The shortest wavelength in the Lyman series of hydrogen spectrum is 912 Å. The shortest wavelength present in Paschen series of spectral lines will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8208\,\mathring{\mathrm{A}} \)
The shortest wavelength of a series is its series limit, the transition from \( n = \infty \) to the lowest level of that series (\( n_f \)):
\( \displaystyle \frac{1}{\lambda_{\min}} = R\left(\frac{1}{n_f^{2}} - 0\right) \;\Rightarrow\; \lambda_{\min} = \frac{n_f^{2}}{R} \)
For the Lyman series \( n_f = 1 \), so \( 1/R = 912\,\mathring{\mathrm{A}} \). For the Paschen series \( n_f = 3 \):
\( \displaystyle \lambda_{\min} = \frac{3^{2}}{R} = 9 \times 912\,\mathring{\mathrm{A}} = 8208\,\mathring{\mathrm{A}} \)
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 – “… 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:
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n^2}\right) \qquad n = 4,5,6\ldots \qquad (12.7) \)”