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Question:

What is the SI unit for the mass of subatomic particles?

Notes:

The weight of a proton is about $1.7 \cdot 10^{-24}\mathrm g$. This is the same as 0.0000000000000000000000017 grams. This is a very small number, so scientists have decided to create a unit so that it can easier calculate the mass of atoms. This is called the atomic mass unit or shortened as "amu".

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  • $\begingroup$ en.wikipedia.org/wiki/Dalton_(unit) $\endgroup$
    – anna v
    Commented Jun 14, 2021 at 18:55
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    $\begingroup$ Your own answer to the question is actually not correct, since you are asking about SI units. There is no such thing as an "SI unit for mass of subatomic particles", but only an SI unit for all masses: The kilogram. $\endgroup$
    – scaphys
    Commented Jun 14, 2021 at 19:04
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    $\begingroup$ Do you want to know the SI unit or the units that scientists typically use in these types of situations? There is no requirement that these questions have the same answer $\endgroup$ Commented Jun 14, 2021 at 19:07
  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – jng224
    Commented Jun 14, 2021 at 19:51

3 Answers 3

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The SI unit for mass is the kilogram.

Most organizations which insist on SI units will make an allowance for the fact that the kilogram is inconveniently large for discussing masses in atomic and subatomic systems. The SI brochure specifically mentions the dalton, which is defined as one-twelfth the mass of a neutral carbon-12 atom in its ground state. The dalton has previously been known as the "atomic mass unit" and abbreviated "amu" or "u."

It's common for working physicists to refer interchangeably to a particle's mass and its equivalent rest-energy, using the relation $E=mc^2$. In this picture the mass unit is the $\mathrm{eV}/c^2$, an electron-volt divided by the square of the speed of light. The electron-volt, like the dalton, is a non-SI unit that's approved for use with the SI. One dalton is a little under $10^9\,\mathrm{eV}/c^2$.

For unstable subatomic particles, the kilogram and the dalton are almost never used; the mass-equivalent energy units are closer to what's actually measured and therefore better documented and tabulated. If you search for the mass of the pion, you'll find $140\,\mathrm{MeV}/c^2$. If you search for "pion mass dalton" you find a talk by a former colleague of mine, whose name happens to be Dalton, in which he mentions the pion mass in energy units.

The non-SI (but SI-consistent) unit used to describe the masses of unstable subatomic particles is the $\mathrm{eV}/c^2$ and its power-of-$10^3$-prefixed multiples; the dalton is mostly used in chemistry.

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The SI unit for mass is kilograms. It doesn't matter what it's the mass of; the units of the mass of anything in SI is still kilograms. That's the point of a coherent system of units: for any physical quantity, there is a unique preferred unit in which to express it, and so long as all your inputs to an equation are expressed in these units, the output will also correctly be expressed in SI units.

The mass of a proton is (as you note) a very small number in these units. But that's OK! That's what scientific notation is for.

It can, of course, be useful to use a non-SI unit to express quantities whose scales differ widely from the human scale, or simply for historical reasons. The CIPM (the international body "in charge" of the SI) recognizes several such units that are particularly useful in this way, such as the astronomical unit, the degree, the litre, the electron-volt, and (most importantly for your purposes) the dalton, which is approximately 10-3 kilograms divided by Avogadro's number.1 Expressing the mass of a molecule in daltons can be useful to tell you its scale relative to other molecules and atoms, in a way that its mass expressed in kilograms makes more opaque. But if you want to (for example) calculate the rms speed of that molecule from the equation $\langle v_\text{rms} \rangle^2 = \frac{1}{3} k T/m$, you still need to convert that $m$ into kilograms.


1 If I understand correctly, it was exactly equal to 1 gram divided by Avogadro's number until the mass of the kilogram was redefined in 2019.

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For subatomic particles, mass is often given in energy units; the rest mass energy divided by $c^2$ gives the mass.

For example, the approximate mass of-
Electron: $m_{\text{Electron}}= 0.511 \mathrm{MeV}/c^2$
Proton: $m_{\text{Proton}}\approx 938 \mathrm{MeV}/c^2$

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