# What is the maximum number of spectral lines emitted by a hydrogen atom when it is in the fourth excited state?

 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. * 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 $$N$$.