4Orbits

IA Dynamics and Relativity

4.5 Rutherford scattering

Finally, we will consider the case where the force is repulsive instead of attractive.

An important example is the Rutherford gold foil experiment, where Rutherford

bombarded atoms with alpha particles, and the alpha particles are repelled by

the nucleus of the atom.

Under a repulsive force, the potential and force are given by

V (r) = +

mk

r

, F (r) = +

mk

r

2

.

For Coulomb repulsion of like charges,

k =

Qq

4πε

0

m

> 0.

The solution is now

u = −

k

h

2

+ A cos(θ − θ

0

).

wlog, we can take A ≥ 0, θ

0

= 0. Then

r =

`

e cos θ − 1

with

` =

h

2

k

, e =

Ah

2

k

.

We know that

r

and

`

are positive. So we must have

e ≥

1. Then

r → ∞

as

θ → ±α, where α = cos

−1

(1/e).

The orbit is a hyperbola, again given by

(x − ea)

2

a

2

−

y

2

b

2

= 1,

with

a

=

`

e

2

−1

and

b

=

`

√

e

2

−1

. However, this time, the trajectory is the other

branch of the hyperbola.

O

b

2α

β

It seems as if the particle is deflected by O.

We can characterize the path of the particle by the impact parameter

b

and

the incident speed

v

(i.e. the speed when far away from the origin). We know

that the angular momentum per unit mass is

h

=

bv

(velocity

×

perpendicular

distance to O).

How does the scattering angle

β

=

π −

2

α

depend on the impact parameter

b and the incident speed v?

Recall that the angle α is given by α = cos

−1

(1/e). So we obtain

1

e

= cos α = cos

π

2

−

β

2

= sin

β

2

,

So

b =

`

√

e

2

− 1

=

(bv)

2

k

tan

β

2

.

So

β = 2 tan

−1

k

bv

2

.

We see that if we have a small impact parameter, i.e.

b k/v

2

, then we can

have a scattering angle approaching π.