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  4. The energy of electron in the nth orbit of hydrogen atom is expressed as En=(-13.6/n2) e V. The shortest and longest wavelength of Lyman series will be

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Q. The energy of electron in the nth orbit of hydrogen atom is expressed as $E_n=\frac{-13.6}{n^2} e V$. The shortest and longest wavelength of Lyman series will be

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Solution:

$\frac{1}{\lambda_{\max }}=R\left[\frac{1}{(1)^2}-\frac{1}{(2)^2}\right] \Rightarrow \lambda_{\max }=\frac{4}{3 R} \approx 1213 \,\mathring{A} $
and $\frac{1}{\lambda_{\min }}= R \left[\frac{1}{(1)^2}-\frac{1}{\infty}\right] \Rightarrow \lambda_{\min }=\frac{1}{ R } \approx 910 \,\mathring{A} $