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When an electron in a higher energy state jumps to a lower energy state in Bohr’s model of hydrogen atom, it emits a photon which has energy equal to the difference in the energy between the final and initial energy states. $$h\nu = E_f - E_i$$ where $\nu$ = frequency of photon emitted

Now what I know is each photon emitted corresponds to an emission line in the spectrum.

So, an electron can make the make the transition from a higher energy state to any lower energy state provided it emits a photon given by the above equation. In the question given, if an electron is in higher energy state; the maximum number of spectral lines is when the electron jumps to the next lowest energy state and so on as then photons will be emitted during each transition.

If an electron moves from $n = 6$ to $5$ we get a spectral line. Then from $n = 5$ to $4$ we get one and so on giving us a maximum of $5$ spectral lines. Now if the electron made the transition directly from $n = 6$ to $n = 4$ or any other lower energy state we would get less than $5$ spectral lines.

However, on looking up answer to the question it was given $15$. How can it produce $15$ spectral lines? When an electron from $n = 6$ jumps it has $5$ options and example it makes the jump to $n = 3$ then it has only two options $n = 2$ or $n = 1$. It can’t go back to $n = 6$ unless energy is supplied and it absorbs energy. So how can more than 5 lines be observed in the emission spectrum?

I am missing a key concept in such problems. Can someone tell me what I am missing?

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    $\begingroup$ The question asks about all possible transitions. Whenever you look into such deexcitation of electrons, obviously we are not concerned about one single electron, but an ensemble of electrons which are present in Hydrogen gas. Thus the total transition from $i^{th}$ state is $i-1$. So, total no. of transitions is $\sum_{i=1}^{6} (i-1) = 15$. $\endgroup$ – Soumik Das Mar 17 at 9:05
  • $\begingroup$ Also see Number of spectral lines. If you expand the sum, you end up with the formula $$N = \frac{n(n-1)}{2} = \frac{6(6-1)}{2} = 15$$ $\endgroup$ – andselisk Mar 17 at 9:15
  • $\begingroup$ @SoumikDas The question asks for an excited electron in a H atom as I have mentioned in the title. Then shouldn’t the answer be 5? $\endgroup$ – Hola Mar 17 at 9:50
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    $\begingroup$ Please note that these answers are only true in the Bohr model, a better quantum model shows that there will be $n^2$ transitions. The orbitals are now labelled with increasing orbital angular momentum, $\ell=0,1,2,3 ...$ as s, p, d, f, g, h.. (e.g 1s, 2s,2p, 3s,3p,3d ...with restriction $\ell\lt n$) and selection rules to govern what transitions are allowed. No doubt you will learn about this soon. $\endgroup$ – porphyrin Mar 17 at 12:10
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Since comments caused certain level of confusion, I guess I'll try to provide a further illustration. You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$. Electron doesn't get stuck forever on any of the levels with $n > 1$.

Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms. In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$. The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this.

I put together a rough drawing in Inkscape to illustrate all possible transitions*:

enter image description here

I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots). To determine $N$, you need to sum the states, as Soumik Das rightfully commented:

$$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$

For $n = 6$:

$$N = \frac{6(6-1)}{2} = 15$$

Obviously the same result is obtained by taking the sum directly.


* Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$.

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  • $\begingroup$ I’m still not able to understand for a single H atom containing 1 electron which is at n=6 as per question, maximum emission lines corresponds to 6 to 5 to 4 so on.. 1, right? Obviously the electron won’t get stuck there hence it jumps to lower energy state. For eg: the electron in n=6 makes jump to n=5, now it can’t go back up to n=6 right (unless you provide energy)?So it has to make jump from n=5 to lower state and hence maximum spectral lines will correspond to 5->4->3->2->1. Can you clarify please? $\endgroup$ – Hola Mar 17 at 13:48
  • $\begingroup$ If it made jump from n=6 to n=3, now it can go to n=2 or to n=1 hence now the number of spectral lines will be less. It can’t go back to n=4 or n=5 or to n=6 from n=4, so why are we including all such lines? Once it makes a jump, it can’t go back it has to move to a lower energy state and so only way to maximise number of lines is maximise number of jumps and hence 6> 5-> 4-> 3> 2-> 1 giving 5 jumps and a line per jump giving 5 emission lines. Please note I am speaking about a Hydrogen atom containing 1 electron as given in the question. $\endgroup$ – Hola Mar 17 at 13:52
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    $\begingroup$ If you artificially limit yourself to a single atom, then yes, maximum number of transitions is 5, but I'm not aware of such experiment for the emission from a single excited hydrogen atom. A resonance fluorescence spectrum from a trapped heavier atom/ion — maybe; but it isn't what the question from your textbook is about. Again, classically you are supposed to review a large number of excited atoms and find the number of the spectral lines by judging all possible transitions en masse. $\endgroup$ – andselisk Mar 17 at 13:59
  • $\begingroup$ Yes I suppose you’re right. The question asks for lines emitted when considering many such H atoms yielding maximum of 15 lines, however upon reading I assumed it asks for one atom and it made me wonder how 15 such lines can be emitted. $\endgroup$ – Hola Mar 17 at 14:06
  • $\begingroup$ Yep, this is pretty common among textbooks' levels of abstraction. Author first talks about a single instance for simplicity, but forgets to switch back to the real world with the superposition of such instances. $\endgroup$ – andselisk Mar 17 at 14:12

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