1000 Solved Problems in Modern Physics

6.3 Solutions 367

6.125 Tthr=[(mk 0 +mk+mΩ−)^2 −(mk−+mp)^2 ]/ 2 mk−

=[(0. 498 + 0. 494 + 1 .675)^2 −(0. 494 + 0 .938)^2 ]/ 2 × 0. 938 = 2 .7GeV

Minimum momentum pK−= (T^2 + 2 Tm)^1 /^2 = (2. 72 + 2 × 2. 7 ×

0 .494)^1 /^2 = 3 .15 GeV/c

Pthr= 3 .15 GeV/c

EK= 2. 7 + 0. 494 = 3 .194 GeV

γK=Ek/mk= 3. 194 / 0. 494 = 6. 46

γc=(γ+m 2 /m 1 )/[1+ 2 γm 2 /m 1 +(m 2 /m 1 )^2 ]^1 /^2

m 2 /m 1 = 938 / 494 = 1. 9

γc=(6. 46 + 1 .9)/(1+ 2 × 6. 46 × 1. 9 + 1. 92 )^1 /^2 = 1. 61

γΩ=γc= 1 .61;βΩ=(γΩ^2 −1)^1 /^2 /γΩ= 0. 79

Proper timet 0 =d/v=d/βc

Observed timet=γt 0 =γd/βc = 1. 61 × 0. 03 / 0. 79 × 3 × 108 =

2 × 10 −^10 s

Probability thatΩ−will travel 3 cm before decay.

=exp(−t/τ)

=exp(− 2 × 10 −^10 / 1. 3 × 10 −^10 )

= 0. 21

6.126 TF(max)p=(9/^32 π

(^2) ) 2 / (^34) π (^2) ( (^2) c (^2) / 2 mpc (^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 .3fm
TF(max)p=(9/ 32 π^2 )^2 /^3 × 4 π^2 (197 MeV−fm)^2 (29/63)^2 /^3 / 2
× 938 ×(1.3)^2 = 26 .886 MeV
PF(max)=(T^2 + 2 Tm)^1 /^2 =[(26.886)^2 + 2 × 26. 886 ×938]^1 /^2
= 226 .2MeV/c
E^2 −(p 1 +p 2 )^2 =E∗^2 (maximum energy will be available whenp 1 and
p 2 are antiparallel)
E∗=[(938+^160 +^938 +27)^2 −(570.^4 −^226 .2)^2 ]^1 /^2
=2034 MeV
md+mπ= 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.