The wavelength of the radiation emitted is $$\lambda_0$$ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $$\frac{20}{x}\lambda_0$$. The value of $$x$$ is _____.

We are given that the wavelength of radiation emitted when an electron jumps from the second excited state to the first excited state of the hydrogen atom is $$\lambda_0$$ and we need to find the wavelength when the electron jumps from the third excited state to the second orbit.

First, the second excited state corresponds to $$n = 3$$ and the first excited state to $$n = 2$$, while the third excited state is $$n = 4$$ and the second orbit is $$n = 2$$.

For the transition $$n = 3 \to n = 2$$, $$\frac{1}{\lambda_0} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \cdot \frac{5}{36}.$$

For the transition $$n = 4 \to n = 2$$, $$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{4} - \frac{1}{16}\right) = R \cdot \frac{3}{16}.$$

Taking the ratio gives $$\frac{\lambda_0}{\lambda} = \frac{R \cdot \tfrac{3}{16}}{R \cdot \tfrac{5}{36}} = \frac{3}{16} \times \frac{36}{5} = \frac{108}{80} = \frac{27}{20},$$ so $$\lambda = \frac{20}{27}\,\lambda_0.$$

Comparing this with the given expression $$\lambda = \frac{20}{x}\,\lambda_0$$ shows that $$x = 27\,. $$