Question

A nucleus \(X\) emits a beta particle to produce a nucleus \(Y\). If their atomic masses are \(M_{x}\) and \(M_{y}\) 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) \(\left(M_{x}-M_{y}+m_{e}\right) c^{2}\)
(C) \(\left(\mathrm{M}_{\mathrm{x}}-\mathrm{M}_{\mathrm{y}}\right) \mathrm{c}^{2}\)
(D) \(\left(M_{x}-M_{y}-2 m_{e}\right) c^{2}\)

Answer 1

Ans: (C)
 \(:{ }_{z} \mathrm{X}^{\mathrm{A}} \rightarrow \underset{\mathrm{z}+1}{\hat{\mathrm{Y}}}+\mathrm{e}^{0}+\overline{\mathrm{v}}+\mathrm{Q}\)
\(m_{x}=M_{x}-z m_{e}\)
\(\frac{m_{y}=M_{y}-(z+1) m_{e}}{m_{x}-m_{y}=M_{x}-M_{y}+m_{e}}\)
\(E=\Delta m C^{2}=\left(m_{x}-m_{y}-m_{e}\right) C^{2}=\left(M_{x}-M_{y}\right) C^{2}\)

Answer 2

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Answer 3

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