Arrange the following lines of Balmer series in the increasing order…

The correct answer is Option (1) → C, A, D, B, E

The Balmer series lines are emitted when the electron falls from a level \( n \ge 3 \) to \( n = 2 \). Their wavelengths are given by

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right) \)

A larger \( n \) gives a larger \( 1/\lambda \), that is a shorter wavelength. The lines are named in order of increasing \( n \):

    \( H_{\alpha}: n = 3 \)
    \( H_{\beta}: n = 4 \)
    \( H_{\gamma}: n = 5 \)
    \( H_{\delta}: n = 6 \)
    \( H_{\infty}: n \to \infty \;(\text{series limit}) \)

So the wavelength is smallest for \( H_{\infty} \) (about 365 nm) and largest for \( H_{\alpha} \) (about 656 nm):

    \( H_{\infty} \lt H_{\delta} \lt H_{\gamma} \lt H_{\beta} \lt H_{\alpha} \)

In terms of the labels the order is C, A, D, B, E.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), but it is still in the CUET 2026 syllabus, so do not skip it.

NCERT (Pre-rationalised edition, 2022-23):
Spectral Series – “… The line with the longest wavelength, 656.3 nm in the red is called H_α; the next line with wavelength 486.1 nm in the blue-green is called H_β, the third line 434.1 nm in the violet is called H_γ; and so on. As the wavelength decreases, the lines appear closer together and are weaker in intensity.”

Spectral Series – “… For n = ∞, one obtains the limit of the series, at λ = 364.6 nm. This is the shortest wavelength in the Balmer series.”