Concept:
For the Balmer series,
\[
\frac{1}{\lambda}
=
R
\left(
\frac{1}{2^2}
-
\frac{1}{n^2}
\right),
\qquad n=3,4,5,\ldots
\]
The maximum wavelength corresponds to the minimum energy transition and the minimum wavelength corresponds to the series limit.
Step 1:Find the maximum wavelength.
For Balmer series, the first line corresponds to
\[
n=3 \rightarrow n=2
\]
Hence,
\[
\frac{1}{\lambda_{\max}}
=
R
\left(
\frac{1}{4}
-
\frac{1}{9}
\right)
=
R
\left(
\frac{5}{36}
\right)
\]
\[
\lambda_{\max}
=
\frac{36}{5R}
\]
Step 2: Find the minimum wavelength.
The minimum wavelength occurs at the series limit,
\[
n=\infty \rightarrow n=2
\]
Therefore,
\[
\frac{1}{\lambda_{\min}}
=
R
\left(
\frac{1}{4}
-
0
\right)
=
\frac{R}{4}
\]
\[
\lambda_{\min}
=
\frac{4}{R}
\]
Step 3: Calculate the ratio.
\[
\frac{\lambda_{\max}}{\lambda_{\min}}
=
\frac{\frac{36}{5R}}
{\frac{4}{R}}
\]
\[
=
\frac{36}{20}
\]
\[
=
\frac{9}{5}
\]
Step 4: State the answer.
\[
{
\frac{\lambda_{\max}}
{\lambda_{\min}}
=
\frac{9}{5}
}
\]
Hence, the correct option is
\[
{(C)}
\]