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Questions  

 238U nucleus decays by emitting an alpha particle of speed vms1. The recoil speed of the residual nucleus is (in  ms1)

a
−4v/234
b
v/4
c
−4v/238
d
4v/238

detailed solution

Correct option is A

Initially 238U nucleus was at rest and after decay its part moves in opposite direction.According to conservation of momentum4v+234V= 238 × 0 ⇒V=−4v234

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