Calculate the wavelength of the spectral line when electron in hydrogen atom undergoes a transition from fourth energy level to second energy level. In which region of electromagnetic radiation, does the radiation fall?

Dear Student,

$$\begin{gathered} wavenumber(\stackrel{\_}{\nu )}=1.09677\times {10}^7\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)m^{-1} \\ =1.09677\times {10}^7\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=1.09677\times {10}^7(0.25-0.0625) \\ (\stackrel{\_}{\nu )}=0.2056\times {10}^7{m}^{-1} \\ orWavelength\left(\lambda \right)=\frac{1}{\stackrel{\_}{\nu }}=\frac{1}{0.2056\times {10}^7{m}^{-1}}=4.86\times {10}^{-7}m \\ or\lambda =486\times {10}^{-9}m=486nm \end{gathered}$$

Since the transition falls to the second energy level it falls in the Balmer series i.e in the Visible spectral region.

Best Wishes !