A photon is emitted in transition from n = 4   to   n = 1 level in hydrogen atom. The…

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

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

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