1. Given data

• Wavelength of radiation: $\lambda = 242\,\text{nm}$
• We need the ionization energy in $\text{kJ mol}^{-1}$.

Since the radiation is just sufficient to ionize sodium, the energy of one photon equals the ionization energy of one sodium atom.

2. Energy of one photon

The energy of a photon is

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

Using:

$h = 6.626 \times 10^{-34}\,\text{J s}$ $c = 3.0 \times 10^8\,\text{m s}^{-1}$ $\lambda = 242\,\text{nm} = 242 \times 10^{-9}\,\text{m}$

So,

$E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{242 \times 10^{-9}}$

$E = \frac{1.9878 \times 10^{-25}}{2.42 \times 10^{-7}}$

$E \approx 8.21 \times 10^{-19}\,\text{J}$

3. Energy per mole

Multiply by Avogadro's number:

$E_{\text{mole}} = (8.21 \times 10^{-19})(6.022 \times 10^{23})\,\text{J mol}^{-1}$

$E_{\text{mole}} \approx 4.94 \times 10^5\,\text{J mol}^{-1}$

4. Convert to kJ mol$^{-1}$

$4.94 \times 10^5\,\text{J mol}^{-1} = 494\,\text{kJ mol}^{-1}$

5. Final answer

The ionization energy of sodium is

$\boxed{494}$