7.3 Solutions
411
7.3.5 Photoelectric Effect
7.63 According to Einsteins’s equation, kinetic energy of the photoelectron
T = h ν − h ν 0
(1)
where ν is the frequency of the incident photon and ν 0 is the threshold
frequency.
λ = 2,536 × 10
−10 m = 253.6 nm
Corresponding energy E = h ν = 1,241/253.6 eV = 4.894 eV
λ 0 = 3,250 × 10
−10 m = 325 nm
E = h ν 0 = 1,241/325 = 3.818
T = 4.894 − 3.818 = 1.076 eV
T = 1 / 2 mv
2
= mc
2
v
2
/2c
2
= 0.511 × 10
6
× v
2
/2 c
2
= 1.076
whence v = 2.05 × 10
−3 c = 6.15 × 10
5 m s
−1
7.64 The momentum p = 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
Put p = 0.925 and m = 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 = h ν − h ν 0
h ν = 1,241/253.7 = 4.89 eV
h ν 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 energy T . The photoelectron must be ejected with
energy in the incident direction. Energy and momentum conservation give
T = h ν
(1)
P = h ν /c
(2)
Equation (1) can be written as the relativistic relation connecting momentum
and kinetic energy
T
2
= c
2 p
2
= T
2
+ 2T mc
2
(3)
Using (1) and (2) in (3), we get
2h ν . mc
2
= 0
Neither h nor mc
2 is zero. We thus end up with an absurd situation. This
only means that both energy and momentum can not be conserved for photoelectric effect with a free electron.