The shortest wavelength of the spectral lines emitted in Balmer Series is: [Rydberg constant \( = 10^{7}\,\mathrm{m^{-1}} \)]
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4000\,\mathring{\mathrm{A}} \)
The Balmer series contains the lines emitted when the electron falls to \( n_1 = 2 \) from higher levels.
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}} - \frac{1}{n_2^{2}}\right), \quad n_2 = 3, 4, 5, \ldots \)
The shortest wavelength is the line of largest energy, the series limit, for \( n_2 \rightarrow \infty \).
\( \displaystyle \frac{1}{\lambda_{\min}} = \frac{R}{4} \;\Rightarrow\; \lambda_{\min} = \frac{4}{R} = \frac{4}{10^{7}}\,\mathrm{m} = 4\times10^{-7}\,\mathrm{m} \)
\( 1\,\mathring{\mathrm{A}} = 10^{-10}\,\mathrm{m} \;\Rightarrow\; \lambda_{\min} = 4000\,\mathring{\mathrm{A}} \)
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 – “… Balmer found a simple empirical formula for the observed wavelengths
\( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right) \qquad (12.5) \)
where λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 107 m–1. This equation is also called Balmer formula.”