1000 Solved Problems in Modern Physics

6.3 Solutions 355

M^2 =2(EpEπ−−PpPπ−cosθ)+mπ+^2 +mπ−^2 (6)

Ep=(P+^2 +mp^2 )^1 /^2 (7)

mp= 0 .938 GeV/c^2 (8)

Using (3), (4), (5), (7) and (8) in (6) and solving forM, we findM =

1 .109 GeV/c^2 which is in good agreement with the massmΛ= 1 .115 GeV/c

Thus, the neutral particle isΛ.

6.88 Let the momenta of photons in the LS bep 1 adp 2 ad energiesE 1 andE 2 .The
invariant massWof the initial state is given by
W^2 =E^2 −p^2 =m^2
In the final state

E^2 −p^2 =(E 1 +E 2 )^2 −|(p 1 +p 2 )|^2

= 2 E 1 E 2 (1−cosφ)= 4 E 1 E 2 sin^2 (φ/2) (becauseE 1 =p 1 and

E 2 =p 2 andp 1 .p 2 =E 1 E 2 cosφ)

Invariance ofE^2 −p^2 gives

Sin (φ/2)=mc^2 /2(E 1 E 2 )^1 /^2

6.89n→p+e−+ν
The proton will carry maximum energy when the neutrino with negligible
mass is at rest.
(qn−qp)^2 =(En−Ep)−Pp^2
=(mn−Ep)^2 −(Ep^2 −mp^2 ); (because neutron is at rest)
=mn^2 +mp^2 − 2 mnEp
ButPp=Pe→Pp^2 =Pe^2
OrEp^2 −mp^2 =Ee^2 −me^2 =(mn−Ep)^2 −me^2
=mn^2 − 2 mnEp+Ep^2 −me^2
∴mn^2 +mp^2 − 2 mnEp=me^2
Thus (qn−qp)^2 =m^2 e
Orqn−qp=mec^2 = 0 .511 MeV/c

6.3.5 Transformation of Angles and Doppler Effect .......

6.90λ=λ

′√(1+β)/(1−β)

β=v/c= 3 × 106 / 3 × 108 = 0. 01

λ′= 6 , 563 A ̊

λ= 6 , 629 A ̊

Δλ=λ−λ′= 6 , 629 − 6 , 563 = 66 A ̊