The wavelengths for emissions in the visible region for Group 1 elements is as follows:

$$ \begin{array}{lc} \hline \text{Element} & λ/\pu{nm} \\ \hline \ce{Li} & 670.8 \\ \ce{Na} & 589.2 \\ \ce{K} & 766.5 \\ \ce{Rb} & 780.0 \\ \ce{Cs} & 455.5 \\ \hline \end{array} $$

Now, considering that the ionisation enthalpies (I.E.s) decrease going down the group, it should be relatively easier to remove an electron, which should reflect in the energy band gaps and ultimately in the wavelengths. Thus, there should be increase in this trend. However, $\ce{Na}$ is an exception. Why so?

I think the $\mathrm{d}$-electrons play a role in the increased band gap for $\ce{Cs}$, which shows in the trend. But I am missing something for $\ce{Na}$. Is its I.E. decreases because size outweighs nuclear charge along with fair amount of screening? But $\ce{Na}$ has a unique balance wherein size and screening are not enough to tip over.

  • 1
    $\begingroup$ It hardly has to do with ionization energy. It's a matter or specific transitions, or more precisely spectral lines associated with them. Find which lines are these^ and then you can get into comparing. $\endgroup$
    – Mithoron
    Commented Oct 27, 2019 at 0:32

1 Answer 1


On the NIST website you can look up the transitions corresponding to the wavelengths you list:

$$ \begin{array}{l c c c c} \hline \text{Element} & λ/\pu{nm}& |i\rangle & |f\rangle & \text{I.E./cm$^{-1}$}\\ \hline \ce{Li} & 670.8 & 1s^22s & 1s^22p & 43\,487\\ \ce{Na} & 589.2 & 2p^63s & 2p^63p & 41\,449\\ \ce{K} & 766.5 & 3p^64s & 3p^64p & 35\,010\\ \ce{Rb} & 780.0 & 4p^65s & 4p^65p & 33\,691 \\ \ce{Cs} & 455.5 & 5p^66s & 5p^67p & 31\,406\\ \hline \end{array} $$

You see that the transition in Cs that you give in the table corresponds to a $7p\leftarrow 6s$ transition and that for consistency you should compare it with a $6p\leftarrow 6s$ transition which has a wavelength around 860 nm (no $d$ electrons involved).

The term values of the electronic states of multi-electron atoms are relatively well described by Rydberg's formula

$$ \tilde{\nu}_{n,\ell}=\text{I.E.} - \frac{\mathcal{R}_M}{(n-\delta_\ell)^2}, $$ with $\mathcal{R}_M$ the mass-corrected Rydberg constant and where the quantum defect $\delta_\ell$ depends on the orbital angular momentum of the electron $\ell$ and on the properties of the ion core. The transition $n'p\leftarrow n''s$ can therefore be approximated by $$ \tilde{\nu}_{n',\ell=1}-\tilde{\nu}_{n'',\ell=0}=\frac{\mathcal{R}_M}{(n''-\delta_{\ell=0})}-\frac{\mathcal{R}_M}{(n'-\delta_{\ell=1})}, $$ and thus depend on the quantum defects for the $s$ and $p$ states. The quantum defects depend on the scattering properties of the Rydberg electron with the ion core and it is difficult to predict their values for different elements. It is therefore difficult to make quantitative predictions. Also for the heavier elements (higher nuclear charge $Z$), relativistic effects become more important, which influences the electronic structure.

Note that these lowest states are not really Rydberg states and the Rydberg formula is a bad approximation. For high values of $n$ the Rydberg formula gets much more accurate. In fact, it allows for a very accurate determination of the ionization energies of these elements.


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