1000 Solved Problems in Modern Physics

3.3 Solutions 245

σ(θ)=

1

4

(

z 1 z 2 e^2

μv^2

) 2

1

sin^4

(θ

2

) (Rutherford scattering formula)

3.119 By Problem 3.116

F(q^2 )=

3

q^2 R^2

(

sinqR

qR

−cosqR

)

(1)

R=roA^1 /^3 = 1. 3 ×(64)^1 /^3 = 5 .2fm

q= 2 posin(θ/2)

qR= 2 cpoRsin(θ/2)/c=^2 ×^300 ×

5. 2 ×sin 6◦

197. 3

= 1 .653 radians (2)

sinqR= 0. 9966 ,cosqR=− 0. 0819 (3)

Inserting (2) and (3) in (1), we findF(q)= 0. 75 ,F^2 ≈ 0 .57. Thus Mott’s

scattering is reduced by 57%.

3.120 F(q^2 )= 1 −q

2

6 ^2 <r

(^2) >+···
q= 2 posin(θ/2)= 2 × 200 ×(sin 7◦)MeV/c= 48 .75 MeV/c
<r^2 >=

6 ^2

q^2

[1−F(q^2 )]

= 6 ×

(197.3)^2

(48.75)^2

(1− 0 .6) fm^2

= 39. 3

∴Root mean square radius= 6 .27 fm

3.121 F(q)=(4π/q)

∫∞

0

ρ(r)sin(qr)rdr

=(4π/π^3 /^2 b^3 q)

∫∞

0

e−r

(^2) /b 2
sin(qr)rdr
=(− 4 /π^1 /^2 b^3 q)

∂

∂q

∫∞

0

e−r

(^2) /b 2
cos(qr)dr
=(− 4 /π^1 /^2 b^3 q)

∂

∂q

[

1

2

(πb^2 )^1 /^2 e−b

(^2) q (^2) / 4

]

F(q)=exp(−b^2 q^2 /4)

<r^2 >=

∫∞

0 r

(^2) ρ(r)4πr (^2) dr
∫
ρ(r)4πr^2 dr

=

∫∞

0 r

(^4) er^2 /b^2 dr
∫∞
0 r
(^2) e−r^2 /b^2 dr
where we have putρ(r)=(1/π^3 /^2 b^3 )e−r
(^2) /b 2
dr
With the change of variabler^2 /b^2 =x, we get
<r^2 >=
b^2

∫∞

0 x

3 / (^2) e−xdx
∫∞
0 x
1 / (^2) e−xdx =

Γ

( 5

2

)

b^2

Γ

( 3

2

) =

3 b^2

2