The longest wavelength in Balmar Series of hydrogen spectrum will be:

The correct answer is Option (4) → 6563 Å

In the Balmer series the electron falls to \( n_f = 2 \). The longest wavelength corresponds to the smallest energy jump, from \( n_i = 3 \).

    \( \displaystyle \frac{1}{\lambda} = R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) = \frac{5R}{36} \)

    \( \displaystyle \lambda = \frac{36}{5R} = \frac{36}{5\times1.097\times10^{7}\,\mathrm{m^{-1}}} \approx 6.563\times10^{-7}\,\mathrm{m} = 6563\,\mathrm{\mathring{A}} \)

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 \( \lambda \) is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc. … Taking n = 3 in Eq. (12.5), one obtains the wavelength of the Hα line:

    \( \displaystyle \frac{1}{\lambda} = 1.097 \times 10^{7}\,\mathrm{m^{-1}}\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.522 \times 10^{6}\,\mathrm{m^{-1}} \)

i.e., \( \lambda = 656.3\,\mathrm{nm} \)”