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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/V`

B

`1/(Ze)`

C

`v^(2)`

D

`1/m`

Text Solution

Verified by Experts

The correct Answer is:
D
`1/2mv^(2)=1/(4piepsilon_(0))((2e)(ze))/r`
` K.E=+(13.6Z^(2))/n^(2)`
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