If the binding energy of ground state electron in a hydrogen atom is 13.6 eV, then, the energy required to remove the electron from the second excited state of Li$$^{2+}$$ will be: $$x \times 10^{-1}$$ eV. The value of $$x$$ is ______.

The binding energy of the ground state electron in a hydrogen atom is 13.6 eV, and we are asked to find the energy required to remove the electron from the second excited state of $$\text{Li}^{2+}$$. Since $$\text{Li}^{2+}$$ is a hydrogen-like ion with $$Z = 3$$ and the second excited state corresponds to $$n = 3$$ (ground state $$n=1$$, first excited $$n=2$$, second excited $$n=3$$), we use the formula for the binding energy in the $$n$$-th orbit of hydrogen-like atoms:

$$E_n = \frac{13.6 \times Z^2}{n^2} \text{ eV}$$

Substituting $$Z = 3$$ and $$n = 3$$ gives

$$E_3 = \frac{13.6 \times 3^2}{3^2} = \frac{13.6 \times 9}{9} = 13.6 \text{ eV}$$

Next, to express this energy in the form $$x \times 10^{-1}$$ eV, we note that

$$13.6 \text{ eV} = x \times 10^{-1} \text{ eV}$$

From the above,

$$x = \frac{13.6}{10^{-1}} = \frac{13.6}{0.1} = 136$$

Therefore, the value of $$x$$ is 136.