1000 Solved Problems in Modern Physics

348 6 Special Theory of Relativity

6.64 (a)π+→μ++νμ
T=Q(Q+ 2 mν)/2(mμ+mν+Q)=(mπ−mμ)^2 / 2 mπ
(Q=mπ−mμ−mν=mπ−mμandmν=0)
mμ= 206. 9 me= 206. 9 × 0. 511 = 105 .7MeV
Therefore,
Tμ= 4. 12 =(mπ− 105 .7)^2 / 2 mπ
Solving formπ
mπ= 141 .39 MeV/c^2 = 141. 39 / 0. 511 me
= 276. 7 me
(b)μ+→e++νe+νμ
Tmaxfor electron is obtained whenνeandνμfly together in opposite direc-
tion. Thus, the three-body problem is reduced to a two-body one.
mμc^2 = 206. 9 × 0. 511 = 105 .72 MeV
Q= 105. 72 − 0. 51 = 105 .21 MeV
Te(max)=Q^2 /2(me+Q)=(105.72)^2 /2(0. 51 + 105 .72)= 52 .6MeV

6.65 Q= 938. 2 + 1 , 875. 5 −(939. 5 + 135 .0)− 2. 2 = 1 ,737 MeV
Tπ=Q(Q+ 2 mn)/2(Q+mπ+mn)
= 1 ,737(1, 737 + 2 × 939 .5)/2(1, 737 + 135 + 939 .5)= 1 ,117 MeV
Total energyEπ=^1 ,^117 +^135 =^1 ,252 MeV

6.66M^2 =m 12 +m 22 +2(E 1 E 2 −P 1 P 2 cosθ)(1)

m 1 = 966 me= 966 × 0 .511 MeV= 493 .6MeV (2)

m 2 =^273 me=^273 ×^0 .511 MeV=^139 .5MeV (3)

p 1 =394 MeV,p 2 =254 MeV (4)

E 1 =(p 12 +m 12 )^1 /^2 =[(394)^2 +(93.6)^2 ]^1 /^2 = 631 .6(5)

E 2 =(p 22 +m 22 )^1 /^2 =[(254)^2 +(139.5)^2 ]^1 /^2 = 289 .5(6)

cosθ=cos 154◦=− 0. 898 (7)

Using (2) to (7) in (1),M= 899 .4MeV

6.67 Using the results of Problem 6.60, the angle between the outgoing particles
after the collision is given by
cosθ=T/(T+ 4 M)
HereT=940 MeV=M
Therefore, cosθ= 0. 2 →θ= 78. 46 ◦

6.68 Using the invariance,E^2 −|

∑

p|^2 =E∗^2

E∗^2 =(E+Ef)^2 −(p^2 +p^2 f+ 2 ppfcosθ)

=(E+Ef)^2 −(E^2 −M^2 +E^2 f−M^2 + 2 ppfcosθ

= 2 M^2 +2(EEf−p.pf)

(a) For parallel momenta,θ= 0 ,p.pf=+ppf

(b) For anti-parallel momentaθ=π,p.pf=−ppf

(c) For orthogonal momentaθ=π/ 2 ,p.pf= 0

6.69 By problem 6.13 it is sufficient to show that tanθtanφ= 2 /(γ+1)
p=γβm=(γ^2 −1)^1 /^2 m