An electron in a hydrogen atom transitions from $n = 3$ to $n = 2$. Calculate the wavelength of the emitted radiation. (Rydberg constant $R = 1.097 \times 10^7$ m$^{-1}$.)

Rydberg formula: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$. $\dfrac{1}{\lambda} = 1.097 \times 10^7 \times \left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 1.097 \times 10^7 \times \dfrac{5}{36} \approx 1.524 \times 10^6$ m$^{-1}$. $\lambda \approx 6.56 \times 10^{-7}$ m $= 656$ nm. This is the **H-α line** of the Balmer series — the deep-red line in the visible hydrogen spectrum. Option D ($1216$ nm) is Lyman-α (UV, $n=2 \to 1$); options A and B are higher Balmer lines (H-δ and H-β).