According to Bohr's theory of hydrogen atom, the ratio of the maximum and minimum wavelength of Lyman series will be

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Aicharya · Accepted Answer

$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
For Lyman series, $\mathrm{n}_1=1$ $\begin{aligned} \quad \frac{1}{\lambda_{\max }} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R \\ \frac{1}{\lambda_{\min }} & =R\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)=R \\ \therefore \quad & \frac{\lambda_{\max }}{\lambda_{\min }} \end{aligned}=\frac{4}{3}$

According to Bohr's theory of hydrogen atom, the ratio of the maximum and minimum wavelength of Lyman series…

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