An alpha nucleus of energy 1/2 mv² bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

Options

(a) 1 / Ze
(b) v²
(c) 1 / m
(d) 1 / v⁴

Correct Answer:

1 / m

Explanation:

Kinetic energy of alpha nucleus is equal to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
1/2 mv² = 1/4π?￢ﾂﾀ . qαZe / r₀
where r₀ is the distance of closest approach
r₀ = 2/4π?￢ﾂﾀ . qαZe / mv²
r₀ ∞ Ze ∞ qα ∞ 1/m ∞ 1/v²
Hence option (c) is correct.