If $$\lambda_1$$ and $$\lambda_2$$ are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of $$\lambda_1 : \lambda_2$$ is:

The third member of the Lyman series corresponds to the transition from $$n = 4$$ to $$n = 1$$. Using the Rydberg formula: $$\frac{1}{\lambda_1} = R\left(\frac{1}{1^2} - \frac{1}{4^2}\right) = R\left(1 - \frac{1}{16}\right) = \frac{15R}{16}$$.

The first member of the Paschen series corresponds to the transition from $$n = 4$$ to $$n = 3$$. Using the Rydberg formula: $$\frac{1}{\lambda_2} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{16 - 9}{144}\right) = \frac{7R}{144}$$.

The ratio $$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2^{-1}}{1/\lambda_1^{-1}}$$ can be computed as: $$\frac{\lambda_1}{\lambda_2} = \frac{7R/144}{15R/16} = \frac{7}{144} \times \frac{16}{15} = \frac{112}{2160} = \frac{7}{135}$$.

Therefore $$\lambda_1 : \lambda_2 = 7 : 135$$.

Hence the correct answer is Option 1: $$7 : 135$$.