{"id":36254,"date":"2022-11-28T13:05:27","date_gmt":"2022-11-28T07:35:27","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=36254"},"modified":"2022-11-28T13:05:27","modified_gmt":"2022-11-28T07:35:27","slug":"ts-inter-2nd-year-physics-study-material-chapter-12","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-12\/","title":{"rendered":"TS Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material<\/a> 12th Lesson Dual Nature of Radiation and Matter Textbook Questions and Answers.<\/p>\n

TS Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter<\/h2>\n

Very Short Answer Type Questions<\/span><\/p>\n

Question 1.
\nWhat are “Cathode rays”? [TS Mar. 19; AP mar. 1 7, may 18, 14]
\nAnswer:
\nCathode rays consists of a steam of fast moving negatively charged particles.<\/p>\n

Speed of cathode rays ranges about 0.1 to 0.2 times light velocity (3 \u00d7 108<\/sup>m\/s)<\/p>\n

Question 2.
\nWhat important fact did Millikan’s experiment establish?
\nAnswer:
\nMillikan 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 ‘\u03a6’ of photo electric surface.<\/p>\n

Question 3.
\nWhat is “work function”? [AP Mar. 19, May 16; TS Mar. 18. 17, 15]
\nAnswer:
\nWork function (\u03a6) :
\nThe mininum energy required by an electron to escape from metal surface is called “work function”.<\/p>\n

Work function depends on nature of metal.<\/p>\n

Question 4.
\nWhat is “photoelectric effect”? [AP Mar. ’18, ’17, ’15, ’14, May ’14; TS May ’17, Mar. ’18. ’16]
\nAnswer:
\nPhotoelectric effect :
\nThe process of liberating an electron from the metal surface due to light energy falling on it is called “photoelectric effect”.<\/p>\n

Question 5.
\nGive examples of “photosensitive substances”. Why are they called so?
\nAnswer:<\/p>\n

    \n
  1. Metals like zinc, cadmium and magnesium will respond to ultraviolet rays.<\/li>\n
  2. Alkalimetals such as sodium, potassium, caesium and rubidium will respond to visible light.<\/li>\n<\/ol>\n

    These substances are called “photo sensitive surfaces” because they will emit electrons when light falls on them.<\/p>\n

    \"TS<\/p>\n

    Question 6.
    \nAn 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]
    \nAnswer:
    \nDe-Broglie wavelength \u03bb = \\(\\frac{h}{p}\\); but KE = \\(\\frac{P^2}{2m}\\)
    \n\"TS<\/p>\n

    For a particle m is more them given other particles.
    \nSo De-Broglie wavelength of a – particle is less.<\/p>\n

    Question 7.
    \nWhat 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]
    \nAnswer:
    \nPhotoelectric effect :
    \nThe process of emitting electron from the metal surface when light energy falling on it is called “photo-electrie effect”.<\/p>\n

    According to Einstein radiation consists of discrete units of energy called quanta of energy radiation.<\/p>\n

    Energy of quanta is called photon in light E = h\u03c5<\/p>\n

    Maximum kinetic energy of photoelectron (Kmax<\/sub>) is the difference of energy of incident radiation (h\u03c5) and work function (\u03a6)
    \n\u2234 Kmax<\/sub> = h\u03c5 – \u03a6 (when \u03c5 > \u03c50<\/sub>)<\/p>\n

    Photoelectric equation can also written as
    \n\"TS<\/p>\n

    Effect of intensity :
    \nAs per Einstein’s photoelectric equation energy of photon E = h\u03c5 decides weather a photon will come out of metal surface or not. If frequency of incident light \u03c5 > \u03c50<\/sub> then electron will come out of that surface.<\/p>\n

    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.
    \n\"TS<\/p>\n

    Effect of voltage on photocurrent :
    \nWhen 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.<\/p>\n

    When \u03c5 >\u03c50<\/sub> 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.
    \n\"TS<\/p>\n

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

    From Einstein\u2019s photoelectric equation
    \nKmax<\/sub> = h\u03c5 – \u03a6<\/p>\n

    i.e., kinetic energy of photoelectron is directly proportional to frequency of incident light.<\/p>\n

    Question 8.
    \nWrite down Einstein’s photoelectric equation. [AP Mar. 19, 15, May 17; TS May 18, 16]
    \nAnswer:
    \nMaximum kinetic energy of photo electron Kmax is the difference of energy of incident radiation (h\u03c5) and work function (\u03a6)
    \nKmax<\/sub>= h\u03c5 – \u03a6 (when \u03c5 > \u03c50<\/sub>)
    \nOR
    \nKmax<\/sub> = eV0<\/sub> = h\u03c5 – \u03a6 Or V0<\/sub> = \\(\\frac{h}{e}\\)\u03c5 – \\(\\frac{\\phi_{0}}{e}\\)<\/p>\n

    Question 9.
    \nWrite down de Broglie’s relation and explain the terms there in. [AP & TS Mar. 18, 16; TS May 17]
    \nAnswer:
    \nde -Broglie assumed that matter will also exhibit wave nature when it is in motion.
    \nde – Broglie wavelength, \u03bb = \\(\\frac{h}{p}=\\frac{h}{m\u03c5}\\)
    \nP = mo, momentum of the body, h = Planck’s constant, \u03bb = wave length of moving particle.<\/p>\n

    \"TS<\/p>\n

    Question 10.
    \nState Heisenberg’s uncertainty principle. [TS Mar. 17; AP Mar. 14]
    \nAnswer:
    \nHeisenberg’s uncertainty principle :
    \nWe cannot exactly find both momentum and position of an electron at the same time. This is called Heisenberg’s uncertainty principle.<\/p>\n

    Short Answer Questions<\/span><\/p>\n

    Question 1.
    \nWhat is the effect of (i) intensity of light (ii) potential on photoelectric current? [TS Mar. ’19, June ’15]
    \nAnswer:
    \nEffect of intensity :
    \nThe 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.<\/p>\n

    Effect of potential :
    \nWhen 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.<\/p>\n

    When negative potential on collector is gradually increased electrons are repelled by collector and photo current decreases.<\/p>\n

    At a particular negative voltage photo current is zero.<\/p>\n

    Stopping potential :
    \nThe minimum negative potential required by collector to stop photo current (or) becomes zero is called cut-off voltage (V0<\/sub>) stopping potential.<\/p>\n

    Question 2.
    \nHow is the de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 volts? [AP Mar. ’15]
    \nAnswer:
    \nde-Broglie wavelength \u03bb = \\(\\frac{h}{\\sqrt{2 \\mathrm{~m}} \\mathrm{eV}}\\)
    \nPotential difference V = 100V; h = 6.63 \u00d7 10-34<\/sup>
    \nMass of electron m = 9.1 \u00d7 10-31<\/sup> kg;
    \ncharge of electron e = 1.6 \u00d7 10-19<\/sup>c
    \n\"TS<\/p>\n

    Question 3.
    \nWhat is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volt? [TS May. ’16]
    \nAnswer:
    \nApplied potential V = 100 V.
    \nde Broglie wavelength
    \n\"TS
    \n\u2234 de Broglie wavelength 1 = 0.1227 nm.<\/p>\n

    Question 4.
    \nDescribe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
    \nAnswer:
    \nExperimental study of photo electric effect :
    \nTo 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.<\/p>\n

    Stopping potential :
    \nThe minimum negative potential required by collector to stop photo current or photo current to become zero is called “cut off voltage V0<\/sub>“.<\/p>\n

    Effect of frequency on stopping potential are<\/p>\n

      \n
    1. Stopping potential varies linearly with frequency of incident light.<\/li>\n
    2. Every photo surface has a minimum cut off frequency \u03c50<\/sub> for which stopping potential V0<\/sub> = 0<\/li>\n<\/ol>\n

      \"TS<\/p>\n

      Question 5.
      \nSummarise the photon picture of electromagnetic radiation.
      \nAnswer:
      \nRadiation consists of discrete units of energy called quanta.
      \nEnergy of quanta (E) = h\u03c5 = \\(\\frac{hc}{\\lambda}\\)<\/p>\n

      In case of light energy quanta is called photon.<\/p>\n

      Properties of photons:<\/p>\n

        \n
      1. Energy of photon E = ho Momentum ho P= \\(\\frac{h\u03c5}{e}\\)<\/li>\n
      2. In interaction of radiation with matter light quanta will behave like particles<\/li>\n
      3. Photons are electrically neutral. So they are not deflected by electric and magnetic fields.<\/li>\n
      4. In photons-particle collision total energy and total mometum of are conserved is in collision photon will totally loose its enery and momentum.<\/li>\n<\/ol>\n

        Question 6.
        \nWhat is the deBroglie wavelength of a ball of mass 0.12Kg moving with a speed of 20ms-1<\/sup>? What can we infer from this result? [AP June ’15]
        \nAnswer:
        \nMass of ball, m = 0.12kg ;
        \nSpeed of ball, v = 20 m\/s
        \nPlancks’ constant, h = 6.63 \u00d7 10-34<\/sup>J
        \nBut de Broglie wave length \u03bb = \\(\\frac{h}{p}=\\frac{h}{mv}\\)
        \n\"TS
        \n\u2234 de Broglie wave length of given moving ball X = 2.762 \u00d7 10-34<\/sup>m<\/p>\n

        Question 7.
        \nThe work function of cesium is 2.14 eV. Find the threshold frequency for cesium. (Take h = 6.6 \u00d7 10-34<\/sup>Js) [IMP]
        \nAnswer:
        \nWork function \u03a60<\/sub> = 2.14 eV; h = 6.6 \u00d7 10-34<\/sup> JS
        \n\"TS<\/p>\n

        Long Answer Questions<\/span><\/p>\n

        Question 1.
        \nHow 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?
        \nAnswer:
        \nEinstein’s photo electric equation :
        \nAccording to Einstein radiation consists of discrete units of energy called quanta of energy radiation.<\/p>\n

        Energy of quanta called photon in light E = h\u03c5<\/p>\n

        Maximum kinetic energy of photo electron (Kmax<\/sub>) is the difference of energy of incident radiation (h\u03c5) and work function (\u03a6)
        \n\u2234 Kmax<\/sub>= h\u03c5 – \u03a6(when \u03c5 > \u03c50<\/sub>)
        \nPhoto electric equation can be written as
        \n\"TS<\/p>\n

        Effect of intensity :
        \nAs 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 \u03c5 > \u03c50<\/sub>, then electron will come out from surface.
        \n\"TS<\/p>\n

        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.<\/p>\n

        Effect of voltage on photo current :
        \nWhen 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.
        \n\"TS<\/p>\n

        When \u03c5 > \u03c50<\/sub> 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.<\/p>\n

        Effect of frequency and stopping potential :
        \n\u21d2 Kinetic energy of photo electron is directly proportional to frequency of incident light.
        \nFrom Einstein\u2019s photo electric equation
        \n\u2234 Kmax<\/sub>= h\u03c5 – \u03a6<\/p>\n

        Stopping potential :
        \nThe minimum negative potential required by the collector to stop photo current is called stopping potential.<\/p>\n

        At this potential even the fastest electron (or) electron with maximum kinetic energy is prevented to reach the collector.
        \n\"TS<\/p>\n

        When frequency incident light increases then Kmax<\/sub> of electron increases. Hence stopping potential V0<\/sub> will also increase.<\/p>\n

        \u2234 The graph between stopping potential V0<\/sub> and frequency o must be a straight line.<\/p>\n

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

        In this way Einstein’s photo electric equation successfully explained photo electric effect.<\/p>\n

        \"TS<\/p>\n

        Question 2.
        \nDescribe the Davisson and Germer experiment. What did this experiment conclusively prove?
        \nAnswer:
        \nDavisson and Germer experiment :
        \nIn this experiment, electrons are produced by heating a tungsten filament coated with barium oxide with the help of a low voltage battery.<\/p>\n

        These electrons are focussed on to a nickel target in the form of a sharp beam.<\/p>\n

        This electron beam is accelerated by the strong positive potential on nickel target.<\/p>\n

        After colliding the target electron beam will get scattered.<\/p>\n

        The scattered electron beam is collected by electron beam detector called collector.<\/p>\n

        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.<\/p>\n

        Intensity is found to be maximum at a target potential of 54V with a scattering angle of 50\u00b0
        \n\"TS<\/p>\n

        This value coincides with the de Broglie wave length of electron.<\/p>\n

        Importance :
        \nThis experiment proved the existence of matter waves practically. It gave a strong support to de Broglie’s hypothesis of matter wave concept.
        \n\"TS<\/p>\n

        Intext Question and Amswers<\/span><\/p>\n

        Question 1.
        \nThe photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
        \nAnswer:
        \nPhotoelectric cut-off voltage, V0<\/sub> = 1.5 V ;
        \nmaximum kinetic energy Ke<\/sub> = eV0<\/sub>
        \nWhere, e = Charge on an electron = 1.6 \u00d7 10-19<\/sup> C
        \n\u2234 ke<\/sub> = 1.6 \u00d7 10-19<\/sup> \u00d7 15 = 2.4 \u00d7 10-19<\/sup>J
        \n\u2234 The maximum kinetic energy of the photoelectrons emitted K = 2.4 \u00d7 10-19<\/sup> J.<\/p>\n

        Question 2.
        \nThe energy flux of sunlight reaching the surface of the earth is 1.388 \u00d7 10\u00b3 W\/m\u00b2. 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.
        \nAnswer:
        \nEnergy flux of sunlight reaching the surface of earth, \u03a6 = 1.388 \u00d7 10\u00b3 W\/m\u00b2
        \nHence, power of sunlight per square metre, P = 1.388 \u00d7 10\u00b3W
        \nSpeed of light, c = 3 \u00d7 108<\/sup> m\/s ; Planck\u2019s constant, h = 6.626 \u00d7 10-34<\/sup> Js
        \nAverage wavelength of photons present in sunlight, \u03bb = 550 nm = 550 \u00d7 10-19<\/sup> m
        \nNumber of photons per square metre incident on earth per second = n
        \n\u2234 Power p = n E
        \n\"TS
        \n\u2234 3.84 \u00d7 1021<\/sup> photons are incident per square metre \/ sec on earth.<\/p>\n

        Question 3.
        \nIn an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 \u00d7 10-15<\/sup> V s. Calculate the value of Planck\u2019s constant.
        \nAnswer:
        \nThe slope of the cut-off voltage (V) – frequency (v) graph is ; \\(\\frac{V}{v}\\) = 4.12 \u00d7 10-15<\/sup> Vs
        \nBut hv = eV
        \nWhere, e = Charge on an electron = 1.6 \u00d7 10-19<\/sup> C; h = Planck\u2019s constant
        \n\u2234 h = e\u00d7\\(\\frac{V}{v}\\) = 1.6 \u00d7 10-19<\/sup> \u00d7 4.12 \u00d7 10-15<\/sup>
        \n= 6.592 \u00d7 10-34<\/sup>Js
        \n\u2234 Planck\u2019s constant h = 6.592 \u00d7 10-34<\/sup> Js.<\/p>\n

        Question 4.
        \nA 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?
        \nAnswer:
        \nPower of the sodium lamp, P = 100 W ;
        \nWavelength of the emitted sodium light,
        \n\u03bb = 589 nm = 589 \u00d7 10-9<\/sup> m
        \nPlanck\u2019s constant, h = 6.626 \u00d7 10-34<\/sup> Js ;
        \nSpeed of light, c = 3 \u00d7 108<\/sup> m\/s
        \n(a) Energy of photon E = \\(\\frac{hc}{\\lambda}\\)
        \n\"TS
        \n(b) Number of photons delivered to the sphere = n
        \nPower P = nE
        \n\"TS
        \nPhotons delivered per second = 2.96 \u00d7 1020<\/sup><\/p>\n

        Question 5.
        \nThe threshold frequency for a certain metal is 3.3 \u00d7 1014<\/sup> Hz. If light of frequency 8.2 \u00d7 1014<\/sup> Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
        \nAnswer:
        \nThreshold frequency of the metal,
        \nv0<\/sub> = 3.3 \u00d7 1014<\/sup> Hz;
        \nCharge on an electron, e = 1.6 \u00d7 10-19<\/sup> C ;
        \nFrequency of light incident on the metal, v0<\/sub> = 8.2 \u00d7 1014<\/sup> HZ;
        \nPlanck\u2019s constant, h = 6.626 \u00d7 10-34<\/sup> Js
        \nCut-off voltage for the photoelectric emission from the metal = V0<\/sub>; But eV0<\/sub> = h(\u03c5 – \u03c50<\/sub>)
        \n\"TS
        \n\u2234 The cut-off voltage for the photoelectric emission V0<\/sub> = 2.0292 V.<\/p>\n

        \"TS<\/p>\n

        Question 6.
        \nThe work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wave-length 330 nm?
        \nAnswer:
        \nNo.
        \nWork function of the metal, \u03a60<\/sub> = 4.2 eV;
        \nCharge on an electron, e = 1.6 \u00d7 10-19<\/sup> C
        \nPlanck\u2019s constant, h = 6.626 \u00d7 10-34<\/sup> Js ;
        \nWavelength of the incident radiation,
        \n\u03bb = 330 nm = 330 \u00d7 10-9<\/sup> m
        \nSpeed of light, c = 3 \u00d7 108<\/sup> m\/s ;
        \nThe energy of the incident photon E = \\(\\frac{hc}{\\lambda}\\)
        \n\"TS<\/p>\n

        Question 7.
        \nlight of frequency 7.21 \u00d7 1014<\/sup> Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 \u00d7 105<\/sup> m\/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
        \nAnswer:
        \nFrequency of photon, v = 488 nm
        \n= 488 \u00d7 10-9<\/sup> m ;
        \nPlanck\u2019s constant, h = 6.626 \u00d7 10-34<\/sup> Js
        \nMaximum speed of the electrons, v = 6.0 \u00d7 105<\/sup> m\/s ;
        \nMass of an electron, m = 9.1 \u00d7 10-31<\/sup> kg
        \nRelation between v and K.E, \\(\\frac{1}{2}\\) mv\u00b2
        \n= h(\u03bd – \u03bd0<\/sub>)\u21d2 \u03bd0<\/sub> = \u03bd –\\(\\frac{mv^2}{2h}\\)
        \n\"TS
        \nThreshold frequency \u03bd0<\/sub> = 4.738 \u00d7 105<\/sup> Hz<\/p>\n

        Question 8.
        \nLight 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.
        \nAnswer:
        \nWavelength of light produced by the argon laser, \u03bb = 488 nm = 488 \u00d7 10-9<\/sup> m
        \nStopping potential, V0<\/sub> = 0.38 V ;
        \nBut leV = 1.6 \u00d7 10-19<\/sup> J
        \n\u2234 V0<\/sub> = \\(\\frac{0.38}{1.6\\times10^{-19}}\\)eV
        \nPlanck\u2019s constant, h = 6.6 \u00d7 10-34<\/sup> Js ;
        \nCharge on an electron, e = 1.6 \u00d7 10-19<\/sup> C
        \nSpeed of light, c = 3 \u00d7 10 m\/s
        \nFrom Einstein\u2019s photoelectric effect,
        \n\"TS<\/p>\n

        Question 9.
        \nCalculate the
        \n(a) Momentum\/and
        \n(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
        \nAnswer:
        \nPotential difference, V = 56 V;
        \nPlanck\u2019s constant, h = 6.6 \u00d7 10-34<\/sup> Js
        \nMass of an electron, m = 9.1 \u00d7 10-31<\/sup> kg ;
        \nCharge on an electron, e = 1.6 \u00d7 10-19<\/sup> C<\/p>\n

        (a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, for
        \n\"TS
        \nThe momentum of each accelerated electron
        \np = mv = 9.1 \u00d7 10-31<\/sup> \u00d7 4.44 \u00d7 106<\/sup>
        \nMomentum of each electron p
        \n= 4.04 \u00d7 10-24<\/sup> kg m s-1<\/sup>
        \n(b)De Broglie wavelength of an electron,
        \n\"TS<\/p>\n

        Question 10.
        \nWhat is the
        \n(a) Momentum,
        \n(b) Speed, and
        \n(c) De Broglie wavelength of an electron with kinetic energy of 120 eV.
        \nAnswer:
        \nKinetic energy of the electron, Ek<\/sub> = 120 eV;
        \nPlanck\u2019s constant, h = 6.6 \u00d7 10-34<\/sup> Js
        \nMass of an electron, m = 9.1 \u00d7 10-31<\/sup> kg ;
        \nCharge on an electron, e = 1.6 \u00d7 10-19<\/sup> C
        \n(a) Kinetic energy of electron Ek<\/sub> = \\(\\frac{1}{2}\\)mv\u00b2
        \nWhere, \u03c5 = Speed of the electron
        \n\"TS
        \nMomentum of the electron, p = mv
        \n\u2234 P = 9.1 \u00d7 10-31<\/sup> \u00d7 6.496 \u00d7 106<\/sup>
        \n= 5.91 \u00d7 10-24<\/sup> kg m s-1<\/sup>.
        \n(b) Speed of the electron, v = 6.496 \u00d7 106<\/sup> m\/s (from eq 1)<\/p>\n

        (c) De Broglie wavelength of an electron having a momentum p, is given as:
        \n\"TS
        \n\u2234 de Broglie wavelength of the electron is 0.112 nm.<\/p>\n

        Question 11.
        \nWhat is the de Broglie wavelength of
        \n(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km\/s,
        \n(b) a ball of mass 0.060 kg moving at a speed of 1.0 m\/s, and
        \n(c) a dust particle of mass 1.0 \u00d7 10-9<\/sup> kg drifting with a speed of 2.2 m\/s?
        \nAnswer:
        \n(a) Mass of the bullet, m = 0.040 kg ;
        \nSpeed of the bullet, v = 1.0 km\/s
        \n= 1000 m\/s
        \nPlanck\u2019s constant, h = 6.6 \u00d7 10-34<\/sup>Js
        \nBut De Broglie wavelength \u03bb = \\(\\frac{h}{mv}\\)
        \n\"TS<\/p>\n

        (b)Mass of the ball, m = 0.060 kg
        \nSpeed of the ball, v = 1.0 m\/s
        \nDe Broglie wavelength \u03bb = \\(\\frac{h}{mv}\\)
        \n\"TS<\/p>\n

        (c) Mass of the dust particle, m =1 \u00d7 10-9<\/sup> kg;
        \nSpeed of the dust particle, v = 2.2 m\/s
        \nDe Broglie wavelength \u03bb = \\(\\frac{h}{mv}\\)
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 12.
        \nAn electron and a photon each have a wavelength of 1.00 nm. Find
        \n(a) their momenta,
        \n(b) the energy of the photon, and
        \n(c) the kinetic energy of electron.
        \nAnswer:
        \nWavelength of an electron (\u03bbe<\/sup>) and a photon (\u03bbp<\/sup>), \u03bbe<\/sup> = \u03bbp<\/sup> = \u03bb = 1nm = 1 \u00d7 10-9<\/sup> m
        \nPlanck\u2019s constant, h = 6.63 \u00d7 10-34<\/sup> Js
        \n(a) The momentum of an elementary partide is given by de Broglie relation:
        \n\"TS<\/p>\n

        (b) The energy of a photon is given by the relation:
        \n\"TS<\/p>\n

        (c) The kinetic energy (A) of an electron having momentum p,is given by the relation:
        \nm = Mass of the electron = 9.1 \u00d7 10-31<\/sup> kg;
        \np = 6.63 \u00d7 10-25<\/sup> kg ms-1<\/sup>
        \n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

        Telangana TSBIE\u00a0TS 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, … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-physics-study-material-chapter-12\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Telangana TSBIE\u00a0TS Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter Textbook Questions and Answers. 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