Question
When a $$U^{238}$$ 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.
$$ - \frac{{4u}}{{238}}$$
Answer :
$$\frac{{4u}}{{234}}$$
Solution :
Here, conservation of linear momentum can be applied
$$\eqalign{ & 238 \times 0 = 4u + 234v \Rightarrow \,\therefore v = - \frac{4}{{234}}u \cr & \therefore {\text{speed}} = \left| {\vec v} \right| = \frac{4}{{234}}u \cr} $$
Here, conservation of linear momentum can be applied
$$\eqalign{ & 238 \times 0 = 4u + 234v \Rightarrow \,\therefore v = - \frac{4}{{234}}u \cr & \therefore {\text{speed}} = \left| {\vec v} \right| = \frac{4}{{234}}u \cr} $$