In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 305 $$\mathring{A}$$. The corresponding difference for the Paschan series in $$\mathring{A}$$ is __________ (nearest integer)

$$\frac{1}{\lambda} = R_{\infty} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

For the Lyman series ($$n_1 = 1$$):

$$\text{Shortest wavelength } (\lambda_{L,\text{min}}) \implies n_2 = \infty \implies \lambda_{L,\text{min}} = \frac{1}{R_{\infty}}$$

$$\text{Largest wavelength } (\lambda_{L,\text{max}}) \implies n_2 = 2 \implies \lambda_{L,\text{max}} = \frac{4}{3R_{\infty}}$$

$$\Delta \lambda_L = \lambda_{L,\text{max}} - \lambda_{L,\text{min}} = \frac{1}{3R_{\infty}} = 304\ \text{Å} \implies \frac{1}{R_{\infty}} = 912\ \text{Å}$$

For the Paschen series ($$n_1 = 3$$):

$$\text{Shortest wavelength } (\lambda_{P,\text{min}}) \implies n_2 = \infty \implies \lambda_{P,\text{min}} = \frac{9}{R_{\infty}}$$

$$\text{Largest wavelength } (\lambda_{P,\text{max}}) \implies n_2 = 4 \implies \lambda_{P,\text{max}} = \frac{144}{7R_{\infty}}$$

$$\Delta \lambda_P = \lambda_{P,\text{max}} - \lambda_{P,\text{min}} = \frac{144}{7R_{\infty}} - \frac{9}{R_{\infty}} = \frac{81}{7R_{\infty}}$$

$$\Delta \lambda_P = \frac{81}{7} \times 912 = \frac{73872}{7} \approx 10553.14\ \text{Å}$$