Work Energy And Power Question 88

Question: A $ ^{238}U $ nucleus decays by emitting an alpha particle of speed $ vm{s^{-1}} $ . The recoil speed of the residual nucleus is (in $ m{s^{-1}} $ ) [CBSE PMT 1995; AIEEE 2003]

Options:

A) $ -4v/234 $

B) $ v/4 $

C) $ -4v/238 $

D) $ 4v/238 $

Show Answer

Answer:

Correct Answer: A

Solution:

Initially 238U nucleus was at rest and after decay its part moves in opposite direction. According to conservation of momentum $ 4v+234V $ = 238 × 0 therefore $ V=-\frac{4v}{234} $



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