O2 undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen requires 498 KJmol−1.what is the maximum wavelength effective for photochemical dissociation of O2 ?
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O
2
→
O
Normal
+
O
Excited
O2→ONormal+OExcited
O
2
→
O
Normal
+
O
Normal
O2→ONormal+ONormal
Energy required for simple dissociation of
O
2
O2
into two normal atoms
=498×
10
3
J/mol
=498×103J/mol
=
498×
10
3
6.023×
10
23
=498×1036.023×1023
J/Molecule
If one atom is excited state has more energy i.e 1.967 eV
=1.967×1.602×
10
−19
J
=1.967×1.602×10−19J
Energy required for photochemical dissociation of
O
2
O2
=
498×
10
3
6.023×
10
23
+1.967×1.602×
10
−19
=498×1036.023×1023+1.967×1.602×10−19
=82.68×
10
−20
+31.51×
10
−20
=82.68×10−20+31.51×10−20
=114.19×
10
−20
J
=114.19×10−20J
∴E=
hc
λ
∴E=hcλ
114.19×
10
−20
=
6.625×
10
−34
×3.0×
10
8
λ
114.19×10−20=6.625×10−34×3.0×108λ
λ=1740.2×
10
−10
m
λ=1740.2×10−10m
λ=1740.2
A
λ=1740.2A
2
→
O
Normal
+
O
Excited
O2→ONormal+OExcited
O
2
→
O
Normal
+
O
Normal
O2→ONormal+ONormal
Energy required for simple dissociation of
O
2
O2
into two normal atoms
=498×
10
3
J/mol
=498×103J/mol
=
498×
10
3
6.023×
10
23
=498×1036.023×1023
J/Molecule
If one atom is excited state has more energy i.e 1.967 eV
=1.967×1.602×
10
−19
J
=1.967×1.602×10−19J
Energy required for photochemical dissociation of
O
2
O2
=
498×
10
3
6.023×
10
23
+1.967×1.602×
10
−19
=498×1036.023×1023+1.967×1.602×10−19
=82.68×
10
−20
+31.51×
10
−20
=82.68×10−20+31.51×10−20
=114.19×
10
−20
J
=114.19×10−20J
∴E=
hc
λ
∴E=hcλ
114.19×
10
−20
=
6.625×
10
−34
×3.0×
10
8
λ
114.19×10−20=6.625×10−34×3.0×108λ
λ=1740.2×
10
−10
m
λ=1740.2×10−10m
λ=1740.2
A
λ=1740.2A
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