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 ^(4))`

Answer

C

Explanation

At the distance of closest approach d, Kinetic energy = Potential energy
`1/2 mv ^(2) = (1)/( 4 pi in _(0)) ((2e)(Ze))/(d)`
where,
Ze = charge of target nucleus
`2e =` charge of alpha nucleus
`1/2 mv ^(2)=` kinetic energy of alpha nucleus of mass m moving wiht velocity v
`or d = (2 Ze ^(2))/(4 pi in _(0) ((1)/(2) mv ^(2))) because dalpha (1)/(m)`