I have 2 questions regarding the solution to the following question:

Alpha decay in SEMF

The solution states that:

Alpha decay solution

Mathematically, I understand how the end result

$$Q=BE\left({}^{A-4}_{Z-2}\mathrm{Y}\right)+BE\left({}^{4}_{2}\mathrm{He}\right)-BE\left({}^{A}_{Z}\mathrm{X}\right)$$ was obtained.

This is because from the definition of nuclear mass the second equality is splitting each of the 3 species into 3 terms, where the first is $$m_N\left({}^{A}_{Z}\mathrm{X}\right)=Zm_p +(A-Z)m_N - BE\left({}^{A}_{Z}\mathrm{X}\right)$$ and the similarly for the 2 other species by virtue of the formula for nuclear mass given below:

Nuclear mass

My 2 questions are:

  1. $Q$ has dimensions of energy, so why are the masses not being multiplied by $c^2$, ie. $$Q=m_N\left({}^{A}_{Z}\mathrm{X}\right)c^2-m_N\left({}^{A-4}_{Z-2}\mathrm{Y}\right)c^2-m_N\left({}^{4}_{2}\mathrm{He}\right)c^2\,?$$
  2. In the algebra that follows $m_N$ is being used, but this is the mass of the nucleus, should it not be $m_n$ (mass of neutron)? I don't see how $m_N$ can possibly be justified here since the proton mass has already been accounted for with $m_p$. Is this a typo, or something I am not understanding correctly?

I have read this similar question and I own a copy of the Williams textbook but neither answer my question here.

  • $\begingroup$ From the book blurb, $m_{N}$ is the ‘nuclear mass’, which could well be taken as energy units already. $\endgroup$
    – Jon Custer
    Jul 20, 2019 at 22:49
  • $\begingroup$ @JonCuster Thanks for your reply, I don't understand what you mean, $m_N$ is mass and hence is measured in kg. We don't use natural units in this course so $c \ne 1$. All I would like to know is if there is one person out there that agrees that in the formula for the $Q$-value it should be $m_n$ and NOT $m_N$? $\endgroup$
    – BLAZE
    Jul 21, 2019 at 11:12
  • 1
    $\begingroup$ Note to potential editors of the question: since part of the question is about a possible typo in the text, be cautious if you decide to convert the screenshots to MathJax. $\endgroup$
    – rob
    Jul 21, 2019 at 14:16

1 Answer 1


Your first question is about dimensional analysis: you say that your text explicitly avoids "natural" units, so an equation like $Q=\cdots + m + \cdots$ is dimensionally incorrect.

You're technically correct here, but it's an opportunity for you to realize something important about "natural" units: they're so natural that physicists use them without thinking about it. As a nuclear physicist, I have zero need to ever remember the proton mass in kilograms. (Without looking it up, I'm pretty sure that the scale is $10^{-27}\rm\,kg$, but I don't even remember the first significant digit.) If I want to think in mass units, I remember that the proton mass is a hair over 1 u; if I wanted to convert to kilograms, I remember the Avogadro constant, and I can calculate. But in general, if you interrupt a nuclear (or particle) physicist who is doing something else and say "what's the proton mass?", she'll say "about a GeV, why?" The equivalence between energy and mass is that deeply ingrained in people, and for good reason.

As you continue in physics, well-intentioned authors who sometimes omit factors of $c^2$ will be the the most common error you discover. And as you get more comfortable with the language of physics, you'll find that you read such errors as typos, which don't interfere with your understanding any more than the doubled "the" in the previous sentence. But as a new learner, it's more of a burden for you. Sorry.

Your second question is another subtle typo question. Your text's notation appears to be

\begin{align} m_N(\text{nuclide}) &= \text{mass of nuclide} \\ m_p &= \text{mass of proton} \\ m_n &= \text{mass of neutron} \end{align}

but then defines the mass of a nuclide as having a term with $m_N$ multiplied by the neutron number. Your written-in-words understanding is correct that this notation is inconsistent. The mass of a nucleus is the proton mass multiplied by the number of protons, plus the neutron mass multiplied by the number of neutrons, minus the (mass equivalent of) the binding energy. (Though in practice it's the binding energy which is defined in relation to the experimentally-measurable masses.)

Good eye. Keep it up.

  • $\begingroup$ Excellent answer, thanks for taking the time to explain that to me. $\endgroup$
    – BLAZE
    Jul 21, 2019 at 18:08

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