1000 Solved Problems in Modern Physics

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+^12 β^2 +^38 β^4 +...−1]m 0 c^2
=[1+^34 β^2 +...](^12 m 0 β^2 c^2 )
where we have expanded the radical binomially
Forv/c1orβ1, obviouslyTm 0 c^2
For small velocitiesT=m 0 β^2 c^2 / 2 +small terms=m 0 v^2 / 2

6.50 (a) The photon energyEγ= 1 , 240 /500 nm= 2 .48 eV
= 2. 48 ×(1. 6 × 10 −^19 )J
= 3. 968 × 10 −^19 J
Effective mass,m=Eγ/c^2 = 3. 968 × 10 −^19 /(3× 108 )^2 = 4. 4 × 10 −^36 kg
(b)E= 1 , 240 / 0 .1nm= 12 ,400 eV= 1. 984 × 10 −^15 J
m= 1. 984 × 10 −^15 /(3× 108 )^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 × 108 )^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 .3MeV/c^2

6.52 Number of Uranium atoms in 1.0 g is
N=N 0 /A= 6. 02 × 1023 / 235 = 2. 56 × 1021
Number in 5 kg= 2. 56 × 1021 × 5 , 000 = 1. 28 × 1025
In each fission∼200 MeV energy is released.
Therefore, total energy released
= 1. 28 × 1025 × 200 = 2. 56 × 1027 MeV
=(2. 56 × 1027 MeV/J)(1. 6 × 10 −^13 J)
= 4 × 1014 J

6.53 The analysis is similar to that for Compton scattering except for some approx-
imations.
Energy conservation gives
E=E′+T (1)
From momentum triangle (Fig 6.6)

p′^2 =p^2 +pe^2 − 2 ppecosθ (2)

Usingc^2 p′^2 =E′^2 −M^2 c^4 (3)

c^2 pe^2 =T^2 + 2 Tmc^2 (4)

p=γMβc (5)

γ= 1 /(1−β)^1 /^2 (6)

Combining (1) – (6) and simplifying and using the fact thatmc^2 E.we

easily obtain the desired result.