The wavelength of second line of Balmer series in terms of Rydberg constant R is

$\frac{16}{3R}$

The correct answer is Option (3) → $\frac{16}{3R}$

Balmer series: $\;\; \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$, where $n = 3,4,5,\dots$

First line (H-α): $n=3 \to 2$

Second line (H-β): $n=4 \to 2$

$\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right)$

$\frac{1}{\lambda} = R \left( \frac{4-1}{16} \right) = \frac{3R}{16}$

$\lambda = \frac{16}{3R}$

Answer: $\;\; \lambda = \frac{16}{3R}$