bei48482_FM

Example 4.8

Find the fraction of a beam of 7.7-MeV alpha particles that is scattered through angles of more

than 45° when incident upon a gold foil 3 10 ^7 m thick. These values are typical of the alpha-

particle energies and foil thicknesses used by Geiger and Marsden. For comparison, a human

hair is about 10^4 m in diameter.

Solution

We begin by finding n, the number of gold atoms per unit volume in the foil, from the relationship

n 

Since the density of gold is 1.93 104 kg/m^3 , its atomic mass is 197 u, and 1 u  1.66

10 ^27 kg, we have

n

5.90 1028 atoms/m^3

The atomic number Zof gold is 79, a kinetic energy of 7.7 MeV is equal to 1.23 10 ^12 J,

and 45°; from these figures we find that

f 7  10 ^5

of the incident alpha particles are scattered through 45° or more—only 0.007 percent! A foil

this thin is quite transparent to alpha particles.

In an actual experiment, a detector measures alpha particles scattered between 

and d, as in Fig. 4.33. The fraction of incident alpha particles so scattered is

found by differentiating Eq. (4.31) with respect to , which gives

1.93 104 kg/m^3



(197 u/atom)(1.66 10 ^27 kg/u)

massm^3



massatom

atoms



m^3

156 Appendix to Chapter 4

Figure 4.33In the Rutherford experiment, particles are detected that have been scattered between 

and d.

Foil

dθ

θ

r r sin θ

rdθ

Area = 4πr^2 sin cos θ 2 2 θdθ

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