Recently, I found that when electrons in an atomic sample de-excite from a higher energy level ($n_2$) to a lower energy level ($n_1$), the number of spectral lines observed in the spectrum is

$$\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$$

Can anyone please tell me how to derive this?


1 Answer 1


When you move from level $n_1$ to level $n_2$, the total number of energy levels are $n_2-n_1+1$ (including $n_1$ and $n_2$).

Now, you want to find the total number of possible spectral lines in the atomic sample.

Note that if you chose any two energy levels (say $n_i$ and $n_j$), you will get a unique spectral line corresponding to those energy levels.

So, the total number of spectral lines possible are $$\binom{n_2-n_1+1}{2}=\dfrac{(n_2-n_1+1)(n_2-n_1)}{2}$$

  • 5
    $\begingroup$ Great answer and is the number of combinations of $n_2-n_1+1$ objects taken 2 at a time. Just for completeness, the brackets mean $\displaystyle \binom{a}{b}=\frac{a!}{b!(a-b)!}$ $\endgroup$
    – porphyrin
    Jul 4, 2021 at 8:09
  • $\begingroup$ This expression is also equivalent to $\Sigma(n_2-n_1)=\Sigma{\Delta n}=n+(n-1)+(n-2)+...+1$ $\endgroup$
    – Apurvium
    Jul 5, 2021 at 5:12
  • $\begingroup$ @Apurvium That’s because that is another way of showing that through rearranging the possible terms and doing summation! :) $\endgroup$
    – Rishi
    Jul 5, 2021 at 5:19

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