The study of hydrogen atom spectrum has mainly five series. Choose th…

The correct answer is Option (3) → A, E, C, D, B

The shortest wavelength of a series is its series limit, the transition from \( n = \infty \) to the lower level \( n_1 \) of that series.

    \( \displaystyle \frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_{1}^{2}} - 0\right) \;\Rightarrow\; \lambda_{\min} = \frac{n_{1}^{2}}{R} \)

So the shortest wavelength grows as \( n_1^{2} \). With \( 1/R = 91.2\,\mathrm{nm} \):

    \( \text{Lyman } (n_1 = 1):\ \lambda_{\min} = 91.2\,\mathrm{nm} \)
    \( \text{Balmer } (n_1 = 2):\ \lambda_{\min} = 365\,\mathrm{nm} \)
    \( \text{Paschen } (n_1 = 3):\ \lambda_{\min} = 821\,\mathrm{nm} \)
    \( \text{Brackett } (n_1 = 4):\ \lambda_{\min} = 1459\,\mathrm{nm} \)
    \( \text{Pfund } (n_1 = 5):\ \lambda_{\min} = 2280\,\mathrm{nm} \)

Increasing order: Lyman, Balmer, Paschen, Brackett, Pfund, that is A, E, C, D, B.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
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. … These are represented by the formulae: Lyman series:

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}\right) \qquad n = 2,3,4\ldots \qquad (12.6) \)

Paschen series:

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^{2}} - \frac{1}{n^{2}}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)

Brackett series:

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{4^{2}} - \frac{1}{n^{2}}\right) \qquad n = 5,6,7\ldots \qquad (12.8) \)

Pfund series:

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{5^{2}} - \frac{1}{n^{2}}\right) \qquad n = 6,7,8\ldots \qquad (12.9) \)”