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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

1/m

B

`1//v^(4)`

C

1/Ze

D

`v^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Distance of closest approach KE = PE
`rArr (1)/(2) mv^(2) = (Ze^(2))/(4pi epsi_(0) r_(0))`
`rArr r_(0)= (Ze^(2))/(2mv^(2) pi epsi_(0)) rArr r_(0) prop (1)/(m)`
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