Chapter 27, Problem 59QAP

To determine

The net energy released from the given steps.

Answer to Problem 59QAP

The net energy released from all three steps is $26.73MeV$

Explanation of Solution

Calculation:

Mass of ${}^{1}H=1.0078251amu$

Mass of Deuterium ${(}^{2}H)=2.010412amu$

Mass of positron $=0.0005486amu$

Now, the 1st step of the reaction splits into two steps
  ${}^{1}H+{ }^{1}H\to He+ \gamma$

Followed by a beta-plus decay
  $He\to H+e^{+} + {\nu }_e$

The mass deficit reaction for the 1st step should look like
  $M.D.(\Delta m)=2m_H -m_D -m_e{}^{+}$

But if you look at your first reaction, you are starting with two ¹H's (1 electron in each atomic mass) and going to one ${}^{2}H$ (1 electron in its atomic mass). So, now when you subtract the initial and final atomic masses, the electrons do not cancel. Hence, you must account for the extra electron in the equation as well.
This gives us the following equation
  $M.D.=2m_H -m_D -m_e{}^{+} -m_e{}_{-}$

  $M.D.=2\left(1.0078251\right)-2.010412-0.0005486-0.0005486=0.000451amu$

The energy from 1 amu mass deficit is $931.5\raisebox{1ex}{$MeV$}\!\left/ \!\raisebox{-1ex}{$amu$}\right.$

The energy for this mass deficit can be calculated to be
  $E_{11} =0.000451\times 931.5=0.42MeV$

The positron and the electron emitted in step-1, immediately interact and annihilate each other.

  $e^{+} +e^{-} \to \gamma +\gamma$

The mass deficit for this is
  $\begin{array}{l}M.D.=m_e{}^{+} +m_e{}_{-} \\ =2\times 0.0005486=0.0010972amu\end{array}$

Energy released due to this mass deficit is
  $\begin{array}{l}{E}_{12} =0.0010972\times 931.5\\ =1.022MeV\end{array}$

For Step 2
  ${}^{2}H+{ }^{1}H\to { }^{3}He+ \gamma$

To calculate the energy released in this reaction we take the difference between the binding energies of ${}^{3}He$ and ${}^{2}H$.

  $BE{(}^{3}He)=7.718MeV$

  $BE{(}^{2}H)=2.225MeV$

The energy released in this reaction,
  $\begin{array}{l}{E}_2 =BE{(}^{3}He)-BE{(}^{2}H)\\ =7.718-2.225\\ =5.493MeV\end{array}$

For Step 3
  ${}^{3}He+{ }^{3}He\to { }^{4}He+2^{1}H+ \gamma$

The energy of this reaction is the difference between the binding energies of ${}^{4}He$ and the two ${}^{3}He$ atoms.

  $BE{(}^{4}He)=28.296MeV$

The energy released for this step is
  $\begin{array}{l}{E}_3 =BE{(}^{4}He)-2BE{(}^{3}He)\\ =28.296-2\times 7.718\\ \text{=}12.86MeV\end{array}$

Now, we have the energies from all the three steps
Step-1 and Step-2 need to occur twice in order to prepare 2 ${}^{3}He$ molecules, which interact and form ${}^{4}He$.

So, the net energy of the proton-proton cycle is
  $\begin{array}{l}E=2E_{11}+2E_{12}+2E_2+E_3\\ =2\times 0.42+2\times 1.022+2\times 5.493+12.86\\ =26.73MeV\end{array}$

Hence, the net energy released from all three steps is $26.73MeV$