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. $\frac{u}{238}$
• C. $-\frac{4u}{234}$
• D. $\frac{u}{234}$

Answer 1

The correct option is C

$-\frac{4u}{234}$

The momentum of the system remains the same before and after decay. This is in accordance with the principle of conservation of momentum.

Now, $4u+234v=238\times 0$
or 234 v = – 4u or $v=–\frac{4u}{234}$
So, (c) is the right choice.