When a $U^{238}$ nucleus originally at rest, decays by emitting an alpha particle having a speed ' $u$ ',…

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
  1. $\frac{4 \mathrm{u}}{238}$
  2. $-\frac{4 \mathrm{u}}{234}$
  3. $\frac{4 \mathrm{u}}{234}$
  4. $-\frac{4 \mathrm{u}}{238}$

Solution

Applying the principle of conservation of linear momentum (4) $(\mathrm{u})=(\mathrm{v})(238) \Rightarrow \mathrm{v}=\frac{4 \mathrm{u}}{238}$

Asked in: JEE Main 2003

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