Kvant Physics Problem 559

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Problem

F559. According to Bethe's theory, the carbon cycle of stellar thermonuclear reactions consists of the following reactions:

$$p + {}^{12}\mathrm{C} \rightarrow {}^{13}\mathrm{N} + \gamma,\qquad {}^{13}\mathrm{N} \rightarrow {}^{13}\mathrm{C} + e^{+} + \nu,$$

$$p + {}^{13}\mathrm{C} \rightarrow {}^{14}\mathrm{N} + \gamma,\qquad p + {}^{14}\mathrm{N} \rightarrow {}^{15}\mathrm{O} + \gamma,$$

$${}^{15}\mathrm{O} \rightarrow {}^{15}\mathrm{N} + e^{+} + \nu,\qquad p + {}^{15}\mathrm{N} \rightarrow {}^{12}\mathrm{C} + {}^{4}\mathrm{He}.$$

Find the energy released in the formation of one mole of helium.

Setup and Assumptions

The physical system is the carbon cycle of thermonuclear reactions in a star. The catalyst nucleus ${}^{12}\mathrm{C}$ is regenerated at the end of the cycle and does not contribute to the net mass balance.

The unknown quantity is the energy released when one mole of helium nuclei ${}^{4}\mathrm{He}$ is produced.

The reactions are

$$p + {}^{12}\mathrm{C} \rightarrow {}^{13}\mathrm{N} + \gamma,$$

$${}^{13}\mathrm{N} \rightarrow {}^{13}\mathrm{C} + e^{+} + \nu,$$

$$p + {}^{13}\mathrm{C} \rightarrow {}^{14}\mathrm{N} + \gamma,$$

$$p + {}^{14}\mathrm{N} \rightarrow {}^{15}\mathrm{O} + \gamma,$$

$${}^{15}\mathrm{O} \rightarrow {}^{15}\mathrm{N} + e^{+} + \nu,$$

$$p + {}^{15}\mathrm{N} \rightarrow {}^{12}\mathrm{C} + {}^{4}\mathrm{He}.$$

We assume that atomic masses may be used throughout. The positrons produced in the two $\beta^+$ decays subsequently annihilate with electrons from the stellar plasma, so their rest energy contributes to the released energy. The neutrinos escape from the star; since the problem asks for the energy released by the reaction itself, the standard mass defect of the net reaction is calculated. The small neutrino kinetic energies are neglected.

The required atomic masses are

$$m(^1\mathrm H)=1.007825,u,$$

$$m(^4\mathrm{He})=4.002603,u.$$

One atomic mass unit corresponds to

$$1,u,c^2 = 931.5\ \mathrm{MeV}.$$

Avogadro's constant is

$$N_A = 6.022\times10^{23}\ \mathrm{mol^{-1}}.$$

Physical Principles

The solution is based on conservation of nucleon number and electric charge.

The energy released in a nuclear reaction is determined by the mass defect,

$$Q=\Delta m,c^2.$$

For a sequence of reactions, the total energy release equals the difference between the total initial and total final masses. Any nuclei appearing on both sides of the overall reaction cancel from the mass balance.

The energy corresponding to one mole of reactions is obtained by multiplying the energy released per reaction by Avogadro's constant:

$$E_{\text{mol}} = N_A Q.$$

Derivation

Adding all six reactions, the intermediate nuclei

$${}^{13}\mathrm N,\quad {}^{13}\mathrm C,\quad {}^{14}\mathrm N,\quad {}^{15}\mathrm O,\quad {}^{15}\mathrm N$$

cancel.

The catalyst nucleus ${}^{12}\mathrm C$ also appears on both sides and cancels.

The net reaction becomes

$$4p \rightarrow {}^{4}\mathrm{He} + 2e^{+} + 2\nu .$$

When atomic masses are used, the electron masses are already included in the atomic masses. The positron masses are automatically accounted for through the use of atomic rather than bare nuclear masses, and the mass defect can be written simply as

$$\Delta m = 4m(^1\mathrm H)-m(^4\mathrm{He}).$$

Substituting the numerical values,

$$\Delta m = 4(1.007825,u)-4.002603,u.$$

Hence

$$\Delta m = 4.031300,u-4.002603,u = 0.028697,u.$$

The energy released per helium nucleus formed is

$$Q = 0.028697,u \times 931.5\ \frac{\mathrm{MeV}}{u}.$$

Thus

$$Q \approx 26.73\ \mathrm{MeV}.$$

Converting to joules,

$$Q = 26.73\times10^{6}\ \mathrm{eV} \times 1.602\times10^{-19}\ \frac{\mathrm J}{\mathrm{eV}},$$

which gives

$$Q \approx 4.28\times10^{-12}\ \mathrm J.$$

For one mole of helium,

$$E_{\text{mol}} = N_AQ = (6.022\times10^{23}\ \mathrm{mol^{-1}}) (4.28\times10^{-12}\ \mathrm J).$$

Therefore,

$$E_{\text{mol}} \approx 2.58\times10^{12}\ \mathrm{J,mol^{-1}}.$$

Result

The net carbon-cycle reaction is

$$4p \rightarrow {}^{4}\mathrm{He}+2e^{+}+2\nu .$$

Its mass defect is

$$\Delta m = 4m(^1\mathrm H)-m(^4\mathrm{He}) = 0.028697,u.$$

The energy released per helium nucleus is

$$Q = \Delta m c^2 = 0.028697\times931.5 \approx 26.73\ \mathrm{MeV}.$$

For one mole of helium nuclei,

$$E_{\text{mol}} = N_AQ \approx 2.58\times10^{12}\ \mathrm J.$$

$$\boxed{ E_{\text{mol}} \approx 2.6\times10^{12}\ \mathrm{J,mol^{-1}} }$$

Sanity Checks

The dimensions are consistent. The mass defect is measured in atomic mass units, multiplication by $c^2$ converts it to energy, and multiplication by $N_A$ gives energy per mole.

The net reaction is the fusion of four hydrogen nuclei into one helium nucleus. The known energy release for this process is approximately $26.7\ \mathrm{MeV}$ per helium nucleus, matching the value obtained here.

Multiplying $26.7\ \mathrm{MeV}$ by Avogadro's constant gives an energy of order

$$10^{24}\times10^{-12}\ \mathrm J \sim 10^{12}\ \mathrm J,$$

which agrees with the calculated result.

The step most susceptible to error is the use of atomic masses. If the positron contribution is handled inconsistently, an incorrect mass defect and hence an incorrect energy release will result. Using atomic masses throughout avoids this difficulty and directly yields the correct net value.