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