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An alpha nucleus of energy (1)/(2)m nu^(...

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

A

`v^(2)`

B

1/m

C

`1//v^(4)`

D

1/Ze

Text Solution

Verified by Experts

The correct Answer is:
B
For the distance of closest approch, we can write,
`(1)/(2)mv^(2)=(Kxx(Ze)(Ze))/(r_(0))rArrr_(0)prop(1)/(m)`
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