# Why do NNLC and NIST appear to give different values for the mass energy of the deuteron?

There is a problem with data that I've obtained over the internet. Here are the two sources of information from which I'm retrieving my data. NNLC and NIST

On NIST, I have read that the mass excess of a deuteron is 13135.72158 keV, the mass of a neutron is 939.565379 MeV, and the mass of a proton is 938.272046 MeV. If you use the simple formula for the total mass of a nuclide,

$$M = M_{Z-A}+M_A-\Delta M,$$

you will arrive at the total mass of the deuteron, which is calculated to be 1890.97 MeV. However, this differs from the value given in NIST, which says that the mass is 1875.612859 MeV.

Why does this difference exist?

• Are you missing some information in the question? What's the binding energy? Nov 25, 2014 at 16:54
• What do you mean? I thought the mass excess took into account the mass defect introduced by the binding energy. Nov 25, 2014 at 18:34
• I think you have a confusion around what 'binding energy' and 'mass excess' mean. The wikipedia entry on Deuterium also links to explanations of those two, which will clear it up. The mass excess relates to the binding energy of the nucleus with respect to the binding energy of C12, since C12 defines the atomic mass unit. Nov 25, 2014 at 19:32
• @JonCuster a slightly expanded version of that comment would probably make a very good answer. Nov 27, 2014 at 5:31
• How would you calculate the mass of deuteron then? Would you do 2 amu (2 nucleons in deuteron) + mass excess (in amu)? Also, how would you calculate the mass excess for a nuclide if you only know the mass excess for its respective atom? Nov 27, 2014 at 22:11

## 1 Answer

I think there can be confusion around what 'binding energy' and 'mass excess' mean. The Wikipedia entry on Deuterium has links to explanation them, which may clear it up if the following doesn't.

IF you could start with isolated protons and neutrons and assemble your own nucleus, the mass (energy) balance of the result would be the sum of the isolated masses (energies) minus the binding energy. A priori without nuclear physics you wouldn't know what the binding energy would be, but you could (and can) measure the difference.

Now, coming the other way, one might look at the standard for atomic mass units, Carbon-12. It has 6 protons and 6 neutrons, so one might think, hey, a deuterium is one proton and one neutron, so the mass of D should be 1/6'th that of C-12, and it will have a mass of 2.00000 amu. That of course ignores the variations in binding energies for different nuclei. So, the nuclear physics folk make it explicit in the tables you reference - no, there is 'mass excess' for a nuclei which tells you how well, relatively, a nucleus is bound compared to C-12. As you scan down the long list you will find nuclei that are bound less well, so have a positive mass excess (such as D) and those that are bound more tightly (say Fe-56 at -60,000 keV). These numbers then also can be used to determine whether colliding two nuclei will result in a stable, larger nucleus (i.e. the Q-value).

So, from the tables, you can calculate the mass of a nucleus from adding up the isolated proton and neutron masses, then accounting for the binding energy to get the mass in MeV. Or I suppose one could take the ratio to C-12, and then correct using the mass excess to get the right mass in amu.

The mass excess relates to the binding energy of the nucleus with respect to the binding energy of C12, since C12 defines the atomic mass unit.