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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?

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    $\begingroup$ 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. $\endgroup$
    – porphyrin
    Feb 10, 2019 at 9:04

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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).

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  • $\begingroup$ How does your $1 \times 6 =n_1(n_2-n_1)$ work with $n_2=7; n_1 =4$? $\endgroup$
    – porphyrin
    Feb 10, 2019 at 9:07
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    $\begingroup$ @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. $\endgroup$
    – Mathew Mahindaratne
    Feb 10, 2019 at 19:22
  • $\begingroup$ Mathew Mahindaratne: I am unable to understand in your answer how did you write s = 21 = 42/2. What was the basis for that? $\endgroup$
    – Somdeb
    Oct 27, 2020 at 21:01
  • $\begingroup$ @Somdeb: I have it as an example. Read the question and my answer completely to understand my concept. $\endgroup$
    – Mathew Mahindaratne
    Oct 28, 2020 at 4:38

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