1000 Solved Problems in Modern Physics

508 9 Particle Physics – I

Q^2 = 2 EiEf(1−cosθ)

Thus, the approximation is valid when the particle mass is much smaller

than the energies.

9.3.4 Decay ...........................................

9.20 (a) If a particle A at rest decays into B and C then the total energy of B will be

EB=

(

mA^2 +mB^2 −mC^2

)

/ 2 mA

Inserting the valuesmA = 494 MeV, mB = 135 MeV andmC =

140 MeV, we find E(π^0 )= 245 .6MeV.

(b) Kinetic energy ofπ^0 isEπ 0 −mπ 0 = 245. 6 − 135 = 110 .6MeV

9.21 IfI 0 is the original intensity andIthe observed intensity, then
I=I 0 e−s/vτ=I 0 e−as
γ= 1 +T/m= 1 + 100 / 139 = 1. 719
β=(γ^2 −1)^1 /^2 /γ=(1. 7192 −1)^1 /^2 / 1. 719 = 0. 813
a= 1 /vτ→τ= 1 /va= 1 /βca= 1 /(0. 813 × 3 × 108 × 9. 1 × 10 −^2 )
τ= 4. 5 × 10 −^8 s
Proper life timeτ 0 =τ/γ= 4. 5 × 10 −^8 / 1. 719 = 2. 62 × 10 −^8 s

9.22mKc^2 =^966.^7 ×^0.^511 =494 MeV
mπc^2 = 273. 2 × 0. 511 = 139 .5MeV
Energy released in the decay
Q=mK− 3 mπ= 494. 0 − 3 × 139. 5 =75 MeV
Maximum kinetic energy of one pion will occur when the other two pions
go together in the opposite direction, that is energy is shared between oneπ
and 2π’s
Non-relativistically
Tπ(max)=Q× 2 mπ/(2mπ+mπ)= 75 × 2 / 3 =50 MeV
Relativistically
Tπ(max)=Eπ−mπ=

[(

mK^2 +mπ^2 −(2mπ)^2

)

/ 2 mK

]

−mπ

Inserting the values ofmKandmπ, we findTπ(max)= 48 .4 MeV. The dif-

ference in the two results is 3.3%.

9.23 In the decayA→B+C,
EB=

(

mA^2 +mB^2 −mC^2

)

/ 2 mA

HereB=ν,A=π,C=μ.Itfollowsthat

E′ν=

(

m^2 π−m^2 μ

)

/ 2 mπ(becausemν=0)

9.24 The maximum energy of neutron is obtained when the neutrino has zero
energy, neutron and muon being emitted in the opposite direction.
The energy released in the decay is
Q=mΣ−(mn+mμ+mν)= 1 , 189 −(939+ 106 +0)=144 MeV
Non-relativistically,
Tn=Qmμ/(mμ+mn)=^144 ×^106 /(106+939)=^14 .6MeV
Relativistically,