Find the ratio of minimum to maximum wavelength of radiations emitted when electron jumps from higher energy state into ground state of hydrogen atom.

$\displaystyle \frac{1}{\lambda}=R\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]$ $\displaystyle \frac{1}{\lambda_{\max }}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$ $\displaystyle \lambda_{\max }=\frac{4}{3 R}$ $\displaystyle \frac{1}{\lambda_{\min }}=R\left[\frac{1}{1^{2}}-\frac{1}{\infty}\right]$ $\displaystyle \lambda_{\min }=\frac{1}{R}$ $\displaystyle \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{3}{4}$For $\displaystyle \lambda_{\min }$: $\displaystyle n_{1}=1, n_{2}=\infty$ $\displaystyle E_{2}-E_{1}=\frac{h c}{\lambda_{\min }}$ $\displaystyle 0-(-13.6)=\frac{h c}{\lambda_{\min }}$ $\displaystyle \lambda_{\min }=\frac{h c}{13.6}$ For $\displaystyle \lambda_{\max }$: $\displaystyle n_{1}=1, n_{2}=2$ $\displaystyle \lambda_{\max }=\frac{h c}{-3.4-(-13.6)}$ $\displaystyle \lambda_{\max }=\frac{h c}{10.2}$ $\displaystyle \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{3}{4}$