The shortest wavelengths emitted in hydrogen spectrum corresponding t…

The correct answer is Option (2) → (A), (C), (B), (D)

The shortest wavelength of a series (the series limit) is emitted when the electron falls from \( n_i = \infty \) to the lower level \( n_f \) of that series:

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

So the shortest wavelength is proportional to \( n_f^{2} \):

    \( \text{(A) Pfund series: } n_f = 5,\ \lambda_{\min} = 25/R \)
    \( \text{(C) Brackett series: } n_f = 4,\ \lambda_{\min} = 16/R \)
    \( \text{(B) Balmer series: } n_f = 2,\ \lambda_{\min} = 4/R \)
    \( \text{(D) Lyman series: } n_f = 1,\ \lambda_{\min} = 1/R \)

In decreasing order: (A), (C), (B), (D).

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. … Other series of spectra for hydrogen were subsequently discovered. These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund 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:

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