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. `1/Z^2`
B. `1/m`
C. `v^2`
D. `1/e^2`

Answer

B

Explanation

This is based on Rutherford's scattering experiment. The `alpha` particle moving with a velocity v approaches a heavy nucleus of atomic number Z. At the distance of closest approach `(r_0)` all its K.E. `(1/2mv^2)` is converted into electrostatic P.E.
`therefore 1/2mv^2 =(1)/(4piepsi_0) . (q_1q_2)/(r) = (1)/(4piepsi_0) ((2e)(Ze))/(r_0)`
[`therefore q_1 = 2e " and " q_2 = Ze` ]
`therefore r_0 = (1)/(piepsi_0) (Ze^2)/(mv^2) therefore r_0 prop 1/m`