1000 Solved Problems in Modern Physics

510 9 Particle Physics – I

Eμ=Eμ∗γπ(1+βμ∗βπcosθ∗)

Eμ∗= 105. 7 + 4. 0 = 109 .7MeV;γμ∗= 1 .0378;βμ∗= 0. 2676

γπ= 5 / 0. 1395 = 35 .84;βπ≈ 1

Eμ(max)= 4 .98 GeV;Eμ(min)= 2 .88 GeV

9.29 The decay equation is

I=I 0 exp (−t/γτ)=I 0 exp(−d/βcγτ)(1)

For K+meson beam

βK= 1 /

[

1 +mK^2 /pK^2

] 1 / 2

SubstitutingmK= 0 .5GeV/c^2 andpK=1GeV/c

βK= 0 .894;γK= 2 .23;τK= 10 −^8 s;d=10 m

Using these values in Eq. (1)

I/I 0 =exp(− 1 .672)= 0. 188

which is tolerable.

ForΛ-hyperon beam

βΛ= 1 /

[

1 +mΛ^2 /pΛ^2

] 1 / 2

= 0. 7

γΛ= 1. 96 ,τΛ= 2. 5 × 10 −^10 s;d=10 m

Using the above values in (1) we find

I/I 0 =exp(−97)≈zero

which is not at all useful.

9.30d=vt (1)

t=γT 0 (2)

p=m 0 vγ (3)

Combining (1), (2) and (3)

d=pT 0 /m 0

9.31 Inverse Lorentz transformations gives

Eν∗=γEν(1−βcosθ)=Eν[γ−(γ^2 −1)^1 /^2 cosθ]

=Eν[γ−γ(1− 1 / 2 γ^2 )(1−θ^2 /2)]

Eν∗≈(Eν/ 2 γ)(1+γ^2 θ^2 )(1)

where we have neglected termsθ^2 / 2 γforγ>>1 and considered smallθ.

ButEν∗=

(

mπ^2 −mμ^2

)

/ 2 mπandEπ=γmπ (2)

Using (2) in (1) we get the desired result.

9.32 I/I 0 = 1 / 100 =exp(−d/βcγτ)
Putd=20 m, c= 3 × 108 m/s andτ = 0. 8 × 10 −^10 s and take logeon
both sides to solve forβγ. We findβγ=181. Therefore, the momentum of
hyperon,
cp=mβγ= 1. 19 × 181 =215 GeV
Thus, the minimum momentum required isP=215 GeV/c
The elastic scattering ofΣ−hyperons with protons can be recorded in the
hydrogen bubble chamber from the kinematical fits of events.