1000 Solved Problems in Modern Physics

244 3 Quantum Mechanics – II

∇^2 V=− 4 πZe^2 ρ

whereZeis the nuclear charge andρis the charge density.

∴ f(θ)=− 8 πμ

(

Ze^2

q^3 ^2

)∫

∞

0

ρ(r)sin(qr)rdr

=

(

2 μZe^2

q^2 ^2

)(

4 π

q

)∫∞

0

ρ(r)sin(qr)rdr

The quantity

(

4 π

q

)∫

∞

0 ρ(r)sin(qr)rdris known as the form factor.

3.118 f(θ)=

(

−

2 μ

q^2

)∫∞

0

V(r)sin(qr)rdr (1)

Substituting,

V(r)=

z 1 z 2 e^2

r

e−ar (2)

Wherea= 1 /ro, (1) becomes

f(θ)=−

(

2 μz 1 z 2 e^2

q^2

)∫∞

0

e−arsin(qr)dr

=

− 2 μz 1 z 2 e^2

q^2

q

q^2 +a^2

=

− 2 μz 1 z 2 e^2

^2

(

q^2 + 1 /r 02

) (3)

But the momentum transfer

q= 2 ksin

(

θ

2

)

(4)

The differential cross-section

σ(θ)=|f(θ)|^2 =

4 μ^2 z^21 z^22 e^4

^4

(

4 k^2 sin^2 (θ/2)+ 1 /r 02

) 2 (5)

The general angular distribution of scattered particles is reminiscent of

Rutherford scattering. However forθ<θ 0 , where

sin(θo/2)≈ 1 / 2 kro (6)

the curve does not rise indefinitely but tends to flatten out because when

qro 1, the angular dependence ofσ(θ) is damped out resulting in the

flattening of the curve. The angleθomay be considered as the limiting angle

below which the Rutherford scattering is inoperative because of the shielding

of the atomic nucleus by the electron cloud.

Rutherford scattering is derived from (5) by lettingro→∞, in which case

the scattering would occur from a bare nucleus. The screening potential (2)

now reduces to Coulomb potential. Furthermore, writingk=p=μv,(5)

becomes