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## Q. 12.14

Show that the relations expressed for the disintegration energy Q in Equations (12.38), (12.41), and (12.44) are correct.

$Q=\left[M\left({ }_{Z}^{A} X\right)-M\left( {}_{z+1}^{\quad A}D\right)\right] c^{2} \quad \beta^{-} \text {decay }$ (12.38)

$Q=\left[M\left({ }_{Z}^{A} X\right)-M\left({ }_{z-1}^{\quad A} D\right)-2 m_{e}\right] c^{2} \quad \beta^{+} \text {decay }$ (12.41)

$Q=\left[M\left({ }_{Z}^{A} X\right)-M\left({ }_{z-1}^{\quad A} D\right)\right] c^{2} \quad \text { Electron capture }$ (12.44)

Strategy We begin with the reaction for each of the beta decays ($\beta^{-}, \beta^{+}$, and EC) and change it to an energy equation. We neglect any neutrino mass and atomic binding energies and eventually use atomic masses.

## Verified Solution

$\beta ^{-}$ decay: We begin with $\beta ^{-}$ decay and write the mass-energy equation for the reaction in Equation (12.37).

${ }_{Z}^{A} X \rightarrow{ }_{Z+1}^{\quad A} D+\beta^{-}+\bar{\nu} \quad \beta^{-} \text {decay }$ (12.37)

$M_{\text {nucl }}\left({ }_{Z}^{A} X\right)=M_{\text {nucl }}\left({ }_{z+1}^{\quad A} D\right)+m_{e}+Q / c^{2}$

where we use $M_{\text {nucl }}$ to indicate the nuclear mass. In order to change to atomic masses we add $Zm _{e}$ to each side above.

$M_{\text {nucl }}\left({ }_{Z}^{A} X\right)+Z m_{e}=M_{\text {nucl }}\left({ }_{Z+1}^{\quad A} D\right)+(Z+1) m_{e}+Q / c^{2}$

Because we are neglecting the difference in atomic binding energies on the two sides of the equation, we now write this equation in terms of atomic masses.

$M\left({ }_{Z}^{A} X\right)=M\left({ }_{z+1}^{\quad A} D\right)+Q / c^{2}$

We solve this equation for Q to determine Equation (12.38).

$Q=\left[M\left({ }_{Z}^{A} X\right)-M\left({ }_{Z+1}^{\quad A} D\right)\right] c^{2}$  $\beta^{-} \text {decay }$ (12.38)

$\beta ^{+}$ decay: We write the mass-energy equation for the reaction in Equation (12.40) and follow a procedure similar to that above.

${ }_{Z}^{A} X \rightarrow{ }_{Z-1}^{\quad A} D+\beta^{+}+\nu$ (12.40)

$\left.M_{\text {nucl }}{ }_{Z}^{A} X\right)=M_{\text {nucl }}\left({ }_{z-1}^{\quad A} D\right)+m_{e}+Q / c^{2}$

We again add $Zm _{e}$ to each side and have

$M_{\text {nucl }}\left({ }_{Z}^{A} X\right)+Z m_{e}=M_{\text {nucl }}\left({ }_{z-1}^{\quad A} D\right)+(Z+1) m_{e}+Q / c^{2}$

In this case we only need $(Z-1) m_{e}$ for the daughter atomic mass, which gives us a remaining mass of $2 m_{e}$.

$M\left({ }_{Z}^{A} X\right)=M_{\text {nucl }}\left({ }_{z-1}^{\quad A} D\right)+2 m_{e}+Q / c^{2}$

We solve this equation for Q to determine Equation (12.41).

$Q=\left[M\left({ }_{Z}^{A} X\right)-M\left({ }_{z-1}^{\quad A} D\right)-2 m_{e}\right] c^{2} \quad \beta^{+} \text {decay }$ (12.41)

Electron capture: The mass-energy equation for the electron capture reaction of Equation (12.43) is

${ }_{Z}^{A} X+e^{-} \rightarrow{ }_{Z-1}^{\quad A} D+\nu \quad \text { Electron capture }$ (12.43)

$M_{\text {nucl }}\left({ }_{Z}^{A} X\right)+m_{e}=M_{\text {nucl }}\left({ }_{z-1}^{\quad A} D\right)+Q / c^{2}$

We add $(Z-1) m_{e}$ to each side above and obtain

$M_{\text {nucl }}\left({ }_{Z}^{A} X\right)+Z m_{e}=M_{\text {nucl }}\left({ }_{Z-1}^{\quad A} D\right)+(Z-1) m_{e}+Q / c^{2}$

In this case we have just the right number of electron masses to change to atomic masses.

$M\left({ }_{Z}^{A} X\right)=M\left({ }_{Z-1}^{\quad A} D\right)+Q / c^{2}$

We solve this equation for Q to find Equation (12.44).

$Q=\left[M\left({ }_{Z}^{A} X\right)-M\left({ }_{Z-1}^{\quad A} D\right)\right] c^{2}$    Electron capture (12.44)