The longest wavelength of light that can be used for the ionisation of lithium ion $$Li^{2+}$$ is $$x \times 10^{-8}$$ m. The value of $$x$$ is ______ (Nearest Integer)
(Given : Energy of the electron in the first shell of the hydrogen atom is $$-2.2 \times 10^{-18}$$ J; $$h = 6.63 \times 10^{-34}$$ Js and $$c = 3 \times 10^{8}$$ ms$$^{-1}$$)

We need to find the longest wavelength of light that can ionise $$Li^{2+}$$ from its ground state. The longest wavelength corresponds to the minimum energy required for ionisation.

$$Li^{2+}$$ is a hydrogen-like ion with atomic number $$Z = 3$$ and a single electron. For such ions, the Bohr model applies directly. The energy of the electron in the $$n$$-th shell is $$E_n = E_1^{(H)} \times \frac{Z^2}{n^2}$$, where $$E_1^{(H)} = -2.2 \times 10^{-18}$$ J is the energy of the electron in the first shell of the hydrogen atom.

For the ground state of $$Li^{2+}$$, $$n = 1$$ and $$Z = 3$$: $$E_1 = -2.2 \times 10^{-18} \times \frac{9}{1} = -19.8 \times 10^{-18}$$ J.

The ionisation energy equals $$|E_1| = 19.8 \times 10^{-18}$$ J (energy to go from $$n = 1$$ to $$n = \infty$$).

Using the relation $$E = \frac{hc}{\lambda}$$, the longest wavelength of light that can ionise $$Li^{2+}$$ is $$\lambda = \frac{hc}{E}$$.

Substituting: $$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{19.8 \times 10^{-18}}$$.

Numerator: $$6.63 \times 3 = 19.89$$, so the numerator is $$19.89 \times 10^{-26}$$.

Therefore: $$\lambda = \frac{19.89 \times 10^{-26}}{19.8 \times 10^{-18}} = \frac{19.89}{19.8} \times 10^{-26+18} = 1.0045 \times 10^{-8}$$ m.

Thus $$x \approx 1$$ (nearest integer).

So, the answer is $$1$$.