# Nuclear binding energy problem

## Main Question or Discussion Point

Hi, could someone please try and figure out where I'm going wrong here.
This is a calculation for a collision between 2 helium-3 nuclei which produces 1 helium-4 nucleus and 2 protons. The numbers are taken from wikipedia.
Mass before = 2(3.0160293 u) = 6.0320586 u
Mass after = 1 helium-4 nucleus + 2 protons = 4.002602 + 2(1.00727646677) = 6.017154932 u

6.0320586 - 6.017154932 = 0.014903668 u
1 u = 931.494 MeV
0.014903668 * 931.494 = 13.88267732 MeV, and yet wiki says it's just 12.9 MeV. Can anyone see the problem? I tried another calculation similar to this for another equation on wikipedia and got almost the exact answer that they had, but this is just a bit off.

## Answers and Replies

Related High Energy, Nuclear, Particle Physics News on Phys.org
jtbell
Mentor
The masses that you find in tables are atomic masses, not nuclear masses. They include the electrons that accompany the nucleus in an un-ionized atom.

The masses that you find in tables are atomic masses, not nuclear masses. They include the electrons that accompany the nucleus in an un-ionized atom.
Thanks for the help but I don't fully understand. What can be done to correctly modify my calculation?

Ahhhh I understand. The protons that are emitted are actually hydrogen nuclei, complete with electrons. It adds up now. Thanks.

jtbell
Mentor
Mass before = 2(3.0160293 u) = 6.0320586 u
This includes the four electrons that two helium-3 atoms contain.

Mass after = 1 helium-4 nucleus [actually atom] + 2 protons = 4.002602 + 2(1.00727646677) = 6.017154932 u
Your proton mass really is the "bare" proton mass (i.e. not the mass of a hydrogen-1 atom). Therefore this total includes only two electrons from the helium-4 atom.

Hi there,

Into a reaction like you mention above, you never take into account the electrons. You are talking about NUCLEAR reactions, which from the name of it implies only the bare nuclei. Therefore, like jtbell said, your calculations need to take into account the nuclear mass and not the atomic mass. Cheers