Question

Electromagnetic radiation of wavelength 663 nm is just sufficient to ionize 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$^{-34}$ J-s, c = 3.00 $\times$ 10$^8$ ms$^{-1}$, N$_A$ = 6.02 $\times$ 10$^{23}$ mol$^{-1}$]

Answer 1

181

Answer 2

The energy of a photon is given by $E = \frac{hc}{\lambda}$. The ionization energy for one mole of atoms is $I.E. = \frac{hc N_A}{\lambda}$. Substituting the given values: $I.E. = \frac{(6.63 \times 10^{-34} \text{ J-s}) \times (3.00 \times 10^8 \text{ ms}^{-1}) \times (6.02 \times 10^{23} \text{ mol}^{-1})}{663 \times 10^{-9} \text{ m}}$ $I.E. \approx 180600 \text{ J mol}^{-1}$ Converting to kJ mol$^{-1}$: $I.E. \approx 180.6 \text{ kJ mol}^{-1}$ Rounded to the nearest integer, the ionization energy is 181 kJ mol$^{-1}$.