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How much energy is released in the first nuclear reaction of the p-p chain which produces energy in lower-mass stars like the Sun? The value of 1.44 MeV is given in many sources, but I don't understand why. The reaction is:

p + p -> 2H + e+ + νe

Mass of proton is 938.783 Mev, mass of deuterium 1876.124 MeV, mass of electron 0.511 MeV. So I would expect the production of

2 * 938.783 - 1876.124 - 0.511 = 0.931 MeV

I get the 1.44 MeV without positron on the right side... In the stellar interior anihilation of the positron with a free electron is expected. Sometimes, it is said that taking the anihilation into account gives the 1.44 MeV value, but I can't see why.

The energy produced in anihilation should be 2 * 0.511 MeV, that is 1.022 MeV, a this plus 0.931 MeV gives 1.862 MeV, not 1.44 MeV.

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Those aren't the proton and deuteron masses; those are the masses of protium and deuterium atoms, which means they include an electron mass (and also the small amount of binding energy between the nucleus and electron, but that's not relevant at this precision).

The CODATA value for the proton mass is 938.272 MeV, and the NIST value for the deuteron mass is 1875.61 MeV. Note how these are both roughly 1 electron-mass below your values. Plugging these into your calculation gives:

$$2\times 938.272 - 1875.61 - 0.511 = 0.423\text{ MeV}$$

Combining this with the $2\times0.511=1.022\text{ MeV}$ released from the positron annihilation gives a total of $1.44\text{ MeV}$.

The reason your calculation almost worked is as follows:

You included an extra electron mass inside the proton mass, and you included an extra electron mass inside the deuterium mass. Let's rewrite your equation with these included:

$$2\times (m_p+m_e) - (m_D+m_e) - m_e=2m_p-m_D$$

which is too high by one electron-mass due to the extra electron included with the second proton. When you added another electron-mass, you were now two electron-masses higher than the difference in binding energy, which is exactly the amount of energy released in electron-positron annihilation and therefore leads to the correct answer.

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  • $\begingroup$ I used Live Chart of Nuclides (www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html), clicked on 1H and used the atomic mass (1.0782503224 amu) as the mass of proton. Clearly, I was wrong, since this gives me 938.783 MeV.... $\endgroup$ – Leos Ondra Feb 22 at 13:47
  • $\begingroup$ Where is possible to get authoritative masses of atomic nuclei (rather than atoms with electrons)? $\endgroup$ – Leos Ondra Feb 22 at 14:06

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