414
7 Nuclear Physics – I
T 2 = hc/λ 2 − W
T 2 − T 1 = hc(1/λ 2 − 1/λ 1 )
(4.0 − 1.8) × 1.6 × 10
−19 J = hc
1
700×10 −10 −
1
800×10 −10
Solving for h, we get h = 6.57 × 10
−34 Js
The accepted value is h = 6.625 × 10
−34 J-s
7.73 hν 0 = W = 4.7 eV
λ 0 = 1,241/4.7 = 264 nm.
7.74 T A = 4.25 − W A
(1)
T B = 4.7 − W B
(2)
T A
T B
=
P
2
A
P
2
B
=
λ
2
B
λ
2
A
= 4
( 3 )
T B = T A − 1.5
( 4 )
Solving the above equations,
T A = 2 eV and T B = 0.5 eV
W A = 2.25 eV and W B = 4.2 eV
7.3.6 Pair Production
7.75 The minimum photon energy required for the e
− e
+ pair production is 2mc,
where m is the mass of electron. Therefore
h ν = 2mc
2
= 2 × 0.511 = 1.022 MeV
The corresponding wavelength
λ = (1, 241/1.022 × 10
6 eV) nm = 1.214 × 10
−12 m.
7.76 In the annihilation process energy released is equal to the sum of rest mass
energy of positron and electron which is 2mc
2 . Because of momentum and
energy conservation, the two photons must carry equal energy. Therefore each
gamma ray carries energy
E γ = mc
2
= 0.511 MeV
The wavelength of each photon
λ = 1, 241/(0.511 × 10
6 )nm = 2.428 × 10
−3 nm = 2.428 × 10
−12 m
7.77 Let us suppose that the e
+ e
− pair is produced by an isolated photon of energy
E = h ν and momentum h ν/c. Let the electron and positron be emitted with
momentum p − and p + , their total energy being E − and E + . Energy conservation gives
h ν 0 = E + + E −
(1)
The momentum conservation implies that the three momenta vectors, p 0 , p +
and p − must form the sides of a closed triangle, as in Fig. 7.17. Now in any
triangle, any side is equal or smaller than the sum of the other sides. Thus