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Question

A U238 nucleus originally at rest, decays by emitting an αparticle, say with a velocity of v m/s. The recoil velocity in (ms1) of the residual nucleus is

A
4v238
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B
4v238
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C
v4
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D
4v234
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Solution

The correct option is D 4v234
Mass of alpha particle is ma=4m where m is mass of a proton.
And mass of residual nucleus will be mr=(2384)m=234m
Momentum conservation mava+mrvr=0 putting values of masses and va=v
we get vr=4v234

Option D is correct.

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