1000 Solved Problems in Modern Physics

(Romina) #1

242 3 Quantum Mechanics – II


But the total cross-section is given by

σt=

4 π
k^2

(2l+1) sin^2 δl.

It follows thatIm f(0)=kσt/ 4 π. The last equation is known as the opti-
cal theorem.

3.116 V(r)=


(


Ze^2
2 R

)(

3 −

r^2
R^2

)

;0<r<R (1)

=−

Ze^2 e−ar
r

;R<r<∞ (2)

Inside the nucleus the electron sees the potential as given by (1) corre-
sponding to constant charge distribution, while outside it sees the shielded
potential given by (2). The scattering amplitude is given by

f(θ)=−(2μ/q^2 )

∫∞

0

V(r)rsin(qr)dr

= 2 μ

Ze^2
q^2

[(

1

2 R

)∫ R

0

(

3 −

r^2
R^2

)

rsin(qr)dr+

∫∞

R

sin(qr)e−ardr

]

(3)

The first integral is easily evaluated and the second integral can be written
as
∫∞

R

sin(qr)e−ardr=

∫∞

0

sin(qr)e−ardr−

∫ R

0

sin(qr)e−ardr (4)

=

q
q^2 +a^2


∫R

0

sin(qr)e−ardr (5)

(Lima→0)=

1

q


∫R

0

sin(qr)dr=

1

q

cos(qr)

We finally obtain

f(θ)=

(


2 μZe^2
q^2 ^2

)(

3

q^2 R^2

)(

sin(qR)
qR

−cosqR

)

σ(θ)finite size=σ(θ)point charge|F(q)|^2
where the form factor is identified as

F(q)=

(

3

q^2 R^2

)(

sin(qR)
qR

−cos(qR)

)

The angular distribution no longer decreases smoothly but exhibits sharp
maxima and minima reminiscent of optical diffraction pattern from objects
with sharp edges. The minima occur whenever the condition tanqR=qR,
is satisfied. This feature is in contrst with the angular distribution from a
smoothly varying charge distribution, such as Gaussian, Yakawa, Wood-
Saxon or exponential, wherein the charge varies smoothly and the maxima
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