Electromagnetic radiation of wavelength $242 \, nm$ is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $kJ \, mol^{-1}$.

(N/A) The energy required to ionise one mole of sodium atoms is given by the formula $E = \frac{N_A h c}{\lambda}$.
Substituting the values:
$N_A = 6.022 \times 10^{23} \, mol^{-1}$
$h = 6.626 \times 10^{-34} \, J \cdot s$
$c = 3 \times 10^8 \, m \cdot s^{-1}$
$\lambda = 242 \times 10^{-9} \, m$
$E = \frac{(6.022 \times 10^{23} \, mol^{-1}) \times (6.626 \times 10^{-34} \, J \cdot s) \times (3 \times 10^8 \, m \cdot s^{-1})}{242 \times 10^{-9} \, m}$
$E \approx 494700 \, J \cdot mol^{-1}$
Converting to $kJ \cdot mol^{-1}$:
$E = 494.7 \, kJ \cdot mol^{-1}$