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One of the basic calculations in nuclear theory is obtaining nuclear mass based on the liquid drop model. One uses Weizsäcker's formula to get the binding energy

$$E_B=a_VA-a_SA^{2/3}-a_C\frac{Z(Z-1)}{A^{1/3}}-a_A\frac{(Z-N)^2}{A}\pm\delta$$

and the mass follows by

$$M=Zm_p+Nm_n-\frac{E_B}{c^2}$$

I was reading about isospin, in particular the triplet $^{12}\text{B}, ^{12}\text{N}$ and $^{12}\text{C}$ with total isospin $t=1$. But it mentions that the $t = 1$ state in $^{12}\text{C}$ is not the ground state, but it is rather $15~\text{MeV}/c^2$ above the ground state.

Looking at the units with which the gap is expressed, they have to be mass. Does this mean that excited nuclear states have higher mass? I think it makes sense in light of Weizsäcker's formula because excited nucleons would have lower binding energy (they are "higher" inside the potential), but I am not sure.

The problem arises when I try to think in analogy to atomic states, because an electron can certainly be in excited atomic states but I have never heard of its mass increasing because of this.

Are nuclei different in this sense? Or were the units of the gap wrongly reported?


EDIT: After looking at the comment by @DJohnM it ocurred to me that maybe the gap for the atomic states was so small that the mass difference in the electron is negligible. So I calculated it.

Using the hydrogen atom model, the largest energy gap is the one between the $n=1$ and $n=2$ states. The energy of each state is given approximately by

$$-13.6 \frac{Z^2}{n^2}\text{eV}$$

for $^{12}\text{C}$ we have $Z=6$, so the gap is $13.6\cdot 36\left(-\frac{1}{4}+1\right)\text{eV}\approx 367~\text{eV}$. The electron mass on the other hand is approximately $511~\text{keV}/c^2$.

I also used Weizsäcker's formula to get the mass of the $^{12}\text{C}$ nucleus. It is $M_{^{12}\text{C}}\approx 11.2~\text{GeV}/c^2$. Now if we compare how big the gaps are relative to each of the particles's mass we get,

$$\frac{c^2M_{^{12}\text{C}}}{\Delta E_{^{12}\text{C}}}\approx\frac{11.2~\text{GeV}}{15~\text{MeV}}\approx745$$

(I used the fractions inverted because it seems clearer) and for the electron

$$\frac{c^2m_e}{\Delta E_e}\approx\frac{511~\text{keV}}{367~\text{eV}}\approx1392$$

This means that the gaps are on the same order of magnitude, so you couldn't argue that the case of the electron is negligible. Then, why is it never mentioned? How would you account for this variability in the Schrödinger equation (there is the factor of $1/2m$ with the momentum operator)?

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    $\begingroup$ This question vastly overcomplicates a relatively simple issue. The answer to the title is yes. This has nothing to do with isospin, the liquid drop model, etc. Other than that, you seem to be confusing yourself by imagining that the electron gains mass because it's in a higher energy state. What gains mass is the atom, not the electron. $\endgroup$
    – user4552
    Nov 8 '19 at 4:30
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    $\begingroup$ The relativistic mass is only useful as a concept in the rest mass system, where it is the same as the rest mass. Your confusion comes because you are using relativistic mass for QM bound particles.. The mass of the atom , even considering electrons and nucleus separately (which can only be a thought experiment in a quantum mechanical setting) comes from the four vectors of the particles, added, and calculating the invariant mass of the system .hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html $\endgroup$
    – anna v
    Nov 8 '19 at 5:36
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    $\begingroup$ S V, Pieter is talking about nuclear isomers, not chemical isomers. These are metastable excited nuclear states, some of them have very long half-lives, the most extreme being Tantalum-180m. $\endgroup$
    – PM 2Ring
    Nov 8 '19 at 5:40
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    $\begingroup$ Yes, atoms in excited states have greater mass. People don't mention it because the difference is very slight, so there are just about no situations where it's relevant. $\endgroup$
    – knzhou
    Nov 8 '19 at 5:51
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    $\begingroup$ @annav Nobody mentioned relativistic mass, apart from yourself. An excited atom (or excited nucleus) has a slightly higher rest mass due to its increased energy. $\endgroup$
    – PM 2Ring
    Nov 8 '19 at 5:59
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Yes, according to special relativity, if you excite any bound system, you increase its mass because of $E_0=mc²$. You can increase its mass by this mechanism until the two constituents are finally separate.

If you have two particles, their rest energy and therefore their mass is always biggest in the free case and lower in any bound state. If two free particles meet there needs to be dissipation which carries this energy away in order for a bound state to be formed.

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  • $\begingroup$ Thank you for your answer @Marc, but I have expanded my question somewhat. Do you know anything about the other questions I formulated? $\endgroup$
    – user137661
    Nov 8 '19 at 2:50
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    $\begingroup$ @SV: If someone's already answered the original form of your question, it would be better to accept this answer, then ask your new question as a separate question. $\endgroup$
    – user4552
    Nov 8 '19 at 4:04
  • $\begingroup$ @S V: I have only glanced at your calculations. One obvious error is that you compare the excitation energy to the electron mass. The analogon to the mass of the nucleus is the mass of the atom which is at least 3 orders of magnitudes bigger than the electron mass. $\endgroup$
    – Marc
    Nov 14 '19 at 18:29
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Yes. In the case of nuclear states, the mass difference can be measured with remarkable precision by high resolution mass spectroscopy. I found this paper by Babcock et al (2018) about different isomers of some isotopes of indium. There is good agreement with the energy differences inferred from gamma spectroscopy.

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