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)

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

$\text{Conservation of Momentum}(M_U-M_{\alpha })v_u+M_{\alpha }{v}_{\alpha }=0$

$\text{(Here}{M}_U\text{=Mass of Uranium}\&M_{\alpha }\text{=Mass of}\alpha \text{Particle)}(v_u=\text{velocity of Uranium}\&v_{\alpha }=\text{u= velocity of}\alpha \text{particle)}$

$234v_u+4u=0$

$\Rightarrow v_u=\frac{-4u}{234}$