Atomic StructureHard

Question

O2 undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom, 1.2 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen requires 482.5 kJ/mol. The maximum wavelength e­ffective for photochemical dissociation of O2 is (1 eV = 96.5 kJ/mol)

Options

A.248 nm
B.1033.3 nm
C.1236.2 nm
D.200 nm

Solution

Energy needed for photochemical dissociation

$= 482.5\text{ KJ/mol + 1.2 eV = }\left( \frac{482.5}{96.5} + 1.2 \right)eV = 6.2\text{ eV} $$$\therefore\lambda \approx \frac{1240}{6.2} = 200\text{ nm}$$

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