Q. A $_{\ }^{238}U$ nucleus decays by emitting an alpha particle of speed $v\ ms^{- 1}$. The recoil speed of the residual nucleus is (in $ms^{- 1}$)
A) $- 4v/234$
B) $v/4$
C) $- 4v/238$
D) $4v/238$
Explanation
(a) Initially $_{\ }^{238}U$ nucleus was at rest and after decay its part moves in opposite direction
According to conservation of momentum
$4v + 234V = 238 \times 0 \Rightarrow V = - \frac{4v}{234}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.