WBJEE · Physics · Nuclear Physics
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)
- A \(\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}-\mathrm{m}_{\mathrm{e}}\right) \mathrm{c}^{2}\)
- B \(\left(M_{x}-M_{y}+m_{e}\right) \mathrm{c}^{2}\)
- C \(\left(M_{x}-M_{y}\right) \mathrm{c}^{2}\)
- D \(\left(M_{x}-M_{y}-2 m_{e}\right) \mathrm{c}^{2}\)
Answer & Solution
Correct Answer
(C) \(\left(M_{x}-M_{y}\right) \mathrm{c}^{2}\)
Step-by-step Solution
Detailed explanation
Hint: \(\mathrm{m}_{\mathrm{x}}=\mathrm{M}_{\mathrm{x}}-\mathrm{zm}_{\mathrm{e}}\)…
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