346
6 Special Theory of Relativity
6.56 T K = [(M D − M K )
2
− M
2 π ]/2M D = [(1865 − 494)
2
− 140
2 ]/2 × 1865
= 498.67 MeV
P K = (T K
2
+ 2M K T K )
1/2
= 861 MeV/c
E K
2
= p K
2
+ m K
2
= (0.861)
2
+ (0.494)
2
= 0.9853 GeV
2
E K = 0.9926 GeV/c
E π
2
= p π
2
+ m π
2
= (0.861)
2
+ (0.140)
2
= 0.7609 GeV
2
E π = 0.8736 GeV
M x = E K + E π = 1.8662 GeV/c
2
It is a D
0 meson.
6.57 The mass of neutrino is zero. Applying conservation laws of energy and
momentum
E μ + E ν = m π c
2
(1)
p μ = p ν
(2)
Multiplying (2) by c and squaring
c
2 p μ
2
= c
2 p ν
2
Or E μ
2
− m μ
2 c
4
= E ν
2
Or E μ
2
− E ν
2
= m μ
2 c
4
(3)
Solve (1) and (3)
γ μ = (m
2 π + m
2 μ )/2m π m μ
β μ = (1 − 1/γ
2
μ )
1/2
= (m π
2
− m μ
2 )/(m μ
2
+ m μ
2 )
6.58 E B + E C = m A c
2 (energy conservation)
(1)
P B = P C (momentum conservation)
(2)
Or c
2 P B
2
= c
2 P C
2
(3)
Using the relativistic equations E
2
= c
2 p
2
+ m
2 c
4 , (3) becomes
E B
2
− m B
2 c
4
= E C
2
− m C
2 c
4
(4)
Eliminating E C between (1) and (4), and simplifying
E B = (m A
2
+ m B
2
− m C
2 )c
2
/2m A
(5)
6.59 K
+
→ e
+
+ π
◦
+ ν e
The maximum energy of positron will correspond to a situation in which the
neutrino is at rest. In that case the total energy carried by electron will be
E e (max) = (m
2
K +m
2
e −m
2
π 0 )/2m K = (494
2
+0.5
2
+135
2 )/2×494 = 228.5MeV
∴ T e (max) = 228 MeV
6.60 Let the incident particle carry momentum p 0 . As the scattering is symmetrical,
each particle carries kinetic energy T /2 and momentum P after scattering, and
makes an angle θ/2 with the incident direction.
Momentum conservation along the incident direction gives
p 0 = p cos θ/2 + p cos θ/2 = 2 p cos θ/2
( 1 )
Or (T
2
+ 2T mc
2 )
1/2
= 2(T
2
/4 + 2T /2mc
2 )
1/2 cos θ/2
( 2 )
Squaring (2), and using the identity, cos
2
θ/2 = (1 + cos θ )/2
We get the result cos θ = T /(T + 4mc
2 )
6 Special Theory of Relativity
6.56 T K = [(M D − M K )
2
− M
2 π ]/2M D = [(1865 − 494)
2
− 140
2 ]/2 × 1865
= 498.67 MeV
P K = (T K
2
+ 2M K T K )
1/2
= 861 MeV/c
E K
2
= p K
2
+ m K
2
= (0.861)
2
+ (0.494)
2
= 0.9853 GeV
2
E K = 0.9926 GeV/c
E π
2
= p π
2
+ m π
2
= (0.861)
2
+ (0.140)
2
= 0.7609 GeV
2
E π = 0.8736 GeV
M x = E K + E π = 1.8662 GeV/c
2
It is a D
0 meson.
6.57 The mass of neutrino is zero. Applying conservation laws of energy and
momentum
E μ + E ν = m π c
2
(1)
p μ = p ν
(2)
Multiplying (2) by c and squaring
c
2 p μ
2
= c
2 p ν
2
Or E μ
2
− m μ
2 c
4
= E ν
2
Or E μ
2
− E ν
2
= m μ
2 c
4
(3)
Solve (1) and (3)
γ μ = (m
2 π + m
2 μ )/2m π m μ
β μ = (1 − 1/γ
2
μ )
1/2
= (m π
2
− m μ
2 )/(m μ
2
+ m μ
2 )
6.58 E B + E C = m A c
2 (energy conservation)
(1)
P B = P C (momentum conservation)
(2)
Or c
2 P B
2
= c
2 P C
2
(3)
Using the relativistic equations E
2
= c
2 p
2
+ m
2 c
4 , (3) becomes
E B
2
− m B
2 c
4
= E C
2
− m C
2 c
4
(4)
Eliminating E C between (1) and (4), and simplifying
E B = (m A
2
+ m B
2
− m C
2 )c
2
/2m A
(5)
6.59 K
+
→ e
+
+ π
◦
+ ν e
The maximum energy of positron will correspond to a situation in which the
neutrino is at rest. In that case the total energy carried by electron will be
E e (max) = (m
2
K +m
2
e −m
2
π 0 )/2m K = (494
2
+0.5
2
+135
2 )/2×494 = 228.5MeV
∴ T e (max) = 228 MeV
6.60 Let the incident particle carry momentum p 0 . As the scattering is symmetrical,
each particle carries kinetic energy T /2 and momentum P after scattering, and
makes an angle θ/2 with the incident direction.
Momentum conservation along the incident direction gives
p 0 = p cos θ/2 + p cos θ/2 = 2 p cos θ/2
( 1 )
Or (T
2
+ 2T mc
2 )
1/2
= 2(T
2
/4 + 2T /2mc
2 )
1/2 cos θ/2
( 2 )
Squaring (2), and using the identity, cos
2
θ/2 = (1 + cos θ )/2
We get the result cos θ = T /(T + 4mc
2 )