Answer:
[tex]\nu=2.92\times 10^{15}\ Hz[/tex]
Explanation:
The expression for the energy of an electron in the nth orbit is:-
[tex]E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]
For transitions:
[tex]Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
[tex]\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
Given, [tex]n_i=3\ and\ n_f=1[/tex]
[tex]\Delta E=2.18\times 10^{-18}(\frac{1}{3^2} - \dfrac{1}{1^2})\ J[/tex]
[tex]\Delta E=-1.94\times 10^{-18}\ J[/tex] (negative sign indicates energy release)
Also, [tex]E=h\times \nu[/tex]
Where,
h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]
[tex]1.94\times 10^{-18}=6.626\times 10^{-34}\times \nu[/tex]
[tex]\nu=2.92\times 10^{15}\ Hz[/tex]
