

When a beam of
a particles showers a thin foil of material, the
a particles undergo what is called the Rutherford
Scattering. This scattering corresponds to the Coulomb's repulsion between the
positively charge
a particle (nucleus of helium which is made out of
two protons and two neutrons) and the nucleus of the atoms that constitute the
thin foil. In the original experiment, the foil was made out of gold.
Assumptions
 Alpha particles and Nucleus can be
considered as point masses and charges.
 Electric repulsion between the charges is
the only force present in the scattering.
 The nucleus is so massive that it does not
move during the scattering. Consequently, the energy of the
a particle does not change during the scattering.
 The foil containing the targets is so thin
that only a single scattering occurs.
 Cross section of adjacent nuclei do not
overlap

r is the
instantaneous distance between the nucleus and the
a
particle.
b is the
impact parameter.
q is the
scattering angle.


The total change in the linear momentum of the
a
particle can be calculated from the impulse produce by the electric
force between the
a
particle and the nucleus,

Because the magnitude of the linear momentum does not
change during the scattering, the triangle formed in the drawing above
is isosceles and
from
where,

Since the nucleus does not move during the scattering, the
kinetic energy before and after of the a
particle are the same,
.
Thus,
or the magnitude of the linear momentum does not
change during the scattering.
Looking at the diagram on the left, the magnitude
of the change in linear momentum and the magnitude of the linear
momentum of the
a
particle are related by
with
and
.
Thus,
which
implies that


The magnitude of the change in linear momentum can
be calculated from
where b is the
angle between the change in linear momentum vector and the vector
representing the instantaneous direction of the Coulomb’s force on the a
particle. Since the limits of integration include from
to
. The contribution of the
second integral vanishes. Therefore,


When the a particle is in an
arbitrary position, as in the case of the drawing (on the leftabove),
the Coulomb’s force makes the arbitrary angle b
with the vector representing the total change in linear momentum. 

The minimal value of b
is obtained when the a particle is at
; at this point,
. Then,


The maximum value of b
is obtained when the a particle is at
; at this point,
. Then,

Therefore,
with the angular
velocity of the a particle.
The derivative of the time with respect to the angle is,
. Substituting
in the integral,
the previous integral, the force is the electric force
(Coulomb’s force) .
Replacing this value in the previous integral result in
The result of the integral is
Thus, the scattering angle is given by
; using
we get,



Cross section: if a particle is directed inside
the corresponding area, the particle will be scattered through and angle
q or more,
. In the
drawing on the left, rays
,
,
, and
represent
particles incidents on the nucleus with impacting parameter smaller than
b,
consequently, the scattering angles are

In particular,
represents the case
when the impact parameter is b corresponding to an angle of scattering
of q. However, the particle trajectory
corresponds to an
impact parameter bigger than b,
, thus, the
scattering angle for this trajectory is smaller than q,
.
with
results in

aParticles
Scattered between and
Thus,

Notice that
decreases with
increasing (see graph below)
The element of surface shown in the previous drawing can be
written as
If represents the total
number of incident
aparticles,
then, the corresponding number scattered in an angle is .
The number of scattered particle in an angle per unit of area, , is
simplification,
Rutherford Scattering Formula is obtained,
