Question

When a $U^{238}$ nucleus originally at rest, decays by emitting an alpha particle having a speed 'u', the velocity of the residual nucleus is-

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

Answer 1

Answer: (D)

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

Applying conservation of momentum $\overrightarrow{p_i}=\overrightarrow{p_f}$
$\overrightarrow{p_i}=0,\overrightarrow{p_f}=m_{\alpha }\overrightarrow{V_{\alpha }}+m_R\overrightarrow{V_R}$
$m_{\alpha }\overrightarrow{V_{\alpha }}+m_R\overrightarrow{V_R}=0$ $V_R=-\frac{m_{\alpha }\overrightarrow{V_{\alpha }}}{m_R}=-\frac{4u}{238-4}=-\frac{4u}{234}$