A nucleus of mass M + Delta m is at rest and | vedclass

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(C) Let the masses of the daughter nuclei be $m_1 = \frac{M}{4}$ and $m_2 = \frac{3M}{4}$.Since the parent nucleus is at rest,by the law of conservati

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$c\sqrt{\frac{6\Delta m}{M}}$

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(C) Let the masses of the daughter nuclei be $m_1 = \frac{M}{4}$ and $m_2 = \frac{3M}{4}$.Since the parent nucleus is at rest,by the law of conservation of linear momentum,the magnitudes of the momenta of the two daughter nuclei must be equal: $p_1 = p_2 = p$.Thus,$m_1 v_1 = m_2 v_2 \implies \frac{M}{4} v_1 = \frac{3M}{4} v_2 \implies v_2 = \frac{v_1}{3}$.The energy released in the decay is $Q = \Delta m c^2$. This energy is shared as kinetic energy between the two nuclei:$\Delta m c^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$.Substituting the values: $\Delta m c^2 = \frac{1}{2} (\frac{M}{4}) v_1^2 + \frac{1}{2} (\frac{3M}{4}) (\frac{v_1}{3})^2$.$\Delta m c^2 = \frac{M v_1^2}{8} + \frac{3M}{4} \cdot \frac{v_1^2}{18} = \frac{M v_1^2}{8} + \frac{M v_1^2}{24} = \frac{3M v_1^2 + M v_1^2}{24} = \frac{4M v_1^2}{24} = \frac{M v_1^2}{6}$.Solving for $v_1$: $v_1^2 = \frac{6 \Delta m c^2}{M} \implies v_1 = c \sqrt{\frac{6 \Delta m}{M}}$.

$A$ nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of mass $\frac{M}{4}$ and $\frac{3M}{4}$. If the speed of light is $c$,the speed of the daughter nucleus of mass $\frac{M}{4}$ is:

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