I'm so confused in the use of nuclear masses and atomic masses. I have two questions.

From the book "Outline of Modern Physics" by Ronald, I understand that the semiempirical mass formula (Weizsäcker's formula) is

$$M=Zm_{p}+(A-Z)m_{n}-b_{1}A+b_{2}A^{2/3}+b_{3}Z^{2}A^{-1/3}+b_{4}(A-2Z)^{2}A^{-1}+b_{5}A^{-3/4} \qquad (1)$$

and this formula is for the dependence of the mass of a nucleus on $A$ (mass number) and $Z$ (atomic number), i.e. is the formula that gives you (approximately) the mass of a nucleus.

1st question $M$ is the nuclear mass or the atomic mass? (I understand that the difference between these two masses are the masses of the electrons. Obviously it's small, but is nonzero.)

Then, he defines the average binding energy per nucleon like

$$BE=[Zm_{p}+(A-Z)m_{n}-M]c^{2}/A \qquad (2)$$

So far so good.

Then, the book tries to do an example, and calculate the binding energy per nucleon for $_{42}^{98}Mo$ and writes

$$BE=[Zm_{p}+(A-Z)m_{n}-M_{nuc}]c^{2}/A \qquad (3)$$

and says: "where the atomic masses are used for $m_p$ and $M_{nuc}$ (so that the electron masses cancel)." And insert the numbers.

$$BE=[(42(1.007825u)+56(1.008665u)-97.905409u)/98] \times 931.5\frac{MeV}{u} \qquad (4)$$

But, according to Wikipedia the isotopic mass of $_{42}^{98}Mo$ (atomic mass) is 97.9054082 u, and the mass of the proton is 1.007276466812 u. So my

2nd question is: why does the book uses the value 1.007825u for the mass of a proton instead of the value 1.007276466812 u, and why does it uses the atomic mass instead of the nuclear mass in $M_{nuc}$?

Note The book uses the value 1.007825u for the mass of a proton and the atomic mass instead of the nuclear mass in $M$, in other examples. Why?


1 Answer 1


The Weizsäcker formula and all similar formulae in nuclear physics are formulae for the masses of the nuclei, not atoms. That's true by design: the models behind the individual terms (droplet, shells etc.) are models for the nucleus only.

At the same moment, the accuracy of similar semiempirical formulae is not that marvelous which means that the errors are comparable to the mass of the electrons near $0.5\,{\rm MeV}$. So if you interpret the Weizsäcker formula as a formula for the whole atomic mass, you will increase the errors just a little bit.

Now, when we care about the full accuracy, there are several amu units. See e.g.


where the second page is on the history. The chemical amu i.e. 1.00732 amu per proton is deprecated, and similarly is the related "dalton", and your "physical" amu, $m_p=1.0076$ based on $1/12$ of the carbon-12 atom (including electrons in this sentence, but not in other sentences), is very much alive.

The figure 1.007825 amu is the mass of the hydrogen atom (proton plus electron; the electronic excitations are negligible). Note that $1.007825\sim 1.007276(1+1/1836)$. So this figure is useful exactly to evaluate the mass of the whole atom – an atom is composed of neutrons and the "proton plus electron" combinations because the number of protons agrees with the number of electrons.

With a great accuracy, the mass of the atom may be calculated simply by adding the rest masses of the electrons i.e. by replacing $m_p$ by $m_p+m_e$ in the nuclear formulae. That's why the formulae with $1.007825$ are omnipresent. The error introduced by this approximation is comparable just to the atomic binding energies which are multiples of 1 electronvolt, about 1 million times smaller than the other terms we care about (and 1 billion times lower than the proton rest mass), and are neglected pretty much in any discussion about nuclear physics.

  • $\begingroup$ So, it's the same to use the value $m_p = 1.007825 u$ and atomic mass in $M$ that using the value $m_p = 1.007276 u$ with the nuclear mass in $M$, am I right? $\endgroup$
    – Ana S. H.
    Commented May 29, 2013 at 11:45
  • 1
    $\begingroup$ Right, if I understand your grammar here. $\endgroup$ Commented May 29, 2013 at 15:59

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