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 $N$.
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
A total number of 15 lines (5 + 4 + 3 + 2 + 1) will be obtained in this hydrogen emission spectrum. The total number of spectral lines emitted when an electron drops to the ground state from the ‘nth‘ level may be computed using the following expression: $\frac{n(n-1)}{2}$ Since n = 6, total no. spectral lines =6(6-1)/{2}=15 Answer Verified
Open in App Suggest Corrections 1 |