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
sin qR
qR
− cos qR
(1)
R = r o A
1/3
= 1.3 × (64)
1/3
= 5.2 fm
q = 2 p o sin(θ/2)
qR = 2cp o R sin(θ/2)/c = 2 × 300 ×
5.2 × sin 6
◦
197.3
= 1.653 radians
(2)
sin qR = 0.9966, cos qR = −0.0819
(3)
Inserting (2) and (3) in (1), we find F(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 p o sin(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 )r dr
= (4π/π
3/2 b
3 q)
∞
0
e
−r
2 /b
2 sin(qr )r dr
= (−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 e
r
2 /b
2 dr
∞
0 r 2 e −r 2 /b 2 dr
where we have put ρ(r ) = (1/π
3/2 b
3 )e
−r
2 /b
2 dr
With the change of variable r
2
/b
2
= x, we get
< r
2
>=
b
2
∞
0 x
3/2 e
−x dx
∞
0 x 1/2 e −x dx
=
Γ
5
2
b
2
Γ
3
2
=
3b
2
2