1000 Solved Problems in Modern Physics

6.3 Solutions 361

6.103 The angle between the twoγ-rays in the LS can be found fom the formula

m^2 c^4 =m 12 c^4 +m 22 c^4 +2(E 1 E 2 −c^2 p 1 p 2 cosφ)

Puttingm 1 =m 2 = 0 ,cp 1 =E 1 ,cp 2 =E 2

m^2 c^4 = 2 E 1 E 2 (1−cosφ)= 4 E 1 E 2 sin^2 (φ/2)

sin (φ/2)=mc^2 /2(E 1 E 2 )^1 /^2 (15)

6.104 For small angleφ,

φ=mc^2 /(E 1 E 2 )^1 /^2 (16)

SetE 2 =γmc^2 −E 1

φ=mc^2 /[E 1 (γmc^2 −E 1 )]^1 /^2 (17)

For minimum angle dφ/dE 1 =0. This givesE 1 =γmc^2 /2.

Using this value ofE 1 in (17), we obtain

φmin= 2 /γ (18)

Measurement ofφminaffords the determination ofEπviaγ.

φmin= 2 mc^2 /Eπ

6.105 E 1 +E 2 =E (energy conservation) (1)

p 1 +p 2 =p (momentum conservation) (2)

Taking the scalar product

(p 1 +p 2 ).(p 1 +p 2 )=p.p

or

p 12 +p 22 + 2 p 1 p 2 cosθ=p^2 (3)

Usingc=1, Eq. (3) becomes

E 12 +E 22 + 2 E 1 E 2 cosθ=E^2 −m^2 (4)

LetE 2 /E 1 =D, (5)

the disparity factor. Then (1) becomes

E 1 (D+1)=E (6)

Combining (4), (5) and (6)

2 DE^2 (1−cosθ)=m^2

Or sinθ/ 2 =[m/ 2 E][

√

D+ 1 /

√

D]

For smallθ,

θ=[m/E][

√

D+ 1

√

D]