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The closed distance of approach of an al...

The closed distance of approach of an `alpha-` particle travelling with velocity V towards a stationary nucleus is `d`.For the closest distance to become `(d)/(3)` towards a stationary nucleus of double of the charge,the velocity of projection of the `alpha-` particle has to be

A

6V

B

`sqrt6V`

C

`(V)/(56)`

D

`(3V)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(1)/(2) mv^(2) = (1)/(4pi in_(0)) (q_(1)q_(2))/(r )`
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