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A nucleus X emits a beta particle to produce a nucleus Y. If their atomic masses are $\mathrm{M}_{\mathrm{x}}$ and $\mathrm{M}_{\mathrm{y}}$ respectively, the maximum energy of the beta particle emitted is (where $m_{e}$ is the mass of an electron and $c$ is the velocity of light)
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Verified Answer
The correct answer is:
$\left(M_{x}-M_{y}\right) \mathrm{c}^{2}$
Hint:
$\mathrm{m}_{\mathrm{x}}=\mathrm{M}_{\mathrm{x}}-\mathrm{zm}_{\mathrm{e}}$
$\frac{\mathrm{m}_{\mathrm{y}}=\mathrm{M}_{\mathrm{y}}-(\mathrm{z}+1) \mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{x}}-\mathrm{m}_{\mathrm{y}}=\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}+\mathrm{m}_{\mathrm{e}}}$
$\mathrm{E}=\Delta \mathrm{mC}^{2}=\left(\mathrm{m}_{\mathrm{x}}-\mathrm{m}_{\mathrm{y}}-\mathrm{m}_{\mathrm{e}}\right) \mathrm{C}^{2}=\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}\right) \mathrm{C}^{2}$
$\mathrm{m}_{\mathrm{x}}=\mathrm{M}_{\mathrm{x}}-\mathrm{zm}_{\mathrm{e}}$
$\frac{\mathrm{m}_{\mathrm{y}}=\mathrm{M}_{\mathrm{y}}-(\mathrm{z}+1) \mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{x}}-\mathrm{m}_{\mathrm{y}}=\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}+\mathrm{m}_{\mathrm{e}}}$
$\mathrm{E}=\Delta \mathrm{mC}^{2}=\left(\mathrm{m}_{\mathrm{x}}-\mathrm{m}_{\mathrm{y}}-\mathrm{m}_{\mathrm{e}}\right) \mathrm{C}^{2}=\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}\right) \mathrm{C}^{2}$
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