Calculate the wavelength of radiation emitted when an electron falls from third excited state to ground state in the Lyman series of Hydrogen atom.
($R_H=1.1\times {10}^7{m}^{-1})$

The correct option is B $\lambda =0.97\times {10}^{-7}m$
The wave number of a photon of any atom experiencing transition from $n_2$ to $n_1$ shell can be expressed as:$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where $\lambda$ is the wavelength. $3^{rd}$ excited state means $n_2=4$ and ground state means $n_1=1$

Substituting the values, we get
$\frac{1}{\lambda }=1.1\times {10}^7{m}^{-1}(\frac{1}{1^2}-\frac{1}{4^2})$
$\therefore$ $\lambda =0.97\times {10}^{-7}m$