Q165 / 225 · JEE Main 2003 · Physics

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

  • Correct answer
    A
    $\frac{4 \mathrm{u}}{238}$
  • B
    $-\frac{4 \mathrm{u}}{234}$
  • C
    $\frac{4 \mathrm{u}}{234}$
  • D
    $-\frac{4 \mathrm{u}}{238}$

Solution

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

Hence, the correct answer is option: 'A'.

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