Understanding Mass Excess and Atomic Mass in Nuclear Physics

[artis]

I see this term comes up in KAERI table and elsewhere.
Now let me guess here and tell me whether I'm correct.
The only element whose atomic mass expressed in AMU is exactly a round number is C12, for all other elements it's either bit more or less so in order not to have to write long numbers in the tables they approximate the element number to a round number while the actual atomic mass is a rather long decimal number (numbers after the decimal point according to how precise one wants to be) , so for all other elements they have their element number which is a rounded up number of the real atomic mass and the difference being expressed as the "mass excess" or "excess mass" ?
Is this so?

 

[vanhees71]

Since 2019 there's a subtle change. The atomic mass unit (or Dalton) is still the same as before (it's a non-SI unit but accepted for use within the SI), i.e., it's 1/12 of the mass of a carbon atom (with the nucleus being the isotope $^{12}\text{C}$), but since the mol as well as the kg has been redefined the molar-mass constant has changed (but only by a tiny amount of course):

$$M_{\text{u}}=0.99999999965(30) \cdot 10^{-3} \;\text{kg} \; \text{mol}^{−1}.$$

https://en.wikipedia.org/wiki/Dalton_(unit)
https://en.wikipedia.org/wiki/Molar_mass_constant

 

[artis]

Can you or anyone please take a look at this video for example, where in about 40:00 time he starts to calculate the Binding energy. So the masses of protons + neutrons are added and then he subtracts the mass excess as well as the element number, but why use the mass excess , why can't one just add the masses of protons + neutrons and subtract the precise atomic decimal mass number and not mess with the mass excess number at all?

 

[vanhees71]

Ok, that's about the binding energies. The nucleus is a bound state of nucleons (protons and neutrons). Just think about bringing the protons and neutrons from infinity together making up the nucleus you get a total energy of the nucleus being the sum of the masses of protons and neutrons (times $c^2$) but due to the attractive interaction you get some energy out, and that's the socalled mass defect, i.e., you get
$$m_{\text{Nucleus}}=Z m_{\text{p}} + (A-Z) m_{\text{n}}-E_{\text{B}}/c^2,$$
where I made the binding energy $E_{\text{B}}>0$ so that I need to subtract it. That's of course because of Einstein's famous mass-energy equivalence: The mass of a composite system is the energy of this system in the reference frame where it is at rest as a whole (more accurately in the center-momentum frame, i.e., the reference frame where the total momentum of the system vanishes) divided by $c^2$.

One should carefully distinguish atomic mass, which is about the mass of neutral atoms, including the electrons around them (if you want to be very accurate you have a mass defect from the binding energy of the electrons around the atomic nucleus too).

I only watched very briefly into the movie. Note that he got the units a bit messed up in the beginning. The conversion between $\text{u}=\text{amu}=\text{Da}$ and (mega) electron volts is of course
$$1 \text{u}=931.494 102 42(28) \text{MeV}/c^2 .$$
A mass unit must be always an energy unit divided by nd not times a velocity squared.

That's why theoretical physicists set $c=1$ ;-))

 

[artis]

ok @vanhees71 , I suppose you always subtract the binding energy since I don't know of any atoms that would have a negative binding energy whatever even that would mean.?

By the way I sort of feel my original question wasn't answered, where I asked whether the mass excess is just a way to show the difference between the rounded element number and the real element mass?And maybe while you or anyone else are at it , can I please ask to see the fragment at the end of the video where he talks about a fission reaction of Boron 10, used in the BNCT (boron neutron capture therapy).
So a epithermal neutron created from a proton hitting beryllium nucleus strikes boron10 (in the cancer cell) creating an alpha (which goes on to damage the cancer cell) and Li7 + a gamma ray.
This reaction seems to be exothermic aka one that releases additional energy instead of consuming energy.
But can it really be considered an exothermic one? Because the neutrons needed to start the reaction had to be created from a proton accelerator (cyclotron) striking beryllium which in the process probably consumes much more energy than the final fission reaction liberates.

In general I get that fission can happen to both heavy elements/isotopes as well as to light elements/isotopes but the difference is that for some heavy elements it can be self sustained while for light elements it always has to be "pushed" by a net input energy to create the conditions necessary for it to happen , even though the reaction itself is exothermic like the neutron boron10 reaction?

 

[vanhees71]

I don't know, what you mean by "mass excess". There is a mass defect in nuclear physics (and if you are very precise also in atomic physics) which means that the bound nucleus's mass is the sum of the masses of the nucleons minus the binding energy$/c^2$.

For scattering processes all you need is the conservation of relativistic energy and momentum and the "mass-shell" conditions for the involved particles/nuclei.

 

[artis]

I know the mass defect @vanhees71 but please see the video I posted in post #3 he shows the KAERI table on the screen there many times, for every element that table gives all kinds of parameters one of which is called mass excess. I think I have a idea what it means I just wanted to be sure.

 

[vanhees71]

I don't want to watch through the entire video. The guy only demonstrates pretty nicely that it's a bad idea to mix units and then not converting them properly into each other when using a pocket calculator. What he does is to calculate the binding energies from given binding energies as measured in terms of binding energy of $^{12}\text{C}$. From this you can get the binding energies of each nuclei. It's easier to directly look at the binding energies (most conveniently in units of $\text{MeV}/c^2$) as it's done in nuclear physics.