If Rydberg's constant is R, the longest wavelength of radiation in Paschen series will be $$\frac{\alpha}{7R}$$, where $$\alpha$$ = _____.

The Paschen series corresponds to transitions to $$n = 3$$. The longest wavelength corresponds to the smallest energy transition: $$n = 4 \to n = 3$$.

$$\frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = R\left(\frac{16 - 9}{144}\right) = \frac{7R}{144}$$

$$\lambda = \frac{144}{7R}$$

So $$\alpha = 144$$.

The answer is $$\boxed{144}$$.