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Question

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy K1 is ro. The distance of the closest approach when the α – particle is fired at the same nucleus with kinetic energy 2K1 will be.

A
ro2
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B
ro4
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C
2ro
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D
4ro
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Solution

The correct option is A ro2
At a distance of closest approach, kinetic energy of α particle = Potential energy of system.
Kq1q2r=12mv2
Initially kinetic energy,
K1=Kq1q2r0 .....(i)
When K2=2K1
2K1=Kq1q2r ....(ii)
From the above equation we get,
r=ro2

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