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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