physical chemistry - What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state?

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?

Answer

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*:

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.

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* 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 $N$.