The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is
Answer & explanation
Correct answer: option 4
Shortest wavelength comes from $n_1=\infty$ to $n_2=1$ and longest wavelength comes from $n_1=6$ to $n_2=5$ in the given case. Hence $\frac{1}{\lambda_{\min }}=R\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)=R$
$\frac{1}{\lambda_{\max }}=R\left(\frac{1}{5^2}-\frac{1}{6^2}\right)=R\left(\frac{36-25}{25 \times 36}\right)=\frac{11}{900} R$
∴ $\frac{\lambda_{\max }}{\lambda_{\min }}=\frac{900}{11}$