10.3 Solutions 559
(v)
Fig. 10.2(c)AEM
interaction
Fig. 10.2(d)Weak
interaction
10.2 (i) Weak interaction because neutrino is involved
(ii) Does not occur because lepton number is violated
(iii) Electromagnetic interaction as gamma ray is involved andΔS= 0
(iv) Allowed as a weak decay if proton is bound but forbidden when proton
is free because proton is lighter than the sum of masses of the product
particles.
(v) Does not occur as a strong or electromagnetic interaction because
ΔS = 0
(vi) Strong interaction becauseΔS=0 and other quantum numbers are
conserved.
(vii) Weak interaction, because a lepton- antilepton pair is involved.
10.3 IfP 0 is the momentum of Kaon,p 1 andp 2 the momenta of the pions, then
momentum conservation requires
p 0 =p 1 −p 2 = 2 p 2 −p 2 =p 2
Energy conservation requires
E 2 +E 1 =E 0
or
√
p^22 +m^2 π+√
p^21 +m^2 π=√
p 02 +m^2 K (1)
Butp 2 =p 0 andp 1 = 2 p 2 = 2 p 0 (2)
Using (2) in (1) and solving the resultant equationp 0 =mK
2(
m^2 K− 4 m^2 π
2 m^2 K+m^2 π