The approximate wavelength of the spectral line emitted when an excited electron in a hydrogen atom

Question

The approximate wavelength of the spectral line emitted when an excited electron in a hydrogen atom undergoes a direct downward transition from the energy level $n=3$ to $n=2$ is

Professor PC Thomas Classes · Answer

(a) : This precise transition actively generates the first line of the Balmer series (H-alpha). Applying the Rydberg formula: $\frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}ight) = 1.097 	imes 10^7 	imes \left(\frac{1}{4} - \frac{1}{9}ight) = 1.097 	imes 10^7 	imes \frac{5}{36} \mathrm{~m}^{-1}$. Inverting this value strictly to solve for the wavelength gives $\lambda = \frac{36}{5 	imes 1.097 	imes 10^7} \mathrm{~m} \approx 6.56 	imes 10^{-7} \mathrm{~m} = 656 \mathrm{~nm}$.