question_answer

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

A)          \[-4v/234\]

B)                       \[v/4\]

C)          \[-4v/238\]                     

D)          \[4v/238\]

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 Þ \[V=-\frac{4v}{234}\]