Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization energy of metal A in kJ mol$$^{-1}$$ is ______. (Rounded-off to the nearest integer)
[$$h = 6.63 \times 10^{-34}$$ Js, $$c = 3.00 \times 10^8$$ ms$$^{-1}$$, $$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$]

The ionization energy of metal A equals the energy of the photon with wavelength 663 nm that is just sufficient to ionise the atom. The energy of a single photon is given by $$E = \frac{hc}{\lambda}$$.

Substituting the given values: $$E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{663 \times 10^{-9}}$$.

Computing the numerator: $$6.63 \times 10^{-34} \times 3.00 \times 10^8 = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}$$ J.

Computing the energy per atom: $$E = \frac{1.989 \times 10^{-25}}{663 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{6.63 \times 10^{-7}} = 3.00 \times 10^{-19}$$ J.

To convert to kJ mol$$^{-1}$$, we multiply by Avogadro's number and divide by 1000: $$IE = \frac{3.00 \times 10^{-19} \times 6.02 \times 10^{23}}{1000}$$.

$$IE = \frac{1.806 \times 10^{5}}{1000} = 180.6$$ kJ mol$$^{-1}$$.

Rounding to the nearest integer, the ionization energy is $$\boxed{181}$$ kJ mol$$^{-1}$$.