The wavelength of the first line of the Balmer series is $656 \ nm$. Calculate the wavelength of the second line of this series. (Note: First line means $n=3 \to n=2$ and second line means $n=4 \to n=2$)

(N/A) The Rydberg formula for the Balmer series is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 2$.
For the first line $(n_2 = 3)$: $\frac{1}{656} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right)$.
For the second line $(n_2 = 4)$: $\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right)$.
Dividing the two equations: $\frac{\lambda_2}{656} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{80}{108} = \frac{20}{27}$.
Therefore,$\lambda_2 = 656 \times \frac{20}{27} \approx 486 \ nm$.