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Here is part of the energy level diagram of hydrogen:

n=4 --> -0.85eV

n=3 --> -1.50eV

n=2 --> -3.40eV

n=1 --> -13.6eV

When an electron of energy 12.1eV collides with this atom, photons of three different energies are emitted. Show on the diagram (with arrows) the transitions responsible for these three photons.

I tried solving this problem by working out the difference of energies that would give 12.1eV. I noticed that -13.6-(-1.5)=12.1. However I wasn't able to progress from here and would appreciate any pointers/solutions. (Note: This isn't homework, it's from a worksheet I found online.)

Thanks

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    $\begingroup$ I added the homework tag because as you can see from the definition of the tag that questions with homework tag need not be an actual homework assignment: „Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright. $\endgroup$ – Gonenc Mogol May 10 '15 at 21:06
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Photons can be emitted when electrons change energy levels. You say that you have worked out where a 12.1 eV difference is. In an ordinary hydrogen atom, the electron will be in the $n_1$ state.

Now, what energy state will the electron be in if an ordinary hydrogen atom absorbs 12.1 eV of energy?

After absorbing that energy, the electron can lose energy and give off a photon. It must go to a lower energy state. What states are lower? Once it gets back to $n_1$ it stops losing energy.

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  • $\begingroup$ "Now, what energy state will the electron be in if an ordinary hydrogen atom if it absorbs 12.1 eV of energy?" What is "the electron"? Sorry, I'm not sure I understand the mechanics of all this. What happens when the atom collides with the atom? It gains energy but then what, what happens to this energy? Thanks! $\endgroup$ – user45220 May 10 '15 at 21:29
  • $\begingroup$ I think I'm starting to get it. We see that the only differences that give $12.1$ are $13.6-1.5=12.1$ and $(13.6-3.4)+(3.4-1.5)=12.1$. Therefore the two possibilities are: (1) An electron falls from $n=3$ to $n=1$. OR (2) An electron falls from $n=3$ to $n=2$, then from $n=2$ to $n=1$. But again, I don't see why this happens. Can you please explain in detail what happens when the electron collides with the atom? What happens within the atom, why do electrons fall rather than get excited? $\endgroup$ – user45220 May 10 '15 at 22:07
  • $\begingroup$ The question asks about an independent electron colliding with a hydrogen atom. The atom itself has an electron bound to it. The energy of the independent electron is transferred to the bound electron, placing it in a higher, but still bound, energy state. Next the bound electron (forget about the independent electron, now) will lose energy, just like a marble dropping down some stairs. When the electron "falls" from a higher (less negative) energy level to a lower (more negative) level, it emits a photon. You have properly outlined the possibilities in the previous comment. $\endgroup$ – Bill N May 11 '15 at 15:56
  • $\begingroup$ The electrons are attracted toward the nucleus. The natural tendency of systems is to move toward a state of lower potential energy. That's what the electrons are doing, via the emission of a photon. The photon is emitted in order to conserve energy. $\endgroup$ – Bill N May 11 '15 at 15:58
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Now I assume that the question is asking the following:

When an electron of energy 12.1eV collides with this atom, photons of three different energies could be emitted. Show on the diagram (with arrows) the transitions responsible for these three photons.

Because from one single collision the emission of three photons doesn't make much sense to me. The use of plural in your question also points to my assumption.

Assuming that this is so you know that the electron bound to hydrogen can make the following transitions:

\begin{align} n=1 \to n=2\\ n=1 \to n=3\\ n=1 \to n=4\\ \end{align}

Calculating the energy differences $\Delta E$ for each case should solve your problem.

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  • $\begingroup$ Thanks. But why only $n=1\rightarrow n\in \{2,3,4\}$. Why not $n=2\rightarrow n=3$ for example? $\endgroup$ – user45220 May 10 '15 at 21:57
  • $\begingroup$ I assumed that the electron is in its ground state, not excited by physicist. $\endgroup$ – Gonenc Mogol May 10 '15 at 21:59

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