1000 Solved Problems in Modern Physics

3.3 Solutions 241

Putα= 1 /aandβ=q/

F(q)≈ 4 πA

∫∞

0

re−αr

sinβr

β

dr

I=−

∂

∂α

∫∞

0

e−αrsinβrdr

=

− 1

β

∂

∂α

β

α^2 +β^2

=

1

β

2 αβ

(α^2 +β^2 )^2

=

2 α

(α^2 +β^2 )^2

=

2

α^3

(

1 +

β^2

α^2

)− 2

= 2 a^3 /(1+q^2 a^2 /^2 )

F(q)= 8 πAa^3 /

(

1 +q^2 /qo^2

)

, whereqo=/a

thusF(q)≈ 1 /

(

1 +q

2

q 02

) 2

(b) The characteristic radius

a=



qo

=

c

qoc

=

197 .3MeV−fm

0. 71 × 1 ,000 MeV

= 0 .278 fm

3.114 f(θ)=− 2 πμ 2

∫

V(r)eiq.rd^3 r

=−

μ

2 π^2

∫∞

r= 0

∫π

θ= 0

∫ 2 π

φ= 0

V(r)eiqrcosθr^2 sinθdθdφdr

=−

μ

2 π^2

∫∞

0

V(r)r^2 dr

∫+ 1

− 1

eiqrcosθd(cosθ

∫ 2 π

0

dφ

=−

2 μ

^2

∫

V(r)r^2 dr

qr

[

eiqr−e−iqr

2 i

]

=−

2 μ

q^2

∫

rsin(qr)V(r)dr

3.115 From the partial wave analysis of scattering the scattering amplitude

f(θ)=

1

k

Σl(2l+1)(ηlexp(2iδl)−1)/^2 i)pl(cosθ).

For elastic scattering without absorptionηl=1, and

f(θ)=

1

k

Σl(2l+1)

[

exp(2iδl)−1)/ 2 i

]

pl(cosθ)

=

1

k

Σl(2l+1) exp(iδl)sinδlpl(cosθ).

Now forθ= 0 ,pl(cosθ)=pl(1)=1 for any value ofl, and exp(iδl)=

cosδl+isinδl. Therefore the imaginary part of the forward scattering

amplitude

Im f(0)=

1

k

∑

l

(2l+1) sin^2 δl.