7.3 Solutions 411
7.3.5 Photoelectric Effect ................................
7.63 According to Einsteins’s equation, kinetic energy of the photoelectron
T=hν−hν 0 (1)
whereνis the frequency of the incident photon andν 0 is the threshold
frequency.
λ= 2 , 536 × 10 −^10 m= 253 .6nm
Corresponding energyE=hν= 1 , 241 / 253 .6eV= 4 .894 eV
λ 0 = 3 , 250 × 10 −^10 m=325 nm
E=hν 0 = 1 , 241 / 325 = 3. 818
T= 4. 894 − 3. 818 = 1 .076 eV
T=^1 / 2 mv^2 =mc^2 v^2 / 2 c^2 = 0. 511 × 106 ×v^2 / 2 c^2 = 1. 076
whencev= 2. 05 × 10 −^3 c= 6. 15 × 105 ms−^17.64 The momentump=300 Br MeV/c
= 300 × 3. 083 × 10 −^3
= 0 .925 MeV/c
E^2 =(T+m)^2 =p^2 +m^2
T=(p^2 +m^2 )^1 /^2 −m
Putp= 0 .925 andm= 0. 511
T= 0 .546 MeV
(a) The kinetic energy of the photoelectrons is 0.546 MeV
(b) The energy of the gamma ray photons is 0. 546 + 0. 116 = 0 .662 MeV
7.65 T=hν−hν 0
hν= 1 , 241 / 253. 7 = 4 .89 eV
hν 0 = 1 , 241 / 325 = 3 .81 eV
T=eV= 4. 89 − 3. 81 = 1. 08
The required potential is 1.08 V
7.66 Suppose the photoelectric effect does take place with a free electron due to the
absorption of a photon of energyT. The photoelectron must be ejected with
energy in the incident direction. Energy and momentum conservation give
T=hν (1)
P=hν/c (2)
Equation (1) can be written as the relativistic relation connecting momentum
and kinetic energy
T^2 =c^2 p^2 =T^2 + 2 Tmc^2 (3)
Using (1) and (2) in (3), we get
2 hν.mc^2 = 0
Neitherhnormc^2 is zero. We thus end up with an absurd situation. This
only means that both energy and momentum can not be conserved for photo-
electric effect with a free electron.