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 need to find the wavelength of the second line of the Paschen series of the hydrogen atom, given that the first line has wavelength 720 nm.

The Paschen series corresponds to transitions to $$n = 3$$, and the Rydberg formula for hydrogen is $$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$.

For the first line of the Paschen series, corresponding to a transition from $$n_2 = 4$$ to $$n_1 = 3$$, $$\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{16 - 9}{144}\right) = \frac{7R}{144}$$, while for the second line, corresponding to a transition from $$n_2 = 5$$ to $$n_1 = 3$$, $$\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{25}\right) = R\left(\frac{25 - 9}{225}\right) = \frac{16R}{225}$$.

Taking the ratio of these expressions gives $$\frac{\lambda_2}{\lambda_1} = \frac{7R/144}{16R/225} = \frac{7 \times 225}{16 \times 144} = \frac{1575}{2304}$$, and hence $$\lambda_2 = 720 \times \frac{1575}{2304} = 720 \times \frac{1575}{2304}$$. Simplifying $$\frac{720}{2304} = \frac{5}{16}$$ yields $$\lambda_2 = \frac{5 \times 1575}{16} = \frac{7875}{16} = 492.1875 \text{ nm}$$, which rounds to the nearest integer as $$\lambda_2 \approx 492$$ nm.

The answer is 492.