The ionization energy of sodium in $$kJ \text{ mol}^{-1}$$, if electromagnetic radiation of wavelength $$242$$ nm is just sufficient to ionize sodium atom is ______. (nearest integer)

We need to find the energy (in kJ/mol) corresponding to a wavelength of 242 nm.

Using $$E = \frac{hc}{\lambda}$$:

$$ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{242 \times 10^{-9}} $$

$$ E = \frac{19.878 \times 10^{-26}}{242 \times 10^{-9}} = \frac{19.878 \times 10^{-26}}{2.42 \times 10^{-7}} $$

$$ E = 8.214 \times 10^{-19} \text{ J} $$

Multiply by Avogadro's number ($$N_A = 6.022 \times 10^{23}$$):

$$ E_{mol} = 8.214 \times 10^{-19} \times 6.022 \times 10^{23} = 49.46 \times 10^{4} \text{ J/mol} $$

$$ E_{mol} = 494.6 \text{ kJ/mol} \approx 494 \text{ kJ/mol} $$

The answer is $$\boxed{494}$$ kJ/mol.