Ratio of the wavelengths of first line of Lyman series and first line of Balmer series is
Answer & explanation
Correct answer: option 3
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
For first line of Lymen series $n_1=1$ and $n_2=2$
For first line of Balmer series $n_2=2$ and $n_2=3$
So, $\frac{\lambda_{\text {Lymen }}}{\lambda_{\text {Balmer }}}=\frac{5}{27}$