Correct optlon: (4) 5:9
Explanatlon:
Since the series involved is Balmer series, the transition terminates at \(n=2\)
$$
\therefore \quad \bar{v}=\frac{1}{\lambda}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)
$$
It is clear that \(\lambda\) will have maximum value corresponding to transition from \(n_{2}=3\) to \(n_{1}=2\)
$$
\therefore \frac{1}{\lambda_{\max }}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=R\left(\frac{5}{36}\right)
$$
For minimum value of \(\lambda\) the transition has to be from \(=a ~ t o ~ }=2\)
$$
\begin{aligned}
&\frac{1}{\lambda_{\operatorname{mis}}}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)=R \frac{1}{4} \\
&\frac{\lambda_{\min }}{\lambda_{\operatorname{mas}}}=\frac{4}{R} \times \frac{5 R}{36}=\frac{5}{9}
\end{aligned}
$$
