344
6 Special Theory of Relativity
6.49 T = (γ − 1)m 0 c
2
= [1/(1 − β
2 )
1/2
− 1]m 0 c
2
= [(1 − β
2 )
−1/2
− 1]m 0 c
2
= [1 +
1
2
β
2
+
3
8
β
4
+ . . . − 1]m 0 c
2
= [1 +
3
4
β
2
+ . . .](
1
2
m 0 β
2 c
2 )
where we have expanded the radical binomially
For v/c 1 or β 1, obviously T m 0 c
2
For small velocities T = m 0 β
2 c
2
/2 + small terms = m 0 v
2
/2
6.50 (a) The photon energy E γ = 1, 240/500 nm = 2.48 eV
= 2.48 × (1.6 × 10
−19 ) J
= 3.968 × 10
−19 J
Effective mass, m = E γ /c
2
= 3.968 × 10
−19
/(3 × 10
8 )
2
= 4.4 × 10
−36 kg
(b) E = 1,240/0.1 nm = 12,400 eV = 1.984 × 10
−15 J
m = 1.984 × 10
−15
/(3 × 10
8 )
2 kg = 2.2 × 10
−32 kg.
6.51 1amu = 1.66 × 10
−27 kg
=
(1.66 × 10
−27 kg)(c
2 )
c 2
= 1.66 × 10
−27
× 2.998 × 10
8 )
2 J/c
2
= 1.492 × 10
−10 J/c
2
= 1.492 × 10
−10 J/MeV.MeV/c
2
= 1.492 × 10
−10
/1.602 × 10
−13
MeV/c
2
= 931.3 MeV/c
2
6.52 Number of Uranium atoms in 1.0 g is
N = N 0 /A = 6.02 × 10
23
/235 = 2.56 × 10
21
Number in 5 kg = 2.56 × 10
21
× 5,000 = 1.28 × 10
25
In each fission ∼ 200 MeV energy is released.
Therefore, total energy released
= 1.28 × 10
25
× 200 = 2.56 × 10
27 MeV
= (2.56 × 10
27 MeV/J)(1.6 × 10
−13 J)
= 4 × 10
14 J
6.53 The analysis is similar to that for Compton scattering except for some approximations.
Energy conservation gives
E = E
+ T
(1)
From momentum triangle (Fig 6.6)
p
2
= p
2
+ p e
2
− 2 pp e cos θ
(2)
Using c
2 p
2
= E
2
− M
2 c
4
(3)
c
2 p e
2
= T
2
+ 2T mc
2
(4)
p = γ Mβc
(5)
γ = 1/(1 − β)
1/2
(6)
Combining (1) – (6) and simplifying and using the fact that mc
2
E. we
easily obtain the desired result.