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I'm suppose to write out reactions where atoms send out alpha radiation and decay. The book uses the 4-2 H, 4 as nucleon number and 2 as proton number, but isn't that wrong? The mass of helium is greater than the alpha particle due to two electrons? Shouldn't they use a different notation for the alpha-particle other than that for helium?

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Hint: The book likely doesn't care about the (relative) error it is making. Why? – Qmechanic May 20 '12 at 11:19
I get that it's a very small error to make. I thought accuracy was a good thing :( – Algific May 20 '12 at 11:47
I think the notation you wrote in your question is not what you intended. Apart from that, the difference is small and an alpha particle is a fully ionized helium atom so using the helium notation is not out of the question. In my classwork, I seem to remember people simply using a lowercase alpha as the symbol. – AdamRedwine May 20 '12 at 15:31

Alpha particles are essentially the same as an ionized Helium atom. The mass of the electron pair can be neglected because the mass of an electron is significantly smaller than that of the protons and the neutrons in the nucleus, so don't have much of an effect for simple purposes. See The Proton to Electron Mass Ratio which shows the scale of the difference.

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I do believe the binding energy of the electrons in helium is greater than the rest mass of two electrons. Either way, you are correct that it makes little difference. – AdamRedwine May 20 '12 at 15:28
@AdamRedwine: the binding energy is of O(13.6 eV), while the rest mass of an electron is .511 MeV/c^{2}. So, the mass of the electrons swamps the binding energy of the electrons. – Jerry Schirmer Jul 31 '14 at 18:38

The mass of a neutral helium atom is 4.002 602 amu.

That mass includes two electrons, each with mass 0.000 5485 amu = 0.511 MeV/$c^2$.

You can ionize helium with ultraviolet light, so the electron binding energy is a few eV. We can neglect the electron binding energy compared to the electron mass.

So the mass of an alpha particle is closer to 4.001 505 amu. This is actually a pretty big difference if you're interested in calculating the energy of an MeV-scale interaction!

However, there's another subtlety that you have to keep in mind. If you have a neutral alpha-emitting nucleus undergo a decay, like $$\mathrm{ ^{238}_{92}U \to {}^{234}_{90}Th + {}^4_2\alpha }$$ don't forget that the initial state on the left already has 92 electrons. In the sudden approximation, the $\alpha$ does come out without any electrons, but the $\mathrm{_{90}Th}$ nucleus is born with all 92 of its parent's electrons, with charge –2. If you want to correct for the mass difference between a bare $\alpha$ and a neutral helium, you must also correct for the mass difference between a neutral thorium and ionic $\mathrm{Th^{2-}}$. Since most neutral atoms have negligible eV-scale binding energies, it's simpler just to use the neutral atom masses everywhere.

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