Answer with Explanation:
[tex]n_1=2, n_2=3[/tex]
Energy of orbital
[tex]E_n=-\frac{2.18\times 10^{-18}}{n^2}[/tex]
Substitute the values
[tex]E_2=-\frac{2.18\times 10^{-18}}{(2)^2}=-\frac{2.18\times 10^{-18}}{4} J[/tex]
[tex]E_3=-\frac{2.18\times 10^{-18}}{3^2}=-\frac{2.18\times 10^{-18}}{9}[/tex] J
[tex]E_2-E_3=\frac{2.18\times 10^{-18}}{9}-\frac{2.18\times 10^{-18}}{4}[/tex]
[tex]E_2-E_3=2.18\times 10^{-18}(-\frac{1}{4}+\frac{1}{9})=2.18\times 10^{1-8}\times \frac{-5}{36} J[/tex]
Energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital=[tex]=E_2-E_3=-\frac{5}{36}\times 2.18\times 10^{-18} J[/tex]
The energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital=[tex]E_3-E_2=2.18\times 10^{-18}(-\frac{1}{9}+\frac{1}{4})[/tex]
The energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital=[tex]\frac{5}{36}\times 2.18\times 10^{-18} J[/tex]
Therefore, in both process the energy of absorbed photon and emitted photon are same but the process are complementary to each other.
