Calculate the energy output in each of the fusion reactions in the proton-proton cycle, and verify the values given in the above summary.

A chain of nuclear reactions known as a proton-proton cycle transforms hydrogen into helium.

\(^1H{ + ^1}H{ \to ^2}H + {e^\dag } + {v_c}\)

Consider the given equation.

In this problem, we compute the energy released for each reaction from Problem 26. The difference in total mass of nuclei before and after the nuclear reaction causes the energy release. The free mass-energy formula can be used to compute the energy difference:

\(E = \Delta m{c^2}\)

To obtain these mass differences, we subtract the masses of the parent nuclei from the masses of the daughter nuclei and all other nuclear reaction by products. Because the unified atomic unit will be used to express all atomic masses, it's important to remember how to convert it to ev:

\(lu = \frac{{931.5{\rm{ }}MeV}}{{{c^2}}}\)

Now we can analyse the reactions:

\(^1H{ + ^1}H{ \to ^2}H + {e^ + } + {v_c}\)

Looking in the Appendix for the atomic masses we get the following:

\(\begin{aligned} ^1H &= 1.007825{\rm{ }}{u^2}\\H &= 2.014102{\rm{ }}u{e^ + }\\ &= 0.00054858{\rm{ }}u{v_e}\\ &= 0.00054858{\rm{ }}u\end{aligned}\)

So we write:

\(\Delta m = (2 \times 1.007825 - 2.014102 - 2 \times 0.00054858)u\)

Which gives the result of:

\(\Delta m = 0.00045084{\rm{ }}u\)

Therefore, for our energy relation we have:

\(\begin{aligned} E &= \Delta m{c^2}\\

&= 0.00045084 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E &= 0.42MeV\end{aligned}\)

\(^1H{ + ^2}H{ \to ^3}He + \gamma \)

For the second reaction we follow the same procedure.

\(^1H{ + ^2}H{ \to ^3}He + \gamma \)

Through the Appendix we look at the atomic masses:

\(\begin{aligned} ^1H &= 1.007825{\rm{ }}u\\^2H &= 2.014102{\rm{ }}u\\^3He &= 3.016030{\rm{ }}u\end{aligned}\)

Photons or\(\gamma \)rays have no mass. And again we calculate the mass difference:

\(\begin{aligned} \Delta m & = (1.007825 + 2.014102 - 3.016030)u\\\Delta m &= 0.005897{\rm{ }}u\end{aligned}\)

And for the total energy we got:

\(\begin{aligned} E &= \Delta m{c^2}\\ &= 0.00045084 \cdot \frac{{931.5MeV}}{{{c^2}}} \cdot {c^2}\\E &= 0.42MeV\end{aligned}\)

Therefore, energy release is \(E = 5.49{\rm{ }}MeV\).

\(^3He{ + ^3}He{ \to ^4}He{ + ^1}H{ + ^1}H\)

Let us solve the given problem.

And for the last reaction of the proton-proton cycle we got:

\(^3He{ + ^3}He{ \to ^4}He{ + ^1}H{ + ^1}H\)

We get the masses trough the Appendix:

\(\begin{aligned} ^1H &= 1.007825{\rm{ }}u\\^4He &= 4.002603{\rm{ }}u\\^3He &= 3.016030{\rm{ }}u\end{aligned}\)

And again we calculate the mass difference:

\(\Delta m = (2 \times 3.016029 - 4.002603 - 2 \times 1.007825)u\)

And for the mass difference, we get the result of:

\(\Delta m = 0.013805{\rm{ }}u\)

And for the total energy we got:

\(\begin{aligned} E &= \Delta m{c^2} &= 0.013805 \cdot \frac{{931.5{\rm{ }}MeV}}{{{c^2}}} \times {c^2}\\E &= 12.859{\rm{ }}MeV\end{aligned}\)

Therefore, the energy release is\(E = 12.859{\rm{ }}MeV\)