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I got taught that if an electron de-excites from any $n_2 $(where $n_2 >3$) to $n=2$, the energy change can be represented by $$\Delta E=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\ \mathrm{eV}$$ What if an electron excites from $n=2$ to $n_2 $(where $n_2 >3$)?

Do we have to reverse the signs in the brackets?

Also, if it excites like I said, will it be absorbing visible light?(Because in de-excitation, it emits visible light)

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    $\begingroup$ The principal quantum numbers for the energy levels go as n = 1, 2, 3, 4 .. ∞. The lowest level is $n$=1. If one goes from a higher level to a lower one a photon is emitted (as fluorescence) and the atom has lost this amount of energy, say $n$=2 to $n$=1. Absorption is the opposite $n $=1 to $n$=2 . As ΔE is positive for absorption it must be negative for emission. Draw out a diagram of energy levels and it becomes clearer. The equation for ΔE is the difference in energy of the two levels, starting at the lower level for absorption and upper for emission. $\endgroup$ – porphyrin Jul 6 '16 at 16:07
  • $\begingroup$ @porphyrin thank you. Put it as an answer, ill accept it. $\endgroup$ – Siddharth Venu Jul 7 '16 at 12:08
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The excitation of electrons are caused by photons while their de-excitation is causes the emission of photons of the same energy as the energy differemce between two states. The general expression for energy of n-th excited state of H-atom is given by $$E_n=-\frac{13.6 eV}{n^2}$$

Hence by the conservation of energy, the energy released/absorbed is given by the formula you have written. $$h\nu=\Delta E=E_{n_2}-E_{n_1}=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) eV$$

Hence your initial state is always $n_1$ and final state is always $n_2$. The energy will be positive if photon is absorbed and negative if it is released.

So, for specific values of $n_1$ and $n_2$ ($n_1<n_2$), if the photon released while going from $n_2$ to $n_1$ is in visible range, the photon absorbed for corresponding excitation will have same frequency, and hence it has to be in visible range.

Please note that there are selection rules for the transitions, determined by the
transition moment integral. Please go through the book in refernece for details.

References:

  1. D.J. Griffiths, Introduction to Quantum Mechanics , Pearson, 2 ed, 2004. ISBN: 9332542899
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  • $\begingroup$ Hi @Mitradip. It is better to make units upright; as for the present context, you could , instead of using $eV,$ use $\rm eV\;.$ $\endgroup$ – user5764 Jul 19 '16 at 7:13
  • $\begingroup$ Good answer. I know you told us to look in Griffiths, but briefly, what are the selection rules that you speak of? Because as far as I know there are no restrictions on either n_1 or n_2. (Except that they are both positive integers, but that is not a selection rule.) $\endgroup$ – orthocresol Jul 19 '16 at 7:44
  • $\begingroup$ @orthocresol, for hydrogen atom there is no such restriction, but for other system models (like harmonic oscillator) there are. $\endgroup$ – Mitradip Das Jul 26 '16 at 18:55

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