The wavelength of radiation emitted, when \( \mathrm{He^{+}} \) makes a transition from the state n = 3 to the state n = 2 will be: (Take Rydberg constant \( R = 1.097\times10^{7}\,\mathrm{m^{-1}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 164.1 nm
\( \mathrm{He^{+}} \) has one electron, so it behaves like a hydrogen-like ion with \( Z = 2 \). The Rydberg formula then carries a factor \( Z^{2} \).
\( \displaystyle \frac{1}{\lambda} = R Z^{2}\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \)
\( Z = 2 \)
\( n_1 = 2 \)
\( n_2 = 3 \)
\( \displaystyle \frac{1}{\lambda} = 1.097\times10^{7} \times 4 \times \left(\frac{1}{4} - \frac{1}{9}\right) = 1.097\times10^{7} \times 4 \times \frac{5}{36} \)
\( \displaystyle \frac{1}{\lambda} = 6.094\times10^{6}\,\mathrm{m^{-1}} \)
\( \displaystyle \lambda = \frac{1}{6.094\times10^{6}}\,\mathrm{m} = 1.641\times10^{-7}\,\mathrm{m} = 164.1\,\mathrm{nm} \)
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. The value of R is 1.097 × 107 m–1. This equation is also called Balmer formula.”