Assignment SolutionsDual Nature September 19 2013 CH 4 β DUAL NATURE OF RADIATION & MATTER SOLUTIONS No. Constants used: , h = 6.625 x 10-34 Js, e = 1.6 x 10-19C, c = 3 x 108 m/s Answers Two metals A, B have work functions 2 eV and 4 eV respectively. Which metal has a lower threshold wavelength for photo electric effect? 1 Given β A < β B πππ = β i.e. π² ππ = β ππͺ ππ = β B π²- constant Since β A < β B , ππ© < ππ¨ The de Broglie wave length associated with an electron accelerated through a potential difference V is Ξ». What will be its wavelength, when the acceleration potential is increased to 4V? 2 For an electron, ππ = ππ.ππ π½ For a potential 4V, ππβ² β Å i.e. π ππ½ π π½ ππ β ππβ² = Ξ»/2 ππ π A particle of a mass M at rest decays into 2 particles of masses m1 and m2 having nonzero velocities. What is the ratio of the de-Broglie wavelength of the two particles? 3 Using the conservation of momentum , π1+ π2= 0 1:1 i.e. π1 = π2 we know that , π = π π ; since π1 = π2 ; π1 = π 2 An electron, alpha particle and a proton have the same K.E. Which of these particles has the shortest de Broglie wavelength? Given : π¬π = π¬πΆ = π¬π ; also ππ < ππ < ππΆ Kinetic energy π¬ = ½ π π 2 ; π¬ = β΄ π= ππ π― π ππ = ππ ππ π π π¬ ------- (1) 4 de-Broglieβs relation gives , π = π π = Ξ± π πππ¬ -------(2) Since E is same for all the particles, (2) can be written as : π = π π Since mass of πΆ particle (ππΆ) is greatest, it has the shortest de-Broglie wavelength. 2 The threshold frequency of metal is ο΅0. When the light of frequency 2ο΅0 is incident on the metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5ο΅0, the maximum velocity of electrons emitted is v2. Find the ratio of v1 to v2. From Einsteinβs photo-electric equation, ππ = β o + ½ π π 2 For the two cases, 5 ππππ = πππ + ½ π π ππ πππ = ½ π π ππ -------------------(3) ππππ = ½ π π ππ ------------------(4) Similarly, (4) / (3), we get, 4= π ππ 1:2 β΄ ππ : ππ = 1 : 2 π ππ An alpha particle and a proton are accelerated thru the same potential difference. Calculate the ratio of linear momentum acquired by the two. We know that, πp < πΞ± ; but πΞ± = ππp -----(5) and πΞ±= ππp -----(6) When a charge q is accelerated through a potential V, its kinetic energy E is given by : 6 E = qV -------(7) β΄ From (1) the momentum of the particle is π = β΄ ππΆ = π ππΆ ππΆ π½ ππ = π π ππ½ π π:1 π ππ ππ π½ From (5) and (6), we get, ππΆ : ππ = π π : 1 When the light of wavelength 400nm is incident on the cathode of a photocell, the stopping potential recorded is 6 V. If the wave of incident light is increased to 600nm. Calculate the new stopping potential. From Einsteinβs photo-electric equation, ππ = β o + ππ½π --------------(8) β΄ ππͺ π = β o + ππ½π -------------(9) For the two cases, ππͺ π 7 = β o + ππ½π --------------(10) and ππͺ πβ² = β o + ππ½β²π -----------(11) (11) β (10), we get, ππ π π πβ² β π π = π½β²π β π½π Substituting the values of h, c, e, Ξ» = 400nm, Ξ»β = 600nm and π½π = 6V, π½β²π = 4.96 V 3 4.96 V 8 9 10 If the intensity of radiation incident on a photocell is doubled, what happens to the number of photoelectrons and energy of the photoelectrons ? As intensity of radiation is increased, the number of photoelectrons increases, but energy of the photons remains the same. Energy will increase only with increase in frequency of radiation, not with increase in intensity. If the frequency of radiation incident on a photocell is doubled, what happens to the number of photoelectrons and energy of the photoelectrons ? Same as above. Light of two different frequencies whose photons have energies 1eV and 2.5eV, respectively, successively illuminate a metal whose work function is 0.5 eV. Find the ratio of maximum speed of the emitted electrons. πππ = β o + ½ π π ππ 1eV = 0.5 ev + ½ π π ππ 0.5 eV = ½ π π ππ ------------------(1) 2.5eV = 0.5 ev + ½ π π ππ 2eV = ½ π π ππ --------------------(2) (1) / (2) we get, π π = π ππ π ππ Increases No Change No Change Increases 1:2 β΄ π π : π π = 1:2 A proton when accelerated through a potential difference of V volt has a de-Broglie wavelength Ξ». What should be the potential required to accelerate an Ξ± particle in order to have the same Ξ» ? Given λα = Ξ»p , The de-Broglie wavelength of a charge accelerated through a potential V is given by: π= 11 π π π ππ½ ππ = π ----(12) and π ππ ππ π½ ππΆ = π π ππΆ ππΆ π½β² ---(13) V/8 volt But πΞ± = ππp, πΞ±= ππp and λα = Ξ»p (given) β΄ equating (12) and (13), we get, π½ 12 π½β² = π When light of wavelength 0.6mm falls on a photocell photoelectrons are emitted for which the stopping potential is 0.5V. With light of wavelength 0.04 mm the stopping potential changes to 1.5V. Find the work-function of the metal in electron-volts. If 5% of energy supplied to a bulb is irradiated as visible light, how many quanta are emitted per second by a 100W bulb ? Assume wavelength of light is 5.6 x 10-5 cm. 13 100W = 100 J/s Only 5% is visible radiation. Ie 5 J/s πͺ Energy of one photon, π¬ = ππ or π¬ = π π ----------(14) Energy of βnβ photons = 5 J β΄π= π πͺ π π = 1.4 x 1019 s-1 4 1.4 x 1019 s-1 What is the ratio of the wavelength of a photon and that of an electron of the same energy ? πͺ 14 πͺ For a photon, π¬ = π π ππ ππ =π πππ¬ π¬ = π i.e. π π = π π¬ ---(15) For an electron, π π = π π = π πππ¬ ---(16) ππ π ππ π¬ π¬ The graph below shows the variation of the max. kinetic energy of photoelectrons with frequency of incident light. Find the threshold frequency and work function of the metal from the graph. E (eV) 8 0 10 20 Ξ½ (x 1014 Hz) 30 1015 Hz 4 eV -4 15 From Einsteinβs photoelectric equation, ππ = β o + ππ½ -----(17) β΄ ππ½ = ππ - β o ----(18) (equation for a straight line with slope = h) But πππ = β o i.e. when π = ππ , ππ½ = π. From the graph ππ = 1015 Hz From the graph and from (18) β π = πππ½ Eby P Kurien, [email protected] 5
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