Assignment Solutions- Dual Nature. September 19

Transcription

1 Assignment Solutions- Dual Nature September 9 03

2 CH 4 DUAL NATURE OF RADIATION & MATTER SOLUTIONS No. Constants used:, = 6.65 x 0-34 Js, e =.6 x 0-9 C, c = 3 x 0 8 m/s Answers Two metals A, B ave work functions ev and 4 ev respectively. Wic metal as a lower tresold wavelengt for poto electric effect? Given A < B ν o = i.e. C = λ o K λ o = K- constant B Since A < B, λ B < λ A Te de Broglie wave lengt associated wit an electron accelerated troug a potential difference V is λ. Wat will be its wavelengt, wen te acceleration potential is increased to 4V? For an electron, λe =.7 V Å i.e. λe V λ/ For a potential 4V, λe 4V λe = λe 3 A particle of a mass M at rest decays into particles of masses m and m aving nonzero velocities. Wat is te ratio of te de-broglie wavelengt of te two particles? Using te conservation of momentum, p + p = 0 i.e. p = p we know tat, λ = ; since p = p p ; λ = λ An electron, alpa particle and a proton ave te same K.E. Wic of tese particles as te sortest de Broglie wavelengt? Given : Ee = Eα = Ep ; also me < mp < mα Kinetic energy E = ½ m v ; E = m v m = p m : p = m E () 4 de-broglie s relation gives, λ = p = m E () α Since E is same for all te particles, () can be written as : λ = k m Since mass of α particle (mα) is greatest, it as te sortest de-broglie wavelengt.

3 5 Te tresold frequency of metal is 0. Wen te ligt of frequency 0 is incident on te metal plate, te maximum velocity of electrons emitted is v. Wen te frequency of te incident radiation is increased to 5 0, te maximum velocity of electrons emitted is v. Find te ratio of v to v. From Einstein s poto-electric equation, ν = o + ½ m v For te two cases, ν o = ν o + ½ m v ν o = ½ m v (3) Similarly, 4ν o = ½ m v (4) (4) / (3), we get, 4 = v v v : v = : : An alpa particle and a proton are accelerated tru te same potential difference. Calculate te ratio of linear momentum acquired by te two. We know tat, m p < m α ; but m α = 4m p -----(5) and q α = q p -----(6) Wen a carge q is accelerated troug a potential V, its kinetic energy E is given by : 6 E = qv (7) From () te momentum of te particle is p = m qv : p α = m α q α V p p = m p q p V From (5) and (6), we get, p α : p p = : Wen te ligt of wavelengt 400nm is incident on te catode of a potocell, te stopping potential recorded is 6 V. If te wave of incident ligt is increased to 600nm. Calculate te new stopping potential. From Einstein s poto-electric equation, ν = o + ev o (8) C λ = o + ev o (9) 7 For te two cases, () (0), we get, c e C λ = o + ev o (0) and C λ = o + ev o () λ λ = V o V o 4.96 V Substituting te values of, c, e, λ = 400nm, λ = 600nm and V o = 6V, V o = 4.96 V 3

4 8 9 0 If te intensity of radiation incident on a potocell is doubled, wat appens to te number of potoelectrons and energy of te potoelectrons? As intensity of radiation is increased, te number of potoelectrons increases, but energy of te potons remains te same. Energy will increase only wit increase in frequency of radiation, not wit increase in intensity. If te frequency of radiation incident on a potocell is doubled, wat appens to te number of potoelectrons and energy of te potoelectrons? Same as above. Ligt of two different frequencies wose potons ave energies ev and.5ev, respectively, successively illuminate a metal wose work function is 0.5 ev. Find te ratio of maximum speed of te emitted electrons. ν o = o + ½ m v ev = 0.5 ev + ½ m v 0.5 ev = ½ m v ().5eV = 0.5 ev + ½ m v ev = ½ m v () () / () we get, 4 = v v v : v = : A proton wen accelerated troug a potential difference of V volt as a de-broglie wavelengt λ. Wat sould be te potential required to accelerate an α particle in order to ave te same λ? Given λ α = λ p, Te de-broglie wavelengt of a carge accelerated troug a potential V is given by: Increases No Cange No Cange Increases : λ = m qv λ p = m p q p V ----() and λ α = m α q α V ---(3) V/8 volt But m α = 4m p, q α = q p and λ α = λ p (given) equating () and (3), we get, 3 V = V 8 Wen ligt of wavelengt 0.6mm falls on a potocell potoelectrons are emitted for wic te stopping potential is 0.5V. Wit ligt of wavelengt 0.04 mm te stopping potential canges to.5v. Find te work-function of te metal in electron-volts. If 5% of energy supplied to a bulb is irradiated as visible ligt, ow many quanta are emitted per second by a 00W bulb? Assume wavelengt of ligt is 5.6 x 0-5 cm. 00W = 00 J/s Only 5% is visible radiation. Ie 5 J/s Energy of one poton, E = ν or E = C λ (4) Energy of n potons = 5 J n = 5 C λ =.4 x 0 9 s -.4 x 0 9 s - 4

5 4 Wat is te ratio of te wavelengt of a poton and tat of an electron of te same energy? For a poton, E = C i.e. λ λ p = C ---(5) For an electron, λ E e = = ---(6) p me λ p λ e = c me E = c m E c m E Te grap below sows te variation of te max. kinetic energy of potoelectrons wit frequency of incident ligt. Find te tresold frequency and work function of te metal from te grap. E (ev) ν (x 0 4 Hz) 0 5 Hz 4 ev From Einstein s potoelectric equation, ν = o + ev -----(7) ev = ν - o ----(8) (equation for a straigt line wit slope = ) But ν o = o i.e. wen ν = ν o, ev = 0. From te grap ν o = 0 5 Hz From te grap and from (8) o = 4eV Eby P Kurien, mail@selltesell.com 5