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Chapter 12

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.

Step-by-Step

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)