The ground state of $O_{2}$ has two unpaired electrons with parallel spins. There are two known low – lying excited states of $O_{2}$ . 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 $95.00 K J / m o l$ and $157.85 kJ / m o l .$ Above the ground state.
What is the wavelength (in $μ m$ ) of the radiation absorbed in the transition from the ground state to the first excited state in [Assume $N_{a} = 6.0 \times 10^{23}, h = 6.6 \times 10^{- 34}, c = 3 \times 10^{8}$ ]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 1.25.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The value of $h = 6.6 \times 10^{- 34}; C = 3 \times 10^{8}$
Avagadro’s number = $6 \times 10^{23}$ and $E = 95.00 K J$
$∴ \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8} \times 6 \times 10^{23}}{95000 J / m o l}$
$\begin{array}{l} = 1.25 \times 10^{- 6} m \\ = 1.25 μ m \end{array}$