6.3 Solutions 343
6.43 (a)γ= 1 +T/m= 1 + 1 , 000 / 940 = 2. 064
β=(γ^2 −1)^1 /^2 /γ=[(2.064)^2 −1]^1 /^2 / 2. 064 = 0. 87
(b)γ= 1 + 1 , 000 / 0. 511 = 1 , 958
β=[1, 9582 −1]^1 /^2 / 1 , 958 = 0. 999973
6.44 (a) There are 6 protons and 6 neutrons in^12 C nucleus.
Δmc^2 =[(1. 007825 × 6 + 1. 008665 × 6 − 12 .000]× 931. 5
= 92 .16 MeV
(b)Δmc^2 =[3× 4. 002603 − 12 .000]× 931. 5
= 7 .27 MeV
(c)Δmc^2 =[2×(mn+mp)−mα]× 931. 5
=[2×(1. 008665 + 1 .007825)− 4 .002603]× 931. 5
= 28 .296 MeV
Total energy required= 3 × 28. 296 + 7. 27 = 92 .16 MeV
which is identical with that in (a)
6.45c^2 p^2 =T^2 + 2 Tmc^2
m=(c^2 p^2 −T^2 )/ 2 T=(368^2 − 2502 )/ 2 × 250 = 145 .85 MeV
Orm= 145. 85 / 0. 511 = 285. 4 me
The particle is identified as the pion whose actual mass is 273me
6.46 E 0 =m 0 c^2 =(9. 1 × 10 −^31 kg)(3× 108 ms−^1 )^2
= 8. 19 × 10 −^14 J
=(8. 19 × 10 −^14 J)(1 MeV/ 1. 6 × 10 −^13 J)
= 0 .51 MeV
6.47γ= 1 /(1−β^2 )^1 /^2 = 1 /(1− 0. 62 )^1 /^2 = 1. 25
Energy acquired by electron
T=(γ−1)mec^2 =(1. 25 −1)× 0. 51 = 0 .1275 MeV
1 eV energy is acquired when an electron (or any singly charged particle) is
accelerated from rest through a P.D of 1 V. Hence the required P.D is 0.1275
Mega volt or 127.5 kV.
6.48 (a)K(relativistic)=(γ−1)m^0 c
(^2) (1)
K(classical)=(1/2)m 0 v^2 =(1/2)m 0 c^2 β^2 =(1/2)m 0 c^2 (γ^2 −1)/γ^2 (2)
ΔK
K
=
K(relativistic)−K(Classical)
K(relativistic)=
γ− 1
2 γ(3)
where we have used (1) and (2)
PuttingΔK/K= 1 /100, we findγ= 50 /49. Using
β=(γ^2 −1)/γ (4)
we obtainβ= 0. 199
Orv= 0. 199 c
(b) PuttingΔK/K= 10 / 100 = 1 /10 in (3), we findγ= 5 /4, Using (4) we
obtainβ= 0. 6
Orv= 0. 6 c