1000 Solved Problems in Modern Physics

366 6 Special Theory of Relativity

6.118 Tthreshold=[(mp+mp+mp)^2 −(mp+mp)^2 ]/ 2 mpby Eq. (6.53)

mp= 1 , 837 me= 1 , 837 × 0. 00 .00051 GeV= 0 .937 GeV

M= 273 × 0 .00051 GeV= 0 .137 GeV

Using the above values we findTthreshold= 0 .167 GeV

6.119 Tthreshold=(mk+mΛ)^2 −(mp+mp)^2 /^2 mp

mk= 0 .498 GeV,mΛ= 1 .115 GeV,mπ= 0 .140 GeV,mp= 0 .938 GeV

Using these values, we findTthreshold= 0 .767 GeV=767 MeV

Note that when pions are used as bombarding particles the threshold for

strange particle production is lowered then in N–N collisions. However, first

a beam of pions must be produced in N–N collisions.

6.120 Consider the reactionP+P→P+P+nπ

Tthreshold=[(mp+mp+nmπ)^2 −(mp+mp)^2 ]/ 2 mp

Simplifying we get the desired result

6.121 Tthreshold=[(mp+mπ 0 )^2 −(mp+0)^2 ]/ 2 mp
Usingmp=940 MeV andmπ 0 =135 MeV, we
findTthreshold=145 MeV
Note that the threshold energy for pion production in collision with gamma
rays is only half of that for N–N collisions. But the cross-section is down by
two orders of magnitude as the interaction is electromagnetic.

6.122 Tthreshold=[(mΞ−+mk+mk 0 )^2 −(mπ−+mp)^2 ]/ 2 mp

=[(1, 321 + 494 +498)^2 −(140+938)^2 ]/ 2 × 938

= 2 ,233 MeV

Note that forΞproduction, the threshold is much higher than that for

∑−

production as it has to be produced in association with two other strange

particles (see Chaps.9 and 10).

6.123 Using the invariance,E^2 −|

∑

p|^2 =E∗^2 −|

∑

p∗|^2

At threshold: (E+Mp)^2 −Eν^2 =(Mp+Mμ+Mw)^2 − 0

(5+ 0 .938)^2 − 52 =(0. 938 + 0. 106 +Mw)^2

Mw= 2 .16 GeV

Since the reaction does not proceed,Mw> 2 .16 GeV

6.124 Tthr=[(mp+mΛ+mk)^2 −(2mp)^2 ]/ 2 mp

=[(0. 938 + 1. 115 + 0 .494)^2 −(2× 0 .938)^2 ]/ 2 × 0. 938

= 1 .58 GeV