1000 Solved Problems in Modern Physics

3.3 Solutions 237

3.109 By Problem 3.104

σ(θ)=

1

k^2

[

sin^2 δ 0 +6sinδ 0 sinδ 1 cos(δ 1 −δ 0 ) cosθ+9sin^2 δ 1 cos^2 θ

]

(1)

We assume that at low energiesδ 1 δ 0. Now in the scattering with a hard

sphere

tanδl=−

(ka)^2 l+^1

(2l+1)(1. 1. 3. 5 ... 2 l−1)^2

δ 0 (H.sphere)=−ka, for allka

Andδ 1 (H.sphere)=−(ka)

3

3 ,forka^1

Neglecting higher powers ofδ′s, we can write (1)

σ(θ)=

1

k^2

[(

δ 0 −

δ 03

3!

) 2

+ 6 δ 0 δ 1 cosδ

]

=

1

k^2

[

δ 02 −

δ^40

3

+ 6 δ 0 δ 1 cosδ

]

=

1

k^2

[

k^2 a^2 −

k^4 a^4

3

+6(−ka)

(

−

k^3 a^3

3

)

cosθ

]

=a^2

[

1 −

k^2 a^2

3

+ 2 k^2 a^2 cosθ

]

σ=

∫(

dσ

dΩ

)

. 2 πsinθdθ= 2 π

∫+ 1

− 1

a^2

(

1 −

k^2 a^2

3

+ 2 k^2 a^2 cosθ

)

d cosθ

= 4 πa^2

[

1 −

(ka)^2

3

]

3.110 A spherical nucleus of radiusRwill be totally absorbing, or appear “black”
when the angular momentuml<R/λ. In that caseηl=0 in the reaction
and scattering formulae.

σr=πλ-^2

∑

l

(2l+1)(1−|ηl|^2 )

σs=πλ-^2

∑

l(2l+1)|^1 −ηl|

2

(|>ηl>0)

Puttingηl= 0

σr=σs=πλ-^2

∑R/λ

l= 0

(2l+1)

The summation can be carried out by using the formula for arithmetic

progression

S=na+

n(n−1)d

2

Herea= 1 ,d= 2 ,n=R/λ-