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Question

The distance of closest approach of an alpha-particle fired towards a nucleus with momentum p is r. If the momentum of the alpha-particle is 2p, the corresponding distance of closest approach is :

A
4r
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B
2r
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C
2r
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D
r4
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Solution

The correct option is C r4
At the distance of closest approach, r
K=14πε02eZe2K
K=P22m
r=44πε02Ze2p2pm
14πε04mZe2p2
Thus, r1p2
r1r2=p22p21=(2p)2p2
r1r2=41r2=r14=r4

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