$$\dfrac{x}{x+4}$$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) third permitted energy level to the second level and (ii) the highest permitted energy level to the second permitted level. The value of $$x$$ will be ______.

The energy of a photon emitted during transition from level $$n_1$$ to level $$n_2$$ in hydrogen is:

$$E = 13.6\left(\dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\right) \text{ eV}$$

Case (i): Transition from $$n = 3$$ to $$n = 2$$:

$$E_1 = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 13.6 \times \dfrac{9 - 4}{36} = 13.6 \times \dfrac{5}{36}$$

Case (ii): Transition from $$n = \infty$$ to $$n = 2$$ (highest permitted level):

$$E_2 = 13.6\left(\dfrac{1}{4} - 0\right) = 13.6 \times \dfrac{1}{4} = 13.6 \times \dfrac{9}{36}$$

The ratio of the energies:

$$\dfrac{E_1}{E_2} = \dfrac{13.6 \times \frac{5}{36}}{13.6 \times \frac{9}{36}} = \dfrac{5}{9}$$

The given ratio is $$\dfrac{x}{x + 4}$$. Comparing:

$$\dfrac{x}{x + 4} = \dfrac{5}{9}$$

$$9x = 5(x + 4) = 5x + 20$$

$$4x = 20$$

$$x = 5$$

Therefore, the value of $$x$$ is $$\boxed{5}$$.