In the Bohr model, an electron transitions from $n = 3$ to $n = 2$ in a hydrogen atom (the H-α line of the Balmer series). The wavelength of the emitted photon is closest to (Take $hc = 1240\,\text{eV\,nm}$.)

1. Photon energy: $E_\text{photon} = E_i - E_f = E_3 - E_2$. 2. $E_3 = -13.6/9 = -1.511\,\text{eV}$ and $E_2 = -13.6/4 = -3.400\,\text{eV}$. 3. $E_\text{photon} = -1.511 - (-3.400) = 1.889\,\text{eV}$. 4. Wavelength: $\lambda = hc/E_\text{photon} = 1240/1.889 = 656.4\,\text{nm}$. 5. This is the well-known H-α RED line of the Balmer series, observable in the spectrum of glowing hydrogen and in stellar atmospheres. 6. Option A ($103\,\text{nm}$) is the Lyman-$\alpha$ line ($n=2 \to 1$). Option B ($486\,\text{nm}$) is H-$\beta$ ($n=4 \to 2$). Option D ($1875\,\text{nm}$) is the Paschen-$\alpha$ line ($n=4 \to 3$). Knowing the series helps distinguish. _Source: NCERT Class 12 Physics Part 2, Ch 12, §12.5 (Line Spectra of the Hydrogen Atom) + Fig. 12.5, p. 7 + p. 11._