The ratio of wavelength of spectral lines $$H_\alpha$$ and $$H_\beta$$ in the Balmer series is $$\frac{x}{20}$$. The value of $$x$$ is _____.

For the Balmer series, the wavelength is given by

$$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n^2}\right)$$

For $$H_\alpha$$, the transition is from $$n = 3$$ to $$n = 2$$:

$$\frac{1}{\lambda_\alpha} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9-4}{36}\right) = \frac{5R}{36}$$

For $$H_\beta$$, the transition is from $$n = 4$$ to $$n = 2$$:

$$\frac{1}{\lambda_\beta} = R\left(\frac{1}{4} - \frac{1}{16}\right) = R\left(\frac{4-1}{16}\right) = \frac{3R}{16}$$

Now taking the ratio of wavelengths,

$$\frac{\lambda_\alpha}{\lambda_\beta} = \frac{1/\lambda_\beta}{1/\lambda_\alpha} = \frac{3R/16}{5R/36} = \frac{3}{16} \times \frac{36}{5} = \frac{108}{80} = \frac{27}{20}$$

We are given that the ratio is $$\frac{x}{20}$$, so

$$\frac{x}{20} = \frac{27}{20}$$

Hence, the answer is $$x = 27$$. So, the answer is $$27$$.