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A nucleus of mass $M+\Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. Speed of light is $\mathrm{C}$.
The speed of daughter nuclei is
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The speed of daughter nuclei is
Solution:
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Verified Answer
The correct answer is:
$c \sqrt{\frac{2 \Delta m}{M}}$
$c \sqrt{\frac{2 \Delta m}{M}}$
Conserving the momentum
$$
\begin{gathered}
0=\frac{\mathrm{M}}{2} \mathrm{~V}_1-\frac{\mathrm{M}}{2} \mathrm{~V}_2 \\
\mathrm{~V}_1=\mathrm{V}_2 \quad \quad \ldots \ldots (1) \\
\Delta \mathrm{mc}^2=\frac{1}{2} \cdot \frac{\mathrm{M}}{2} \mathrm{~V}_1^2+\frac{1}{2} \cdot \frac{\mathrm{M}}{2} \cdot \mathrm{V}_2^2 \quad \quad \ldots \ldots (2)\\
\Delta \mathrm{mc}^2=\frac{\mathrm{M}}{2} \mathrm{~V}_1^2 \\
\frac{2 \Delta \mathrm{mc}^2}{\mathrm{M}}=\mathrm{V}_1^2 \\
\mathrm{~V}_1=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}
\end{gathered}
$$
$$
\begin{gathered}
0=\frac{\mathrm{M}}{2} \mathrm{~V}_1-\frac{\mathrm{M}}{2} \mathrm{~V}_2 \\
\mathrm{~V}_1=\mathrm{V}_2 \quad \quad \ldots \ldots (1) \\
\Delta \mathrm{mc}^2=\frac{1}{2} \cdot \frac{\mathrm{M}}{2} \mathrm{~V}_1^2+\frac{1}{2} \cdot \frac{\mathrm{M}}{2} \cdot \mathrm{V}_2^2 \quad \quad \ldots \ldots (2)\\
\Delta \mathrm{mc}^2=\frac{\mathrm{M}}{2} \mathrm{~V}_1^2 \\
\frac{2 \Delta \mathrm{mc}^2}{\mathrm{M}}=\mathrm{V}_1^2 \\
\mathrm{~V}_1=\mathrm{c} \sqrt{\frac{2 \Delta \mathrm{m}}{\mathrm{M}}}
\end{gathered}
$$
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