The wavelength of second line of Balmer series in terms of Rydberg constant R is
Answer & explanation
Correct answer: option 3
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: $\;\; \lambda = \frac{16}{3R}$