Question

${O_2}$ undergoes photochemical dissociation into 1 normal oxygen atom (O) and more energetic oxygen ${O^*}$ . If ( ${O^*}$ ) has $1.967eV$ more energy than (O) and normal dissociation energy of ${O_2}$ is $498kJ/mol$ , what is the maximum wavelength effective for the photochemical dissociation of ${O_2}$ ?

Answer 1

Hint : To find the maximum wavelength, we’ll have to find the energy of the excited Oxygen Atom. The energy and wavelength can be related by the equation $E = \dfrac{{hc}}{\lambda }$
Where h is the Planck's Constant, C is the speed of light and $\lambda$ is the wavelength.

Complete Step By Step Answer:
We are given the chemical reaction of Dissociation of Oxygen Molecules. Consider the first case where Oxygen Molecule dissociates into two normal oxygen atoms:
${O_2} \to {O_{normal}} + {O_{normal}}({E_{Diss}} = {E_1} = 498kJ/mol)$
${E_1} = 498 \times {10^3}J/mol$
We are given information related to molecules and not molecules. To find the energy in Joules per molecule we’ll divide it by Avogadro’s number.
${E_1} = \dfrac{{498 \times {{10}^3}}}{{6.022 \times {{10}^{23}}}}J/molecule$
In the second case we have a reaction where the Oxygen Molecule is dissociated into one normal and one excited oxygen atom. The reaction can be given as:
${O_2} \to {O_{normal}} + O_{excited}^*({E_{diss}} = {E_2})$
The information given to us is that the energy of the excited atom is $1.967eV$ more than the normal oxygen atom, it can be said that the difference between ${E_1}\& {E_2}$ is $1.967eV$
${E_2} - {E_1} = 1.967eV$
For converting eV into Joules we’ll have to multiply it by $1.6 \times {10^{ - 19}}$
${E_2} - {E_1} = 1.967 \times 1.6 \times {10^{ - 19}}J$
Substituting the value for ${E_1}$ we’ll get:
${E_2} - \dfrac{{498 \times {{10}^3}}}{{6.022 \times {{10}^{23}}}} = 1.967 \times 1.6 \times {10^{ - 19}}$
${E_2} - 8.27 \times {10^{ - 19}} = 3.15 \times {10^{ - 19}}$
${E_2} = 11.42 \times {10^{ - 19}}J/molecule$
To find the wavelength of one molecule of Oxygen, we’ll use the Plank’s Equation.
${E_2} = \dfrac{{hc}}{\lambda } = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda }$
$\lambda = \dfrac{{6.626 \times 3 \times {{10}^{ - 7}}}}{{11.42}} = 1.740 \times {10^{ - 7}}m$
Converting metres into nanometres; $\lambda = 174 \times {10^{ - 9}}m = 174nm$
Therefore, the wavelength for photochemical dissociation of ${O_2}$ is found to be 174nm.

Note :
Remember the following conversions:
Kilojoules to joules: $1kJ = {10^3}J$
Meters to Nanometres/Armstrong: $1m = {10^9}nm = {10^{10}}\mathop A\limits^0$
Electron Volts to Joules: $1eV = 1.6 \times {10^{ - 19}}J$
If we asked the frequency instead of wavelength we can modify the Plank’s Equation to find the relationship Energy and Frequency: $E = h\nu$ (where $\nu = frequency$ ). The relationship between wavelength and frequency can be given as: $\nu = \dfrac{c}{\lambda }$
Where, $\nu$ is the frequency and $\lambda$ is the wavelength and c is the velocity of light.