A nucleus X emits a beta particle to produce | vedclass

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(C) The nuclear reaction for beta-minus decay is given by: $_{Z}X^{A} ightarrow _{Z+1}Y^{A} + _{-1}e^{0} + \bar{
u} + Q$.The $Q$-value of the react

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The nuclear reaction for beta-minus decay is given by: $_{Z}X^{A} ightarrow _{Z+1}Y^{A} + _{-1}e^{0} + \bar{
u} + Q$.The $Q$-value of the reaction is the energy released, which is given by the mass defect multiplied by $c^{2}$.The atomic mass $M_{x}$ of nucleus $X$ is related to its nuclear mass $m_{x}$ by $M_{x} = m_{x} + Zm_{e}$. Thus, $m_{x} = M_{x} - Zm_{e}$.The atomic mass $M_{y}$ of nucleus $Y$ is related to its nuclear mass $m_{y}$ by $M_{y} = m_{y} + (Z+1)m_{e}$. Thus, $m_{y} = M_{y} - (Z+1)m_{e}$.The energy released $Q$ is given by $Q = (m_{x} - m_{y} - m_{e})c^{2}$.Substituting the expressions for nuclear masses:$Q = [(M_{x} - Zm_{e}) - (M_{y} - (Z+1)m_{e}) - m_{e}]c^{2}$$Q = [M_{x} - Zm_{e} - M_{y} + Zm_{e} + m_{e} - m_{e}]c^{2}$$Q = (M_{x} - M_{y})c^{2}$.Therefore, the maximum energy of the emitted beta particle is $(M_{x} - M_{y})c^{2}$.

$A$ 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):

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