Using Bohr's model, calculate the wavelength of the photon emitted when an electron in a hydrogen atom drops from the level n=3 to n=2. State the spectral region of this photon.

Step-by-step solution

1. Use Bohr's energy formula $E_n=-13.6\text{ eV}/n^2$.
2. Compute the energy difference: $\Delta E=E_2-E_3=-13.6/4-(-13.6/9)=13.6(1/4-1/9)$ eV.
3. Evaluate the difference: $\Delta E=13.6\times \frac{5}{36}=1.8889$ eV.
4. Convert to joules: $\Delta E_J=1.8889\times 1.602\times 10^{-19}=3.027\times 10^{-19}$ J.
5. Apply the wavelength formula $\lambda =hc/\Delta E_J$ with $h=6.626\times 10^{-34}$ J·s and $c=3.00\times 10^8$ m/s.
6. Calculate $\lambda =(6.626\times 10^{-34}\times 3.00\times 10^8)/(3.027\times 10^{-19})=6.57\times 10^{-7}$ m = 656.7 nm.
7. Recognise that 656.7 nm lies in the red part of the visible spectrum.

Concept tested: Bohr model, energy levels

Published 7 October 2026 · How this solution was prepared and checked