O2 undergoes photochemical dissociation into one normal and one excited atoms. The excited atom has 1.967 eV more energy than normal. The dissociation of O2 into two normal atoms requires 498 KJ mole-1. Then the maximum wavelength required for photochemical dissociation of O2 is
O2 undergoes photochemical dissociation into one normal and one excited atoms. The excited atom has 1.967 eV more energy than normal. The dissociation of O2 into two normal atoms requires 498 KJ mole-1. Then the maximum wavelength required for photochemical dissociation of O2 is
The correct option is
C
1740A0
We know as per the Mole concept, one mole of atoms =6.022×1023 atoms
So, energy of one atom can be found out by dividing one by this number.
Eper atom = 498 × 1036.023 × 1023J
= 8.26 × 10−19
Efor exitation = 1.967 × 1.6 × 10−19
= 3.1392 × 10−19
total E = 8.26 × 10−19
3.13 × 10−1911.39 × 10−19J
E = hcλ
λ = hcE = 6.625 × 10−34 3 × 10811.39 × 10−19
= 1.74 × 10−7 × 10−10
= 1740Å
For the calculation part, first, we multiply the upper term and then divide it by the lower term, that is how we are getting this answer, which is correct.
The correct option is
C
1740A0
We know as per the Mole concept, one mole of atoms =6.022×1023 atoms
So, energy of one atom can be found out by dividing one by this number.
Eper atom = 498 × 1036.023 × 1023J
= 8.26 × 10−19
Efor exitation = 1.967 × 1.6 × 10−19
= 3.1392 × 10−19
total E = 8.26 × 10−19
3.13 × 10−1911.39 × 10−19J
E = hcλ
λ = hcE = 6.625 × 10−34 3 × 10811.39 × 10−19
= 1.74 × 10−7 × 10−10
= 1740Å
For the calculation part, first, we multiply the upper term and then divide it by the lower term, that is how we are getting this answer, which is correct.
