7.3 Solutions
401
Fig. 7.14 Collision of a
heavy charged particle with
an electron
From the velocity triangle which is isosceles as in Fig. 7.14,
v 2 = 2v c cos ϕ = 2m 1 v cos ϕ/(m 1 + m 2 ) ≈ 2v cos ϕ
(Because m 2 = m << m 1 )
The energy acquired by the electron
W = (m/2) (2v cos ϕ)
2
= 2mv
2 cos
2
ϕ
(1)
If the recoil angle of the electron is ϕ
∗ in the CMS then ϕ = ϕ
∗
/2 and
ϕ
∗
= π –θ
∗ , where θ
∗ is the scattering angle of the incident particle in the
CMS. And so
cos
2
ϕ = sin
2 (θ
∗
/2)/2
( 2 )
W = 2mv
2 sin
2 (θ
∗
/2)
(3)
dW = mv
2 sin θ
∗ d θ
∗
(4)
But Rutherford’s formula for scattering in the CMS is
σ (θ
∗ ) = d σ/d Ω
∗
= z
2 e
4
/4μ
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
∗
= 2π sin θ
∗ d θ
∗ .
Formula (5) becomes
d σ = (2π sin θ
∗ dθ
∗ z
2 e
4 )/(4m
2
v
4 sin
4 (θ
∗
/2))
(6)
Using (3) and (4) in (6)
d σ/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 energy W per unit energy interval.
7.23 When the charged particle just grazes the nucleus
r min = R = 1 / 2 R 0 [1 + (1 + 4b
2
/R 0
2 )
1/2 ]
( 1 )
Solving for b, we obtain
b
2
= R
2 –R R 0
(2)
Denoting the cross-section by σ = πb
2 ,
σ/σ g = π b
2
/π R
2
= 1–R 0 /R
7.24 The closest distance of approach r is given by
r min = (R 0 /2)(1 + cosec (θ/2)) = (R 0 /2)(1 + 2) = 3R 0 /2