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)/(ze)`
B. `v^(2)`
C. `(1)/(m)`
D. `(1)/(v^(2))`

Answer

C

Explanation

As we know, at the distance of closet approch is d.
kinetic energy = Potential energy
` (1)/(2) mv^(2) =(1)/(4pi epsi_(0)) ((2e)(Ze))/(d)`
where Ze = charge of target nucleus,
2e = charge of alpha nucles
`(1)/(2)mv^(2)` =kinetic energy of alpha nuclesus of mass m moving with velocity v.
or `d = (2Ze^(2))/(4pi epsi_(0)((1)/(2)mv^(2)))`
`therefore " "d prop (1)/(m)`