1000 Solved Problems in Modern Physics

6.3 Solutions 353

Butγ= 10 / 0. 14

Therefore,φmin= 0 .028 rad or 1. 6 ◦

6.83 Rest mass energy ofω^0 =Total available energy – (total kinetic energy+
mass energy ofπ+andπ−)

mωc^2 = 2. 29 −(1. 22 + 0. 14 + 0 .14)= 0 .79 GeV

6.84 Energy conservation gives

m 1 γ 1 +m 2 =Mγ (1)

Momentum conservation gives

m 1 γ 1 β 1 =Mγβ (2)

Squaring (1)

m 12 γ 12 +m 22 + 2 γ 1 m 1 m 2 =M^2 γ^2 (3)

Squaring (2)

m 12 γ 12 β 12 =M^2 γ^2 β^2 (4)

Usingβ 1 =(1− 1 /γ 12 )^1 /^2 andβ=(1− 1 /γ^2 )^1 /^2

(4) becomes

m 12 (γ 12 −1)=M^2 (γ^2 −1) (5)

Subtracting (5) from (3)

m 12 +m 22 + 2 m 1 m 2 /(1−v^2 /c^2 )^1 /^2 =M^2

6.85 E

0 =Ep+Eπ (energy conservation) (1)

Q=m 0 −(mp+mπ)(2)

P 02 =Pp^2 +Pπ^2 + 2 PpPπcosθ (3)

OrE 02 −m 02 =Ep^2 −mp^2 +Eπ^2 −mπ^2 + 2 PpPπcosθ (4)

Using (1) in (4) and simplifying

2 EpEπ− 2 PpPπcosθ+mp^2 +mπ^2 =m 02 =(Q+mp+mπ)^2 (5)

OrQ=(mp^2 +mπ^2 + 2 EpEπ− 2 PpPπcosθ)^1 /^2 −(mp+mπ)(6)