The distance of the closest approach of an alpha particle fired at a nucleus with momentum P is r_o. The distance of the closest approach when the alpha particle is fired at the same nucleus with momentum 2P will be

Explanation

The distance of the closest approach (r₀) of an alpha particle is determined by the particle's kinetic energy and the electrostatic potential between the alpha particle and the nucleus. When an alpha particle with momentum P is fired at a nucleus, the closest approach occurs when the kinetic energy is entirely converted into electrostatic potential energy.

Kinetic Energy (KE) of the alpha particle:

KE = ^P2 / 2m

Where P is the momentum and m is the mass of the alpha particle.

Electrostatic Potential Energy at the closest approach

Potential Energy = kZ_e² / r₀

Where:

• k is Coulomb’s constant,
• Z is the atomic number of the nucleus,
• e is the charge of the alpha particle (which is 2e, because it consists of two protons).

Equating Kinetic Energy and Electrostatic Potential Energy

P² / 2m = kZ_e² / r₀

Now, if the momentum is increased to 2P, the kinetic energy becomes:

KE' = ^(2P)2 / 2m = 4P² / 2m

For the new closest approach r₀', the equation becomes:

4P² / 2m = kZ_e² / r_0'

By comparing this with the original equation, we can find the new distance of closest approach:

r_0' = r₀ / 4

Final Answer

The distance of the closest approach when the alpha particle is fired with momentum 2P is r₀/4.