Question

When $U^{238}$ 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 1

Answer: (B)

According to principle of conservation of linear momentum the momentum of the system remains the same before and after the decay.
Atomic mass of uranium = 238 and after emitting an alpha particle
$= 238 − 4 = 234$
$∴ 238 × 0 = 4u + 234 v$
$\therefore v=-\frac{4u}{234}$