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Q.

The ground state of  O2 has two unpaired electrons with parallel spins. There are two known low – lying excited states of O2. State (A) has the two π* electrons with the spins antiparallel but in different orbitals. State (B) has the two π* electrons paired in the same orbital. The energies of the excited states are  95.00 KJ/mol and  157.85 kJ/mol. Above the ground state.
 What is the wavelength (in μm) of the radiation absorbed in the transition from the ground state to the  first excited state in [Assume Na=6.0×1023,h=6.6×1034,c=3×108 ]

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answer is 1.25.

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Detailed Solution

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The value of   h=6.6×1034; C=3×108
Avagadro’s number = 6×1023  and  E=95.00 KJ
     6.6×1034×3×108×6×102395000J/mol
 =1.25×106m=1.25μm
 

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