1000 Solved Problems in Modern Physics

6.3 Solutions 345

Fig. 6.6 (a)Scattering of a
proton with a stationary
electron(b)Momentum
triangle

6.54 The energy released in the decay of pion is
Q=mπc^2 −(mμc^2 +mνc^2 )
=(273− 207 −0)mec^2
= 66 × 0. 511 = 33 .73 MeV
Energy conservation gives
Tμ+Tν= 33. 73 (1)
In order to conserve momentum, muon neutrino must move in opposite direc-
tion
pν=pμ (2)

Multiplying (2) bycand squaring

c^2 p^2 ν=T^2 ν=Kμ+ 2 Kμmμc^2 (3)

Solving (1) and (3) and using

mμc^2 = 207 mec^2 = 207 × 0. 511 = 105 .77 MeV

Kμ= 4 .08 MeV,Kν= 29 .65 MeV

Observe that the lighter particle carries greater energy.

6.55 Let the mass of the final single body beMwhich moves with a velocityβc.
Momentum conservation gives
m× 0. 6 c/(1− 0. 62 )^1 /^2 =Mβc/(1−β^2 )^1 /^2

Or 3m/ 4 =Mβ/(1−β^2 )(1)

Since the total energy is conserved

mc^2 /(1− 0. 62 )^1 /^2 +mc^2 =Mc^2 /(1−β^2 )^1 /^2 (2)

(a) Using (2) in (1),β= 1 / 3

(b) Usingβ= 1 /3in(2),M= 2 .12 m