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

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

A

`(Ze)^(-1)`

B

`v^(2)`

C

`m^(-1)`

D

`v^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Distance of closest approach to `r_(0)=(Ze^(2))/(4piin_(0)xx(1)/(2)mv^(2))`
`:.r_(0)prop(1)/(m)` [All other terms are constant]
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