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 (Hint: $β^{-}$ decay)

(where $m_{e}$ is the mass of an electron and c is the velocity of light)

(M_{x} - M_{y} - m_{e}) c^{2}

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(M_{x} - M_{y} + m_{e}) c^{2}

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(M_{x} - M_{y}) c^{2}

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(M_{x} - M_{y} - 2 m_{e}) c^{2}

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Solution

The correct option is C $(M_{x} - M_{y}) c^{2}$
For $β - d e c a y$
Energy is given by
$E = Δ m c^{2}$
$= (M_{x} - M_{y}) c^{2}$

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A nucleus X emits a beta particle to produce a nucleus Y. If their atomic masses are Mx and My respectively. The maximum energy of the beta particle emitted is (Hint: β− decay) (where me is the mass of an electron and c is the velocity of light)

The correct option is C (Mx−My)c2For β−decay Energy is given by E=Δm c2=(Mx−My)c2

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 (Hint: $β^{-}$ decay)

(where $m_{e}$ is the mass of an electron and c is the velocity of light)

The correct option is C $(M_{x} - M_{y}) c^{2}$
For $β - d e c a y$
Energy is given by
$E = Δ m c^{2}$
$= (M_{x} - M_{y}) c^{2}$