1000 Solved Problems in Modern Physics

7.3 Solutions 401

Fig. 7.14Collision of a

heavy charged particle with

an electron

From the velocity triangle which is isosceles as in Fig. 7.14,

v 2 = 2 vccosφ= 2 m 1 vcosφ/(m 1 +m 2 )≈ 2 vcosφ

(Becausem 2 =m<<m 1 )

The energy acquired by the electron

W=(m/2) (2vcosφ)^2 = 2 mv^2 cos^2 φ (1)

If the recoil angle of the electron isφ∗in the CMS thenφ=φ∗/2 and

φ∗=π–θ∗, whereθ∗is the scattering angle of the incident particle in the

CMS. And so

cos^2 φ=sin^2 (θ∗/2)/2(2)

W= 2 mv^2 sin^2 (θ∗/2) (3)

dW=mv^2 sinθ∗dθ∗ (4)

But Rutherford’s formula for scattering in the CMS is

σ(θ∗)=dσ/dΩ∗=z^2 e^4 / 4 μ^2 v^2 sin^4 θ∗/2(5)

where we have put Z=−1 for electron. Since the electron mass is negligible

compared to that of the incident particle,μ≈m. Further, the element of solid

angle dΩ∗=^2 πsinθ∗dθ∗.

Formula (5) becomes

dσ=(2πsinθ∗dθ∗z^2 e^4 )/(4m^2 v^4 sin^4 (θ∗/2)) (6)

Using (3) and (4) in (6)

dσ/dW= 2 z^2 e^4 /mv^2 W^2 (differential energy spectrum)

This gives us the cross-section for finding the delta rays(emitted electrons)

of energyWper unit energy interval.

7.23 When the charged particle just grazes the nucleus

rmin=R=^1 / 2 R 0 [1+(1+ 4 b^2 /R 02 )^1 /^2 ](1)

Solving forb, we obtain

b^2 =R^2 – RR 0 (2)

Denoting the cross-section byσ=πb^2 ,

σ/σg=πb^2 /πR^2 =1–R 0 /R

7.24 The closest distance of approachris given by

rmin=(R 0 /2)(1+cosec (θ/2))=(R 0 /2)(1+2)= 3 R 0 / 2