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.