bei48482_FM

The quantity d dtis just the angular velocity of the alpha particle about the nucleus

(this is evident from Fig. 4.31).

The electric force exerted by the nucleus on the alpha particle acts along the radius

vector joining them, so there is no torque on the alpha particle and its angular

momentum mr^2 is constant. Hence

mr^2 constantmr^2 mb

from which we obtain



Substituting this expression for dtd in Eq. (4.27) gives

2 m^2 bsin 

() 2

() 2

Fr^2 cos d (4.28)

As we recall, Fis the electric force exerted by the nucleus on the alpha particle. The

charge on the nucleus is Ze, corresponding to the atomic number Z, and that on the

alpha particle is 2e. Therefore

F

and sin 

() 2

() 2

cos d 2 cos

The scattering angle is related to the impact parameter bby the equation

cot  b

It is more convenient to specify the alpha-particle energy KE instead of its mass and

velocity separately; with this substitution,

Scattering angle cot  b (4.29)

Figure 4.32 is a schematic representation of Eq. (4.29); the rapid decrease in as b

increases is evident. A very near miss is required for a substantial deflection.

Rutherford Scattering Formula

Equation (4.29) cannot be directly confronted with experiment because there is no way

of measuring the impact parameter corresponding to a particular observed scattering

angle. An indirect strategy is required.

4  0 KE



Ze^2





2

2  0 m^2



Ze^2





2





2





2

4  0 m^2 b



Ze^2

2 Ze^2



r^2

1



4  0





2

r^2



b

dt



d

d



dt

154 Appendix to Chapter 4

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