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} $
