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

An alpha nucleus of energy `(1)/(2)mv^(2)` bomobards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

A

1/m

B

`1//v^(2)`

C

1/ze

D

`v^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Distamnce of closet apporach , ` r_(0)=(Ze^(2))/(2mv^(2)piepsi_(0))`
` rArr " " r_(0)prop (1)/(m)`
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