Question

Given that the excitation energy of oxygen atoms amounts to 190 kJ mol-1 (over and above ground state energy), find the wavelength and frequency of the weakest radiation able to dissociate molecular oxygen into excited atoms, by the reaction:

O2 + hn ¾® O* + O*.

Hence, state the region of the electromagnetic spectrum where the radiation belongs.

Answer 1

The energy of the photon needs to be  equal to the excitation energy of one molecule for the energy to be just enough for the dissociation.

Given that the excitation energy is . which mean 190 kJ of energy is needed to dissociate 1 mol of O2 molecules.

Now, energy needed to dissociate 1 molecule of O2 is

190 kJ mol-1 6.022 x 1023 mol1 3.155 x 10-22 kj-3.155 x 10-19 j

Hence, we need to equate the above energy using the planck's relation for the energy of a photon carrying out the excitation.

E = 3.155 × 10-19 J-hv 3.155 10 6.626 × 10-34 J s × 10-19 J = 4.76 × 1014 s-1 4.76 x 1014 s-

Hence, the frequency of the weakest radiation able to dissociate a molecule of O2 is
4.76 × 1014 s- 1
.

Now, the wavelength can be calculated as

$\lambda = \frac{c}{\nu}= \frac{3.00 \times 10^8 \ m\ s^{-1}}{4.76 \times 10^{14} \ s^{-1}} = 6.30 \times 10^{-7} \ m = 630 \ nm$

Hence, the wavelength of the weakest radiation is about 630 nm.

The region from 380 nm to 750 nm is the visible wavelength region. Since our wavelength 630 nm falls in this range, the region the photon belongs to is the visible region of electromagnetic spectrum.