight is nearly 50 times more than α – particle. So mass of nucleus used to bombard gold atom has no significant effect.

Question 11.
The Lyman series of hydrogen spectrum line in the ultraviolet region Why?
Answer:
For Lyman series \(\frac{1}{\lambda}\) = R(\(\frac{1}{l^2}-\frac{1}{n^2}\))
n = 2, 3, 4 ….
Wavelength of 1st member of Lyman series is 1216A°. It is in ultraviolet region.

Question 12.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:

Question 13.
The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 1216Å, 6463 Å and 9546Å. Which one of these wavelengths belongs to the Paschen series?
Answer:
λ = 9546Å belongs to Paschen series. Paschen series is in infrared region. So wavelength λ = 9546Å is in infrared region. So it belongs to Paschen series.

Question 14.
Give two drawbacks of Rutherford’s atomic model. [TS Mar. ’19]
Answer:

1. Electron revolving around the nucleus must be continuously accelerated. An accelerating electron must lose energy continuously due to radiation and finally, atom must destroy.
2. When electron radiates energy it will spiral around nucleus. As a result its angular velocity and frequency of spectral lines must change continuously. But these two things are not taking place.

Short Answer Questions

Question 1.
What is impact parameter and angle of scattering? How are they related to each other?
Answer:
Impact parameter :
It is the perpendicular distance of the initial velocity vector of α-particle from centre of nucleus.

In case of head-on collision, impact parameter is minimum and α – particle rebounds back (θ = π). For a large impact parameter α – particle goes undeviated. The chance of head on collision is very small. It, in turn, suggested that mass of atom is much concentrated in a small volume.

Angle of scattering ‘0’ :
It is the angle between the direction of incident α – particle and scattered α – particle.

Relation between impact parameter and angle of scattering, when Impact parameter is less ⇒ angle of scattering is high and vice-versa.

Question 2.
Derive an expression for potential and kinetic energy of an electron in any orbit of a hydrogen atom according to Bohr’s atomic model. How does P.E. change with increasing n? [TS June 15; Mar. 15]
Answer:
According to Bohr model, electrons are revolving around the nucleus in certain permitted orbits.

For electron to revolve in orbit, the electrostatic force and centrifugal force must be equal i.e., F_c = F_e

∴ Kinetic energy of electron (K) = \(\frac{1}{2}\) mv²

Potential energy between electron and nucleus is

– ve sign indicates force of attraction. Relation between potential energy and radius of orbit.
From the equation (2) U ∝ \(\frac{1}{r}\) …………… (3)

From eq. (3) & (4) potential energy
U ∝ \(\frac{1}{n^2}\)
∴ Potential energy of orbit is inversely proportional to square of number of orbit (n²).

Question 3.
What are the limitations of Bohr’s theory of hydrogen atom? [TS May 18, 17; AP Mar. 17, 14, May 17]
Answer:
Limitations of Bohr model:
i) Bohr model is applicable to hydrogen atom only. It ean not be extended even to a two electron system such as helium.

Because it involves force between + vely charged nucleus and electron.

Electrical forces between electrons are not taken into account.

ii) It is not able to explain the intensity variation in spectral lines of different frequencies and why some transitions are more preferred than others.

Question 4.
Explain the distance of closest approach and impact parameter.
Answer:
Distance of closest approach :
In α- particle scattering experiment the α – particle will move near to gold nucleus until it is just stopped.

Kinetic energy of α – particle just before stopped is equal to electrostatic potential.

where = Z atomic number of gold, ’2e’ charge on α – particle.
distance of closest approach d = \(\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{k}}\)

Impact parameter :
It is the perpendicular distance of the initial velocity vector of α – particle from centre of nucleus.

In case of head-on collision impact parameter is minimum and α – particle rebounds back (θ = π). For a large impact parameter α – particle goes undeviated. The chance of head on collision is very small. It in turn suggested that mass of atom is much concentrated in a small volume.

Question 5.
Give a brief account of Thomson model of atom. What are its limitations?
Answer:
J.J. Thomson Model :
J.J.Thomson thought that the positive charge of the atom is uniformly distributed through out the atom and the negatively charged electrons are embedded in it like seeds in a watermelon.

Limitations:

1. Rutherford α – particle scattering experiment showed that in an atom positive charge is concentrated at nucleus of atom.
2. In Thomson model the radiation emitted by solids and gases is due to oscillations of atoms and molecules are governed by the interactions between them.

But experiments on rarefied gases showed that hydrogen always gives rise to a set of lines with fixed wavelength.

Balmer experiments and his formula for Balmer series of wavelength of a group of lines are emitted by atomic hydrogen only. Not by interaction of atoms or molecules.

Question 6.
Describe Rutherford atom model. What are the drawbacks of this model? [AP & TS Mar. ’16]
Answer:
Rutherford’s nuclear model:

1. According to Rutherford the entire positive charge and most of the mass of the atom is concentrated at nucleus. Electrons are revolving around the nucleus as planets revolve around the sun.
2. Rutherford experiments on α – particle suggested that size of atom is about 10^-15 to 10^-14 and size of nucleus is about 10^-10 m.

Drawbacks :

1. When electron is revolving round the nucleus it is in continuously accelerated state. As per classical mechanics, it must emit energy continuously. So electron must follow spiral path instead of circular path. Finally electron must fall on nucleus i.e., + ve charge and atom will destroy which is not happening.
2. This model is not able to explain why electron in an orbit is not radiating any energy.

Question 7.
Distinguish between excitation potential and ionization potential.
Answer:
Excitation potential :

1. Generally gases or vapours are excited at low pressure by passing high current through them. For this we have to apply a very high voltage.
2. When excited atoms or molecules will absorb certain amount of energy from applied potential and gives rise to series of spectral lines called emission spectrum.

Ionisation potential :
It is the amount of minimum energy required to release an electron from the outer most orbit of the nucleus.

From Bohr’s model, energy of the orbit is the ionisation energy of electron in that orbit.
Ex: Energy of 1st orbit in hydrogen is 13.6 eV.

Practically ionisation potential of hydrogen is 13.6 eV.

Difference:
→ Excitation potential will permit electrons to transit between various energy levels whereas ionisation potential will liberate an electron from the influence of nucleus of that atom.

Question 8.
Explain the different types of spectral series of hydrogen atom.
(OR)
Write the different types of Hydrogen Spectral series. The Lyman series of Hydrogen spectrum lies in the ultraviolet region. [TS May 16, Mar. 16, AP Mar. 19, 15, May 18, 16, June 15]
Answer:
Lyman series :
When electrons are jumping from higher energy levels to the first orbit then that series of spectral lines emitted are called “Lyman series”.
In Lyman series, \(\frac{1}{\lambda}\) = R(\(\frac{1}{l^2}-\frac{1}{n^2}\))
where n = 2, 3, …………… etc.

These spectral lines are in ultraviolet region.

Balmer series :
When electrons are jumping the from higher levels to 2nd orbit then that series of spectral lines are called “Balmer series”.

For Balmer series, \(\frac{1}{\lambda}\) = R(\(\frac{1}{2^2}-\frac{1}{n^2}\))
where n = 3, 4, ………… etc.

Spectral lines of Balmer series are in visible region.

Paschen series :
When electrons are jumping on to the 3rd orbit from higher energy levels then that series of spectral lines are called “Paschen series”.
For Paschen series \(\frac{1}{\lambda}\) = R(\(\frac{1}{3^2}-\frac{1}{n^2}\))
where n = 4, 5, …………..

These spectral lines are in near infrared region.

Brackett series :
When electrons are jumping on to the 4th orbit from higher levels then that series of spectral lines are called “Brackett series”.
For Brackett series, \(\frac{1}{\lambda}\) = R(\(\frac{1}{4^2}-\frac{1}{n^2}\))
where n = 5, 6, ………… etc.

Brackett series are in middle infrared region.

Pfund series :
When electrons are jumping on to the 5th orbit from higher energy levels then that series of spectral lines are called “Pfund series”.
For Pfund series \(\frac{1}{\lambda}\) = R(\(\frac{1}{5^2}-\frac{1}{n^2}\))
n = 6, 7, ……….. etc.

These spectral lines are in far infrared region.

Reason for Lyman series in ultraviolet region :
The wavelength of Lyman series is nearly 1200 A° and less. So Lyman series is ultraviolet region.

Question 9.
Write a short note on de Broglie’s explanation of Bohr’s second postulate of quantization. [TS Mar. ’17]
Answer:
De – Broglie’s explanation for Bohr 2nd postulate :
According to de Broglie, the electron in an orbit must be seen as a particle wave. Particle waves can lead to standing waves under resonance condition.

A standing wave can be formed in a wire when length of the wire is equal to wave-length λ or integral multiple of wavelength λ (i.e., nλ). In the same way for an electron moving in a circular orbit, if its circum-ference 2πr_n is equal to nλ then standing waves will be formed in that orbit.
∴ From de-Broglie’s explanation 2πr_n = nλ
But λ = h/p = \(\frac{h}{mv_n}\)
∴ de Broglie wavelength 2πr_n = nh / mv_n
mv_nr_n = \(\frac{nh}{2 \pi}\)
Where mv_nr_n is angular momentum of electron in nth orbit.
In this way de – Broglie hypothesis provided an explanation to Bohr’s 2nd postulate.

Long Answer Questions

Question 1.
Describe Geiger – Marsden Experiment on scattering of α – particles. How is the size of the nucleus estimated in this experiment?
Answer:
Geiger – Marsden experiment :
Description :
In this experiment α – particles from ²¹⁴₈₃Bi are passed through lead blocks containing coaxial holes to emit narrow beam of α – particles. This α – particle beam falls on gold foil and scatters. The scattered α – particles are detected. By rotating the microscope at different angles (θ) number of scattered particles are measured.

Explanation :
In this experiment a beam of α – particles of energy 5.5 MeV are made to fall on gold foil.

α – particle carries 2 units of positive charge and mass is equal to that of helium nucleus.

→ Assume that the gold foil is very thin and α – particle will suffer not more than one scattering during its passage through gold foil.

For gold Z = 79. Its mass is nearly 50 times more than that of α – particle. So during collision it remains almost stationary.

Magnitude of force between α – particle and gold nuclei (F) = \(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}^2}\), ‘r’ = the distance between them.

The magnitude and direction of force changes continuously as it approaches the nucleus.

Impact parameter :
It is the perpendicular distance of the initial velocity vector of α – particle from centre of nucleus.

In case of head – on collision impact parameter is minimum and α – particle rebounds back (θ = π). For a large impact parameter α – particle goes undeviated. The chance of head on collision is very small.

Estimation of size of nucleus :
In this experiment a graph is plotted between number of scattered particles (n) and angle of scattering ‘θ’. It suggested that size of nucleus is about 10^-15 to 10^-14 m. From kinetic theory size of atom is about 10^-10 m. So size of atom is 10,000 to 100,000 times more than nucleus.

Question 2.
Discuss Bohr’s theory of the spectrum of hydrogen atom. [AP Mar. ’16]
Answer:
From Bohr’s model of hydrogen atom

1. Electrons are revolving in certain per-mitted non-radiating orbits around nucleus.
2. For non-radiating orbits, orbital angular momentum L = mvr = nh / 2π.
3. Electrons may jump from one orbit to another orbit. While doing so they will emit or absorb the energy equal to diffe-rence of those orbital energies.
   E = hν = E_i – E_f

When electrons are revolving in the orbit centrifugal force and centripetal force on it are equal F_c = F_e.

For hydrogen atom n = 1
∴ r₁ = h²ε₀/π me² ……………. (5)
This is also called Bohr radius
a₀ = 5.29 × 10^-11 m
Energy of the orbit E = K + U = Kinetic energy + Potential Energy of electron.
∴ E_n = –\(\frac{e^2}{8\pi \epsilon_0 r}\)
By using value of r1 from eq. 4.

From 3rd postulate :
when electrons are jumping from higher to lower orbit.

Where R = Rydberg’s constant. (∵ Electron is initially at orbit n₂ and finally at n₁)

In this way Bohr atom model successfully explained energy of the orbits and origin of spectral lines.

Question 3.
State the basic postulates of Bohr’s theory of atomic spectra. [AP Mar. ’16]
Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom.
Answer:
Bohr’s postulates :
Bohr model of hydrogen atom consists of three main postulates.
i) Electrons in an atom could revolve in certain permitted stable orbits. Electrons revolving in these stable orbits do not emit or radiate any energy.

ii) The stable orbits are those whose orbital angular momentum is an integral multiple of h/2π.
i.e., L = nh / 2π where n = 1, 2, 3 etc. (an integer.)
These stable orbits are also called as non – radiating orbits.

iii) An electron may take a transition between non-radiating orbit.

when electron transition takes place a photon of energy equals to the energy difference between initial and final states will be radiated.
E = hν = E_i – E_f.

Bohr Model of Hydrogen atom :
When electrons are revolving in the orbit electrostatic force and centripetal force on it are equal F_c = F_e.

This is also called Bohr radius
a₀ = 5.29 × 10^-11 m
Energy of the orbit E = K + U = Kinetic energy + Potential Energy of electron.

Intext Question and Answer

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 × 10^-11m. What is the radius of the second orbit?
Answer:
Radius of first orbit r₁ = 5.3 × 10^-11m.
Radius of Bohr orbit r =
For 2nd orbit n = 2,
∴ r = n₂ a₀ =4 × 5.3 10^-11 = 2.12 × 10^-11 m.

Question 2.
Determine the radius of the first orbit of the hydrogen atom. What would be the velocity and frequency of the electron in the first orbit ? Given: h = 6.62 × 10^-34J s, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, k = 9 × 10⁹ m²C.
Answer:
Given h = 6.62 × 10^-34 J,
Mass m = 9.1 × 10^-31 kg
Charge e = 1.6 × 10^-19 C;

Question 3.
The total energy of an electron in the first excited state of the hydrogen atom is – 3.4 eV. What is the potential energy of the electron in this state?
Answer:
Total energy, T.E. = – 3.4 eV.
Potential energy U = 2 x T.E. = – 3.4 × 2
= – 6.8 eV in same orbit.

Question 4.
The total energy of an electron in the first excited state of the hydrogen atom is – 3.4 eV. What is the kinetic energy of the electron in this state?
Answer:
In 1st excited state
Total energy T.E. = – 3.4 eV.
Kinetic energy in 1st excited state = T.E. – U
Where U = 2 × T.E.
∴ K = U – 2U = – U = – (-3.4) = 3.4 eV.

Question 5.
Find the radius of the hydrogen atom in its ground state. Also calculate the velocity of the electron in n = 1 orbit. Given h = 6.63 × 10^-34 Js, m = 9.1 × 10^-31kg, e = 1.6 × 10^-19 C, k = 9 × 10⁹ N m²C^-2.
Answer:
Planck’s constant h = 6.63 × 10^-34 Js.;
Mass of electron m = 9.1 × 10^-31 Kg.
Charge on electron e = 1.6 × 10^-19 C;
K = 9 × 10⁹ m²C²
a) Radius of orbit, r \(\frac{n^2h^2\epsilon_0}{\pi me^2}\) where \(\frac{h^2\epsilon_0}{\pi me^2}\) = a₀ = 5.3 × 10^-11
For 1st orbit n = 1 ⇒ r = 1 × 5.3 × 10^-11
= 5.3 × 10^-11m

Question 6.
Prove that the ionization energy of hydrogen atom is 13.6 eV.
Answer:
From Bohr atom model energy radiated when electrons are jumping from one orbit
to another orbit is E = hυ = \(\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}\) = E₁ – E₂

By definition, ionization potential is the minimum amount of energy required to remove the electron from the orbit. Now for an electron of nearest orbit n₁ = 1 and n₂ = ∞ (because electron is free from influence of nucleus).

Question 7.
Calculate the Ionization energy fora lithium atom.
Answer:

For lithium Z = 3 (But outer most orbit n = 2 in lithium)
∴ lonization energy = Energy of the orbit = \(\frac{13.6\times9}{4}\) = 30.6 eV.

Question 8.
The wavelength of the first member of Lyman series is 1216 Å. Calculate the wavelength of second member of Balmer series.
Answer:
Wavelength of 1st member of Lyman series = 1216 A°.

2nd member of Balmer series ⇒ \(\frac{1}{\lambda_2}\) = R(\(\frac{1}{2^2}-\frac{1}{4^2}\))
∵ For 2nd member of Balmer series n = 4.

Question 9.
The wavelength of first member of Balmer series is 6563 A. Calculate the wavelength of second member of Lyman series.
Answer:
Wavelength of 1 st member of Balmer series = 6563.

Question 10.
The second member of Lyman series in hydrogen spectrum has wavelength 5400 Å. Find the wavelength of first member.
Answer:
Wavelength of 2nd member of Lyman series λ = 5400

Question 11.
Calculate the shortest wavelength of Balmer series. Or Calculate the wavelength of the Balmer series limit.
Given : R = 10970000m^-1.
Answer:
Given Rydberg constant R = 10970000 m^-1
= 1.097 × 10⁷ m^-1
For Balmer series \(\frac{1}{\lambda_2}\) = R(\(\frac{1}{2^2}-\frac{1}{n^2}\))
Short wavelength ⇒ n₂ = ∞

Question 12.
Using the Rydberg formula, calculate the wavelength of the first four spectral lines in the Balmer series of the hydrogen spectrum.
Answer:
For Balmer series \(\frac{1}{\lambda_2}\) = R(\(\frac{1}{2^2}-\frac{1}{n^2}\)) where
n = 2, 3, 4, 5 etc.
Rydberg constant R = 1.097 × 10⁷ m^-1


= 4102 A°

Telangana TSBIE TS Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter Textbook Questions and Answers.

TS Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Type Questions

Question 1.
What are “Cathode rays”? [TS Mar. 19; AP mar. 1 7, may 18, 14]
Answer:
Cathode rays consists of a steam of fast moving negatively charged particles.

Speed of cathode rays ranges about 0.1 to 0.2 times light velocity (3 × 10⁸m/s)

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan proved the validity of Einstein’s photo electric equation and his successful explanation of photo electric effect using light quanta. He experimentally found the value of Planck’s constant ‘h’ and work function on ‘Φ’ of photo electric surface.

Question 3.
What is “work function”? [AP Mar. 19, May 16; TS Mar. 18. 17, 15]
Answer:
Work function (Φ) :
The mininum energy required by an electron to escape from metal surface is called “work function”.

Work function depends on nature of metal.

Question 4.
What is “photoelectric effect”? [AP Mar. ’18, ’17, ’15, ’14, May ’14; TS May ’17, Mar. ’18. ’16]
Answer:
Photoelectric effect :
The process of liberating an electron from the metal surface due to light energy falling on it is called “photoelectric effect”.

Question 5.
Give examples of “photosensitive substances”. Why are they called so?
Answer:

1. Metals like zinc, cadmium and magnesium will respond to ultraviolet rays.
2. Alkalimetals such as sodium, potassium, caesium and rubidium will respond to visible light.

These substances are called “photo sensitive surfaces” because they will emit electrons when light falls on them.

Question 6.
An electron, an a particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength? [TS May, ’18, Mar. ’15]
Answer:
De-Broglie wavelength λ = \(\frac{h}{p}\); but KE = \(\frac{P^2}{2m}\)

For a particle m is more them given other particles.
So De-Broglie wavelength of a – particle is less.

Question 7.
What is Photo-electric effect? How did Einstein’s photo-electric equation explain the effects of intensity (of light) and potential on photo-electric current? [AP May 18, TS June 15]
Answer:
Photoelectric effect :
The process of emitting electron from the metal surface when light energy falling on it is called “photo-electrie effect”.

According to Einstein radiation consists of discrete units of energy called quanta of energy radiation.

Energy of quanta is called photon in light E = hυ

Maximum kinetic energy of photoelectron (K_max) is the difference of energy of incident radiation (hυ) and work function (Φ)
∴ K_max = hυ – Φ (when υ > υ₀)

Photoelectric equation can also written as

Effect of intensity :
As per Einstein’s photoelectric equation energy of photon E = hυ decides weather a photon will come out of metal surface or not. If frequency of incident light υ > υ₀ then electron will come out of that surface.

Number of electrons liberated depends on the number of photons striking the surface i.e., on intensity of light. So as per Einstein’s equation photocurrent liberated must be linearly proportional to intensity of light which is a practically proved fact.

Effect of voltage on photocurrent :
When positive potential on collector is gradually increased then photocurrent i.e., also gradually increased upto certain limit called saturation current all the photoelectrons liberated from. Photosurface reached the collector.

When υ >υ₀ photoelectron is released. The positive potential on collector will accelerate the electron. So it reaches the collector quickly. So photocurrent increases. But collector potential deals nothing with liberation of electron from photosurface.

In this way Einstein’s photoelectric equation explained the saturation of photocurrent with increasing collector +ve potential. Effect of frequency and stopping potential.

From Einstein’s photoelectric equation
K_max = hυ – Φ

i.e., kinetic energy of photoelectron is directly proportional to frequency of incident light.

Question 8.
Write down Einstein’s photoelectric equation. [AP Mar. 19, 15, May 17; TS May 18, 16]
Answer:
Maximum kinetic energy of photo electron Kmax is the difference of energy of incident radiation (hυ) and work function (Φ)
K_max= hυ – Φ (when υ > υ₀)
OR
K_max = eV₀ = hυ – Φ Or V₀ = \(\frac{h}{e}\)υ – \(\frac{\phi_{0}}{e}\)

Question 9.
Write down de Broglie’s relation and explain the terms there in. [AP & TS Mar. 18, 16; TS May 17]
Answer:
de -Broglie assumed that matter will also exhibit wave nature when it is in motion.
de – Broglie wavelength, λ = \(\frac{h}{p}=\frac{h}{mυ}\)
P = mo, momentum of the body, h = Planck’s constant, λ = wave length of moving particle.

Question 10.
State Heisenberg’s uncertainty principle. [TS Mar. 17; AP Mar. 14]
Answer:
Heisenberg’s uncertainty principle :
We cannot exactly find both momentum and position of an electron at the same time. This is called Heisenberg’s uncertainty principle.

Short Answer Questions

Question 1.
What is the effect of (i) intensity of light (ii) potential on photoelectric current? [TS Mar. ’19, June ’15]
Answer:
Effect of intensity :
The number of photo electrons liberated is directly proportional to intensity of incident radiation. So photo current increases linearly with increase of intensity of incident light.

Effect of potential :
When positive potential given to collector, photocurrent is gradually increased up to certain limit called saturation current. In this stage all the photo electrons liberated from photo surface reached the collector.

When negative potential on collector is gradually increased electrons are repelled by collector and photo current decreases.

At a particular negative voltage photo current is zero.

Stopping potential :
The minimum negative potential required by collector to stop photo current (or) becomes zero is called cut-off voltage (V₀) stopping potential.

Question 2.
How is the de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 volts? [AP Mar. ’15]
Answer:
de-Broglie wavelength λ = \(\frac{h}{\sqrt{2 \mathrm{~m}} \mathrm{eV}}\)
Potential difference V = 100V; h = 6.63 × 10^-34
Mass of electron m = 9.1 × 10^-31 kg;
charge of electron e = 1.6 × 10^-19c

Question 3.
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volt? [TS May. ’16]
Answer:
Applied potential V = 100 V.
de Broglie wavelength

∴ de Broglie wavelength 1 = 0.1227 nm.

Question 4.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of photo electric effect :
To study photo electric effect a photo sensitive surface and a metallic plate called collector are arranged on an evacuated glass tube. An arrangement is made to give required positive or negative potential to collector. By changing the filters placed in the path of incident light we will allow light rays of required frequency to fall on given photo surface.

Stopping potential :
The minimum negative potential required by collector to stop photo current or photo current to become zero is called “cut off voltage V₀“.

Effect of frequency on stopping potential are

1. Stopping potential varies linearly with frequency of incident light.
2. Every photo surface has a minimum cut off frequency υ₀ for which stopping potential V₀ = 0

Question 5.
Summarise the photon picture of electromagnetic radiation.
Answer:
Radiation consists of discrete units of energy called quanta.
Energy of quanta (E) = hυ = \(\frac{hc}{\lambda}\)

In case of light energy quanta is called photon.

Properties of photons:

1. Energy of photon E = ho Momentum ho P= \(\frac{hυ}{e}\)
2. In interaction of radiation with matter light quanta will behave like particles
3. Photons are electrically neutral. So they are not deflected by electric and magnetic fields.
4. In photons-particle collision total energy and total mometum of are conserved is in collision photon will totally loose its enery and momentum.

Question 6.
What is the deBroglie wavelength of a ball of mass 0.12Kg moving with a speed of 20ms^-1? What can we infer from this result? [AP June ’15]
Answer:
Mass of ball, m = 0.12kg ;
Speed of ball, v = 20 m/s
Plancks’ constant, h = 6.63 × 10^-34J
But de Broglie wave length λ = \(\frac{h}{p}=\frac{h}{mv}\)

∴ de Broglie wave length of given moving ball X = 2.762 × 10^-34m

Question 7.
The work function of cesium is 2.14 eV. Find the threshold frequency for cesium. (Take h = 6.6 × 10^-34Js) [IMP]
Answer:
Work function Φ₀ = 2.14 eV; h = 6.6 × 10^-34 JS

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation Explain the effect of intensity and potential on photoelectric current? How did this equation account for die effect of frequency of incident light on stopping potential?
Answer:
Einstein’s photo electric equation :
According to Einstein radiation consists of discrete units of energy called quanta of energy radiation.

Energy of quanta called photon in light E = hυ

Maximum kinetic energy of photo electron (K_max) is the difference of energy of incident radiation (hυ) and work function (Φ)
∴ K_max= hυ – Φ(when υ > υ₀)
Photo electric equation can be written as

Effect of intensity :
As per Einstein’s photo electric equation energy of photon decides weather a photon will come out (or) not from metal surface. If frequency of incident light υ > υ₀, then electron will come out from surface.

Number of electrons liberated depends on the number of photons striking the surface i.e., on intensity of light. So as per Einstein’s equation photo current liberated must be linearly proportional to intensity of light which is practically proved.

Effect of voltage on photo current :
When positive potential on collector is gradually increased then photo current i.e., also gradually increased up to certain limit called saturation current. In this stage all the photo electrons liberated from photo surface reached the collector.

When υ > υ₀ photo electron Is released. The positive potential on collector will accelerate the electron. So it reaches the collector quickly. So photo current increases. But collector potential deals nothing with liberation of electron from photo surface.

Effect of frequency and stopping potential :
⇒ Kinetic energy of photo electron is directly proportional to frequency of incident light.
From Einstein’s photo electric equation
∴ K_max= hυ – Φ

Stopping potential :
The minimum negative potential required by the collector to stop photo current is called stopping potential.

At this potential even the fastest electron (or) electron with maximum kinetic energy is prevented to reach the collector.

When frequency incident light increases then K_max of electron increases. Hence stopping potential V₀ will also increase.

∴ The graph between stopping potential V₀ and frequency o must be a straight line.

Then slope of the line is \(\frac{h}{e}\). This is experimentally proved by Millikan.

In this way Einstein’s photo electric equation successfully explained photo electric effect.

Question 2.
Describe the Davisson and Germer experiment. What did this experiment conclusively prove?
Answer:
Davisson and Germer experiment :
In this experiment, electrons are produced by heating a tungsten filament coated with barium oxide with the help of a low voltage battery.

These electrons are focussed on to a nickel target in the form of a sharp beam.

This electron beam is accelerated by the strong positive potential on nickel target.

After colliding the target electron beam will get scattered.

The scattered electron beam is collected by electron beam detector called collector.

In this experiment intensity of electron beam I for different angles of scattering is measured. A graph is plotted between angle of scattering 0 and intensity I with different target potentials of 44V to 68V.

Intensity is found to be maximum at a target potential of 54V with a scattering angle of 50°

This value coincides with the de Broglie wave length of electron.

Importance :
This experiment proved the existence of matter waves practically. It gave a strong support to de Broglie’s hypothesis of matter wave concept.

Intext Question and Amswers

Question 1.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Photoelectric cut-off voltage, V₀ = 1.5 V ;
maximum kinetic energy K_e = eV₀
Where, e = Charge on an electron = 1.6 × 10^-19 C
∴ k_e = 1.6 × 10^-19 × 15 = 2.4 × 10^-19J
∴ The maximum kinetic energy of the photoelectrons emitted K = 2.4 × 10^-19 J.

Question 2.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². How many photons (nearly) per square metre are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:
Energy flux of sunlight reaching the surface of earth, Φ = 1.388 × 10³ W/m²
Hence, power of sunlight per square metre, P = 1.388 × 10³W
Speed of light, c = 3 × 10⁸ m/s ; Planck’s constant, h = 6.626 × 10^-34 Js
Average wavelength of photons present in sunlight, λ = 550 nm = 550 × 10^-19 m
Number of photons per square metre incident on earth per second = n
∴ Power p = n E

∴ 3.84 × 10²¹ photons are incident per square metre / sec on earth.

Question 3.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10^-15 V s. Calculate the value of Planck’s constant.
Answer:
The slope of the cut-off voltage (V) – frequency (v) graph is ; \(\frac{V}{v}\) = 4.12 × 10^-15 Vs
But hv = eV
Where, e = Charge on an electron = 1.6 × 10^-19 C; h = Planck’s constant
∴ h = e×\(\frac{V}{v}\) = 1.6 × 10^-19 × 4.12 × 10^-15
= 6.592 × 10^-34Js
∴ Planck’s constant h = 6.592 × 10^-34 Js.

Question 4.
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Answer:
Power of the sodium lamp, P = 100 W ;
Wavelength of the emitted sodium light,
λ = 589 nm = 589 × 10^-9 m
Planck’s constant, h = 6.626 × 10^-34 Js ;
Speed of light, c = 3 × 10⁸ m/s
(a) Energy of photon E = \(\frac{hc}{\lambda}\)

(b) Number of photons delivered to the sphere = n
Power P = nE

Photons delivered per second = 2.96 × 10²⁰

Question 5.
The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:
Threshold frequency of the metal,
v₀ = 3.3 × 10¹⁴ Hz;
Charge on an electron, e = 1.6 × 10^-19 C ;
Frequency of light incident on the metal, v₀ = 8.2 × 10¹⁴ HZ;
Planck’s constant, h = 6.626 × 10^-34 Js
Cut-off voltage for the photoelectric emission from the metal = V₀; But eV₀ = h(υ – υ₀)

∴ The cut-off voltage for the photoelectric emission V₀ = 2.0292 V.

Question 6.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wave-length 330 nm?
Answer:
No.
Work function of the metal, Φ₀ = 4.2 eV;
Charge on an electron, e = 1.6 × 10^-19 C
Planck’s constant, h = 6.626 × 10^-34 Js ;
Wavelength of the incident radiation,
λ = 330 nm = 330 × 10^-9 m
Speed of light, c = 3 × 10⁸ m/s ;
The energy of the incident photon E = \(\frac{hc}{\lambda}\)

Question 7.
light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Frequency of photon, v = 488 nm
= 488 × 10^-9 m ;
Planck’s constant, h = 6.626 × 10^-34 Js
Maximum speed of the electrons, v = 6.0 × 10⁵ m/s ;
Mass of an electron, m = 9.1 × 10^-31 kg
Relation between v and K.E, \(\frac{1}{2}\) mv²
= h(ν – ν₀)⇒ ν₀ = ν –\(\frac{mv^2}{2h}\)

Threshold frequency ν₀ = 4.738 × 10⁵ Hz

Question 8.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Answer:
Wavelength of light produced by the argon laser, λ = 488 nm = 488 × 10^-9 m
Stopping potential, V₀ = 0.38 V ;
But leV = 1.6 × 10^-19 J
∴ V₀ = \(\frac{0.38}{1.6\times10^{-19}}\)eV
Planck’s constant, h = 6.6 × 10^-34 Js ;
Charge on an electron, e = 1.6 × 10^-19 C
Speed of light, c = 3 × 10 m/s
From Einstein’s photoelectric effect,

Question 9.
Calculate the
(a) Momentum/and
(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Potential difference, V = 56 V;
Planck’s constant, h = 6.6 × 10^-34 Js
Mass of an electron, m = 9.1 × 10^-31 kg ;
Charge on an electron, e = 1.6 × 10^-19 C

(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, for

The momentum of each accelerated electron
p = mv = 9.1 × 10^-31 × 4.44 × 10⁶
Momentum of each electron p
= 4.04 × 10^-24 kg m s^-1
(b)De Broglie wavelength of an electron,

Question 10.
What is the
(a) Momentum,
(b) Speed, and
(c) De Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer:
Kinetic energy of the electron, E_k = 120 eV;
Planck’s constant, h = 6.6 × 10^-34 Js
Mass of an electron, m = 9.1 × 10^-31 kg ;
Charge on an electron, e = 1.6 × 10^-19 C
(a) Kinetic energy of electron E_k = \(\frac{1}{2}\)mv²
Where, υ = Speed of the electron

Momentum of the electron, p = mv
∴ P = 9.1 × 10^-31 × 6.496 × 10⁶
= 5.91 × 10^-24 kg m s^-1.
(b) Speed of the electron, v = 6.496 × 10⁶ m/s (from eq 1)

(c) De Broglie wavelength of an electron having a momentum p, is given as:

∴ de Broglie wavelength of the electron is 0.112 nm.

Question 11.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10^-9 kg drifting with a speed of 2.2 m/s?
Answer:
(a) Mass of the bullet, m = 0.040 kg ;
Speed of the bullet, v = 1.0 km/s
= 1000 m/s
Planck’s constant, h = 6.6 × 10^-34Js
But De Broglie wavelength λ = \(\frac{h}{mv}\)

(b)Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength λ = \(\frac{h}{mv}\)

(c) Mass of the dust particle, m =1 × 10^-9 kg;
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength λ = \(\frac{h}{mv}\)

Question 12.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer:
Wavelength of an electron (λ^e) and a photon (λ^p), λ^e = λ^p = λ = 1nm = 1 × 10^-9 m
Planck’s constant, h = 6.63 × 10^-34 Js
(a) The momentum of an elementary partide is given by de Broglie relation:

(b) The energy of a photon is given by the relation:

(c) The kinetic energy (A) of an electron having momentum p,is given by the relation:
m = Mass of the electron = 9.1 × 10^-31 kg;
p = 6.63 × 10^-25 kg ms^-1