The energy of electron in the nth orbit of hydrogen atom is expressed as $E_n=\frac{-13.6}{n^2} ~eV$. The shortest and longest wavelength of Lyman series will be
Answer & explanation
Correct answer: option 1
$\frac{1}{\lambda_{\max }}=R\left[\frac{1}{(1)^2}-\frac{1}{(2)^2}\right] \Rightarrow \lambda_{\max }=\frac{4}{3 R} \approx$ 1213 Å
and $\frac{1}{\lambda_{\min }}=R\left[\frac{1}{(1)^2}-\frac{1}{\infty}\right] \Rightarrow \lambda_{\min }=\frac{1}{R} \approx$ 910 Å