Physics - Modern Physics Question with Solution | TestHub

For Lyman series:

For first member, $\frac{1}{\lambda }=\frac{13.6Z^2}{hc}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$ ….. (i)

For second member, $\frac{1}{\lambda \text{'}}=\frac{13.6Z^2}{hc}\left[\frac{1}{1^2}-\frac{1}{3^2}\right]$ ….. (ii)

On dividing (i) and (ii), we get

$\lambda \text{'}=\frac{27}{32}\lambda$