An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be proportional to

Options

A. `v^(2)`
B. `1//m`
C. `1//v^(4)`
D. `1//Ze`

Answer

B

Explanation

Charge on `alpha`-particle `= 2e`
Change on target nucles = Ze
When the `alpha`-particle is projected towards the target nucleus, then at `r = r_(0)`, the `alpha`-particle comes to momentary rest. This position `r_(0)` from target nucleus is known as distance of closest approach. Applying law of conservation of energy, we get
`(1)/(2) mv^(2) = (K (ze)(2e))/(r_(0)) rArr r_(0) prop (1)/(v^(2)), r_(0) Ze, r_(0) prop (1)/(m)`