The ratio of maximum frequency and minimum frequency of light emitted in Balmer series of hydrogen spectrum is

9 : 5

The correct answer is Option (2) → 9 : 5

Balmer series corresponds to transitions to $n_f = 2$.

Frequency of emitted light: $f = \frac{R c}{h} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, here $n_f = 2$, $n_i = 3, 4, 5, ...$

Maximum frequency occurs for $n_i \to \infty$:

$f_\text{max} = R c \left( \frac{1}{2^2} - 0 \right) = \frac{R c}{4}$

Minimum frequency occurs for $n_i = 3$:

$f_\text{min} = R c \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R c \left( \frac{1}{4} - \frac{1}{9} \right) = R c \cdot \frac{5}{36}$

Ratio: $\frac{f_\text{max}}{f_\text{min}} = \frac{R c / 4}{5 R c / 36} = \frac{36}{20} = \frac{9}{5}$

Ratio of maximum to minimum frequency: 9 : 5