If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _____ nm. (Nearest integer)

We begin by considering the Paschen series, which involves transitions to $$n = 3$$.

First, for the 1st line (4→3), we write $$\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = R \times \frac{7}{144}$$, and hence $$\lambda_1 = 720$$ nm.

Next, for the 2nd line (5→3), we have $$\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{25}\right) = R \times \frac{16}{225}$$.

Then, substituting these into the ratio, we find $$\frac{\lambda_2}{\lambda_1} = \frac{7/144}{16/225} = \frac{7 \times 225}{144 \times 16} = \frac{1575}{2304}$$,
and therefore $$\lambda_2 = 720 \times \frac{1575}{2304} = 492 \text{ nm}$$.