I've run into a bit of a problem on this weeks coursework.

A proton and an electron initially at rest combine to form hydrogen. Find the wavelength of the emitted photon?

So, as far as I can see, there are two ways to do this: Find the mass of an electron + a proton, find the mass of hydrogen, work out Δm and use E=mc^2 and work out wavelength from E

Secondly, I'd assume, since I thought ionised hydrogen is the same as a proton, just use E=13.6 eV then work out wavelength from that.

The problem is they give very different answers.

Now all my friends are telling me to go with the first, so I reckon that's probably the right one. But why? Is a proton different to a H+ ion?

  • 2
    $\begingroup$ Your two approaches are exactly the same. 13.6 eV is the binding energy of hydrogen. What binding energy means is the difference between the mass of the bound state and the mass of the constituent parts. $\endgroup$
    – Matt Reece
    Nov 23, 2014 at 21:29
  • 2
    $\begingroup$ In terms of why these 2 approaches are giving you different answers, you are looking at a difference in energy $\sim 10eV$. the mass of a proton $\sim 10^9 eV$. In order to get back that energy difference of $13.6eV$ to 3sf you will need the mass of the proton and the hydrogen atom to 10sf, and a similar level of precision with any unit conversion factors. I'm betting that is where your error occurred. Going back to $E=Mc^2$ is not the best approach to this calculation. $\endgroup$ Nov 23, 2014 at 21:56
  • 1
    $\begingroup$ Sorry, I should add here that the answer is two orders of magnitude out - 2800 eV compared to 13.6. I used 9 sig figs: (9.10938291*10^-31)+(1.67262178*10^-27)-1.67352766*10^-27 = 5.05*10^-33 = Δm (5.05*10^-33)*(3*10^8)^2 = 4.55*10^-16 joules, or 2800 MeV $\endgroup$
    – Jack Maggs
    Nov 23, 2014 at 23:31

1 Answer 1


The answers in the comments are both correct - a couple of points to add.

First think about the reverse process - how would you work out the energy required to break up the proton and electron in hydrogen to form a proton and electron. This is the same as the energy that needs to be released on formation of a hydrogen atom. I think you have figured this out from your question, but the point in the comment about it being easier to work things out this way is very good - the change in mass between H and a proton plus and electron is very small and this is not such a good way to work things out.

Secondly, the question is not particularly well put. In reality the hydrogen atom could be formed with the electron in any level; n=1, 2, 3 etc. and the wavelength of the photon would be different in each case. The answers described in the comments and the methods you suggest both assume that you are forming a hydrogen atom in the ground state.

Finally $H^+$ is equivalent to a proton - unless you are thinking of a hydrogen nucleus of mass 2 with one proton and one neutron, but we normally write that as $D^+$.

  • $\begingroup$ From what is said in these comments, may I infer that an excited hydrogen atom is HEAVIER than a ground-state atom? $\endgroup$
    – Sofia
    Nov 23, 2014 at 23:02
  • $\begingroup$ @Sofia - yes this is true! It would be nice to be able to build a mass spectrometer that could tell the difference between say $He^+$ and $He^{+*}$ - ground state Helium ion and excited Helium ion, but I don't think we have enough mass resolution for that at the moment. More about this topic in answer to question at physics.stackexchange.com/questions/145972/… $\endgroup$
    – tom
    Nov 23, 2014 at 23:09

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