Physics - Modern Physics Question with Solution | TestHub

Using the relation between energy of photon and wavelength for the transition in the hydrogen atom, we get

$\frac{hc}{\lambda }=13.6\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

$$\begin{gathered} \Rightarrow \frac{4\times 10^{-15}\times 3\times 10^8}{\lambda }=13.6\left[1-\frac{1}{16}\right] \\ \Rightarrow \lambda =\frac{12\times 10^{-7}\times 16}{13.6\times 15}=94.1\times 10^{-9}=94.1nm \end{gathered}$$