bullet Rutherford Scattering

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

  1. Alpha particles and Nucleus can be considered as point masses and charges.
  2. Electric repulsion between the charges is the only force present in the scattering.
  3. 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.
  4. The foil containing the targets is so thin that only a single scattering occurs.
  5. 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 Coulombís 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 left-above), the Coulombís 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 torque on the a particle is .  Thus, the angular momentum of the particle is conserved with respect to the position of the nucleus. Therefore, and constant. On the other side, with Consequently, equating these two results for the angular momentum,

The derivative of the time with respect to the angle is,   . Substituting in the integral,

the previous integral, the force is the electric force (Coulombís 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,

bullet Cross Section Definition

 

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, .

Calling N the number of nuclei on a volume V and n the number of nuclei per unit of volume, from where the number of nuclei in the foil is  .  Therefore, the foil effective cross section is . Defining the fraction of a particles scattered by the angle q or more as follows,

which in terms of the parameters of the problem is

with results in

a-Particles 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 a-particles, 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,

 

 

 

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by Luis F. SŠez, Ph. D.    Comments and Suggestions: LSaez@dallaswinwin.com