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Question

# The closest distance of approach of an α - particle travelling with a velocity v towards a stationary nucleus is d. For the closest distance to become d2 towards a stationary nucleus of double the charge, the velocity of projection of the α- particle has to be

A
2v
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B
2v
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C
4v
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D
6v
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Solution

## The correct option is B 2vFrom work-energy theorem, (KE)α = P.E of Nucleus and α− particle 12mv2=14πε0(qnqαd) ........(1) For, q′n=2qnd′=d2 12mv′2=14πε0(q′nqαd′) ........(2) On dividing (2) ÷ (1) (v′v)2=qn′qn×dd′ (v′v)2=2qnqn×d(d2)=4 ∴ v′=2v

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