AIIMS 2018 Chemistry Hydrogen Spectrum MCQ Question

In the Balmer series, the transition from n₂ = 3 to n₁ = 2 results in the emission of light with a wavelength calculated using the formula $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Substituting the values, we find $\lambda = 654 \, \text{nm}$, which corresponds to option D, not B. The other options (A, B, C) represent wavelengths for different transitions in the Balmer series, specifically for transitions from higher energy levels to n₁ = 2, but do not match the calculated wavelength for the 3 → 2 transition.