6.3 Solutions
367
6.125 T thr = [(m k0 + m k + m Ω− )
2
− (m k− + m p )
2 ]/2m k−
= [(0.498 + 0.494 + 1.675)
2
− (0.494 + 0.938)
2 ]/2 × 0.938 = 2.7 GeV
Minimum momentum p K − = (T
2
+ 2T m)
1/2
= (2.7
2
+ 2 × 2.7 ×
0.494)
1/2
= 3.15 GeV/c
P thr = 3.15 GeV/c
E K = 2.7 + 0.494 = 3.194 GeV
γ K = E k /m k = 3.194/0.494 = 6.46
γ c = (γ + m 2 /m 1 )/[1 + 2γ m 2 /m 1 + (m 2 /m 1 )
2 ]
1/2
m 2 /m 1 = 938/494 = 1.9
γ c = (6.46 + 1.9)/(1 + 2 × 6.46 × 1.9 + 1.9
2 )
1/2
= 1.61
γ Ω = γ c = 1.61; β Ω = (γ
2
Ω − 1)
1/2
/γ Ω = 0.79
Proper time t 0 = d/v = d/βc
Observed time t = γ t 0 = γ d/βc = 1.61 × 0.03/0.79 × 3 × 10
8
=
2 × 10
−10 s
Probability that Ω
− will travel 3 cm before decay.
= exp(−t/τ )
= exp(−2 × 10
−10
/1.3 × 10
−10 )
= 0.21
6.126 T F (max) p = (9/32π
2 )
2/3 4π
2 (
2 c
2
/2m p c
2 r
2
0 )(Z /A)
2/3
R = r 0 A
1/3
r 0 = R/A
1/3
= 5.17/(63)
1/3
= 1.3 fm
T F (max) p = (9/32π
2 )
2/3
× 4π
2 (197 MeV − fm)
2 (29/63)
2/3
/2
× 938 × (1.3)
2
= 26.886 MeV
P F (max) = (T
2
+ 2T m)
1/2
= [(26.886)
2
+ 2 × 26.886 × 938]
1/2
= 226.2 MeV/c
E
2
− ( p 1 + p 2 )
2
= E
∗2 (maximum energy will be available when p 1 and
p 2 are antiparallel)
E
∗
= [(938 + 160 + 938 + 27)
2
− (570.4 − 226.2)
2 ]
1/2
= 2034 MeV
m d + m π = 938 + 939 − 2.2 + 139.5 = 2, 014 MeV
As the energy available in the CMS is in excess of the required energy, we
do expect the pions to be produced.