A beam of monochromatic light is used to excite the electron in $$Li^{++}$$ from the first orbit to the third orbit. The wavelength of monochromatic light is found to be $$x \times 10^{-10}$$ m. The value of $$x$$ is ______.
[Given $$hc = 1242$$ eV nm]

A beam of monochromatic light excites the electron in $$Li^{++}$$ from the first orbit (n=1) to the third orbit (n=3). For a hydrogen-like atom with atomic number Z, the energy levels are: $$E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$$. For $$Li^{++}$$, Z = 3, so $$E_1 = -\frac{13.6 \times 9}{1} = -122.4 \text{ eV}$$ and $$E_3 = -\frac{13.6 \times 9}{9} = -13.6 \text{ eV}$$. Therefore, the energy required for the transition is $$\Delta E = E_3 - E_1 = -13.6 - (-122.4) = 108.8 \text{ eV}$$.

Using $$E = \frac{hc}{\lambda}$$, the wavelength of the photon is $$\lambda = \frac{hc}{\Delta E} = \frac{1242 \text{ eV} \cdot \text{nm}}{108.8 \text{ eV}}$$, which yields $$\lambda = 11.4 \text{ nm} = 114 \times 10^{-10} \text{ m}$$.

The value of x is 114.