1000 Solved Problems in Modern Physics

506 9 Particle Physics – I

9.12 (a) In elastic scattering, total kinetic energy is conserved. Life-time means

mean-life-time i.e., the time in which the population of unstable particles

is reduced by a factore. The important interactions that the muons undergo

are elastic scattering and the reaction,μ−+p→n+νμ.

Let the muons travel a distancedmetres. Then their intensity will be

reduced due to decay by a factor

I/I 0 =exp(−d/vγτ 0 )(1)

γ= 1 /(1−v^2 /c^2 )^1 /^2 =^1 /

[

1 −(10^6 / 3 × 108 )^2

] 1 / 2

= 1. 0033

I/I 0 =exp−(d/(1. 0033 × 106 × 2. 2 × 10 −^6 ))

=exp−(0. 453 d)(2)

The intensity reduction due to interaction will be

I/I 0 =exp(−dΣ)(3)

Σ=σn= 0. 1 × 10 −^28 × 2. 69 × 1025 = 2. 69 × 10 −^4

I/I 0 =exp(− 2. 69 × 10 −^4 )d (4)

Comparing (2) and (4), reduction due to decay will be by far greater than

by interaction for any value ofd.

(b) Repetition of calculation forT =4GeV,γ= 38. 7 ,β≈1, andd =

12 ,000 m, givesI/I 0 =e−^0.^47 = 0 .625, so that a large number of muons

survive at the ground level. The distance at which muons of energy 4 GeV

will be reduced byeis given byd=vγτ 0 = 3 × 108 × 38. 7 × 2. 2 × 10 −^6 =

2. 55 × 104 m= 25 .5km.

9.13 The geometrical cross-section is
σg=πR^2 =π(r 0 A^1 /^3 )^2 =πr 02 A^2 /^3
=π(1. 3 × 10 −^13 )^2 (208)^2 /^3 = 1. 86 × 10 −^24 cm^2
Number of lead atoms/cm^3 , n = N 0 ρ/A = 6 × 1023 × 11. 3 / 207 =
3. 275 × 1022
Macroscopic cross-section
Σ=nσ=(3. 275 × 1022 )(1. 86 × 10 −^24 )= 0 .06 cm−^1
The interaction length,λ= 1 /Σ= 1 / 0. 06 = 16 .7cm

9.14 Number of interactions per second in volumeV
I=ΣVI 0
=σN 0 ρVI 0 /A
=(40× 10 −^27 cm^2 )(6. 02 × 1023 × 0. 071 /2)× 125 × 2 × 103
= 213 .7 neutral pions/s are produced.
Each pion decays into two photons. Therefore number of photons produced
per second= 427

9.15 I/I 0 =exp(−Σx)
Σ=nσ=Nρσ/A