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Q.

A  238U nucleus decays by emitting an alpha particle of speed v ms−1. The recoil speed of the residual nucleus is (in  ms−1)

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a

−4v/234

b

v/4

c

−4v/238

d

4v/238

answer is A.

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Detailed Solution

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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A  238U nucleus decays by emitting an alpha particle of speed v ms−1. The recoil speed of the residual nucleus is (in  ms−1)