# Formulas for the number of spectral lines

While reading through Narendra Avasthi's Problems in Physical Chemistry, I came across two formulas on p. 64 (Scanned page):

• When electrons de-excite from higher energy level ($$n_2$$) to lower energy level ($$n_1$$) in atomic sample, then number of spectral line observed in the spectrum is

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

• When electron de-excites from higher energy level ($$n_2$$) to lower energy level ($$n_1$$) in isolated atom, then number of spectral line observed in the spectrum is

$$n_2-n_1$$

Can anyone explain how the process of emission differs in these two cases?

• the formula gives the total number of transitions provided the cascade of all possible transitions are added together, thus Balmer+ Paschen+Brackett+Pfund gives (6-2)(6-2+1)/2 =10 transitions. – porphyrin Feb 10 '19 at 9:04

To my understanding, it is simply single atom versus many number of atoms. For example, suppose one atom with an electron at energy level 7 ($$n_2=7$$). That electron can "de-excite" from $$n_2=7$$ to $$n_1=6, 5, 4, 3, 2,$$ or $$1$$. All those transitions give one spectral line for each. Thus, total of $$1 \times 6 = n_1(n_2-n_1)$$ (foot note 1) spectral lines would be present in the spectrum.
Similarly, when there were more than one atom in the sample, excited electrons ($$n_2$$) would be in different states $$(n_2=2, 3, 4, 5, 6,....,\infty)$$. For example, suppose we have aom population having electrons in all levels up to energy level 8 ($$n_2=8, 7, 6,...$$). Suppose those electrons "de-excite" to energy level 2 ($$n_1=2$$). Thus, electrons in $$n_2=8$$ can "de-excite" to energy levels $$7, 6, 5, 4, 3,$$ and $$2$$ meaning total of 6 spectral lines $$(8-2=n_2-n_1)$$. Some atoms with electrons in energy level $$n_2-1=7$$ can also "de-excite" to energy levels $$6, 5, 4, 3,$$ and $$2$$ meaning total of 5 spectral lines $$(7-2=n_2-1-n_1)$$, etc., etc. Thus, total numbers of spectral lines ($$s$$) in this case would be: \begin{align} s&=6+5+4+3+2+1=21=\frac{42}{2}=\frac{7\times6}{2}\\ &=\frac{(8-2+1)(8-2)}{2}\\ &=\frac{(n_2+1-n_1)(n_2-n_1)}{2} \end{align}
Foot note 1: Total number of spectral lines for single atom where $$n_2=7$$ should be: $$1 \times 6 = (n_2-n_1)$$ in the spectrum, not $$n_1(n_2-n_1)$$ as I originally suggested (Thanks @porphyrin for careful reading).
• How does your $1 \times 6 =n_1(n_2-n_1)$ work with $n_2=7; n_1 =4$? – porphyrin Feb 10 '19 at 9:07
• @porphyrin: It was just mental mistake, which I have corrected. There was no $n_1$ in the expression in the question, anyway. Now, to answer your question, total number of spectral lines for $n_2=7$ to $n_1=4$ transition in single atom would be $1 \times 3=7-4=(n_2-n_1)$, representing transitions of $7\rightarrow 6$, $7\rightarrow 5$, and $7\rightarrow 4$. Thank you for your careful reading. – Mathew Mahindaratne Feb 10 '19 at 19:22