Question

In Rutherford’s scattering experiment when an alpha particle (charge = +2 e, mass = 4 m_p) approaches a gold nucleus (Z = 79), it is continuously repelled, so it loses its kinetic energy (K) and its potential energy increases. Finally, an α-particle comes to rest momentarily when whole of the kinetic energy is changed into the potential energy of the charge at that distance from the nucleus. Let this distance be r₀ after which the α-particle returns back again due to electrostatic repulsion. Using the above information, derive an expression for r₀.

Answer 1

Gold nucleus charge = +Ze

Alpha particle charge = +2e

So potential energy at closest approach:

U = `1/(4 pi epsilon_0) xx ((2e)(Ze))/r_0`

= `1/(4 pi epsilon_0) (2Ze^2)/r_0`

At closest approach,

K = U

K = `1/(4 pi epsilon_0) (2Ze^2)/r_0`

Rearranging:

r₀ = `1/(4 pi epsilon_0) (2Ze^2)/K`

For gold, Z = 79:

r₀ = `1/(4 pi epsilon_0) (158 e^2)/K`