[Astro] Proton-Proton Chain Energy

[tylerc1991]

Homework Statement

Find the energy released for the reactions in the Proton-Proton chain.

Homework Equations

Proton-Proton Chain:
¹H + ¹H -> ²H + e⁺ + v
e⁺ + e^- -> γ + γ
²H + ¹H -> ³He + γ
³He + ³He -> ⁴He + 2 ¹H

The Attempt at a Solution

To find the energy released in each of these reactions, I am calculating the mass deficit.

For the first reaction, I am finding a mass deficit of
2(1.007825 u) - 2.014102 u - 0.0005486 = 0.0009994 u
because v (neutrino) is massless.

Then the energy in MeV of this reaction is 0.0009994 u x 931.5 MeV/u = 0.9309411 MeV.

However, I checked http://en.wikipedia.org/wiki/Proton–proton_chain_reaction#The_proton.E2.80.93proton_chain_reaction, and it looks like the energy released is 0.42 MeV.

Did I do the calculation correctly?

Thank you for any help!

 

[TSny]

Looks like you're using atomic masses (which include the mass of the electrons in the atom as well as the mass of the nucleus.) So, you'll need to take account of those atomic electrons.

 

[tylerc1991]

Looks like you're using atomic masses (which include the mass of the electrons in the atom as well as the mass of the nucleus.) So, you'll need to take account of those atomic electrons.

I'm not exactly sure what you mean by this. I've calculated the energies of the other reactions and they seem to correspond to the correct values. It is only the first reaction that is getting me confused.

That is, I found a mass deficit of 2(0.0005486 u) = 0.0010972 u for the second reaction, which corresponds to a release of 1.0220418 MeV of energy (that matches the result online), and I used atomic masses here too.

 

[TSny]

For the second reaction you are just using the mass of the electron rather than an atomic mass.

If you go to the third reaction, you would get the correct answer using atomic masses. That's because the atomic masses of ²H and ¹H each contain the mass of one electron. So, together the two atomic masses include two electron masses. In the ³He product, the atomic mass will include two electrons. Since you end up subtracting the initial and final masses, the 2 electron masses in the hydrogen atomic masses are canceled out by the two electrons in the helium atomic mass. So, you "luck out".

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 ²H (1 electron in its atomic mass). So, now when you subtract the initial and final atomic masses, the electrons do not cancel. You'll need to take care of that.

 

[tylerc1991]

For the second reaction you are just using the mass of the electron rather than an atomic mass.

If you go to the third reaction, you would get the correct answer using atomic masses. That's because the atomic masses of ²H and ¹H each contain the mass of one electron. So, together the two atomic masses include two electron masses. In the ³He product, the atomic mass will include two electrons. Since you end up subtracting the initial and final masses, the 2 electron masses in the hydrogen atomic masses are canceled out by the two electrons in the helium atomic mass. So, you "luck out".

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 ²H (1 electron in its atomic mass). So, now when you subtract the initial and final atomic masses, the electrons do not cancel. You'll need to take care of that.

I see. So there is an electron left over in the first reaction. Now if I calculate the mass deficit using the atomic masses I find

2(1.007825 u) - 2.014102 u - 2(0.0005486 u) = 0.0004508 u,

which corresponds to an energy release of 0.4199 MeV. Thank you for your help!

 

[TSny]

Nice going!