question_answer

When a \[{{\operatorname{U}}^{238}}\] nucleus originally at rest, decays by emitting an alpha particle having a speed ?u?, the recoil speed of the residual nucleus is

A) \[\frac{4u}{238}\]         

B)        \[\frac{4u}{234}\]

C) \[\frac{4u}{234}\]         

D) \[-\frac{4u}{238}\]

Correct Answer: C

Solution :

Here, conservation of linear momentum can be applied   \[238\times 0=4\,u+234v\] \[\therefore \,\,\,\nu =-\frac{4}{234}u\] \[\therefore \,\,\,\,speed=\,\,\left| {\vec{\nu }} \right|=\,\,\frac{4}{234}\,u\]