Question

What is the frequency of light emitted when an electron in a hydrogen atom jumps from the 3^rd orbit to the 2^nd orbit?

Answer 1

$\frac{-2{\mathrm{\pi }}^2{\mathrm{me}}^4{\mathrm{Z}}^2}{{\mathrm{n}}^2{\mathcal{h}}^2}$is the equation for the calculation of the energy of the nth orbit of the hydrogen atom derived by Bohr.

Where m= mass of electron=$(9.1\times {10}^{-28}\mathrm{g})$

e= charge of electron$(1.6\times {10}^{-19}\mathrm{C})$

n= number of orbits

$\mathcal{h}$= Plank's constant$(6.625\times {10}^{-34}\mathrm{Js})$

For hydrogen:- atomic number (Z)=1

By applying all these values, the equation for the calculation of the energy of the nth orbit of the hydrogen atom is derived by Bohr is$\frac{-13.6}{{\mathrm{n}}^2}\mathrm{eV}=\frac{-21.37\times {10}^{-19}}{{\mathrm{n}}^2}\mathrm{Joule}$

Difference in energy= ${\mathrm{E}}_3-{\mathrm{E}}_2=-21.37\times {10}^{-19}\left[\frac{1}{3^2}-\frac{1}{2^2}\right]\mathrm{Joule}=-21.37\times {10}^{-19}\left[\frac{1}{9}-\frac{1}{4}\right]\mathrm{Joule}=3.01\times {10}^{-19}\mathrm{joule}$

According to planks equation $∆\mathrm{E}=\mathcal{h}\nu$

$$\begin{gathered} {\mathrm{E}}_3-{\mathrm{E}}_2=∆\mathrm{E}=3.01\times {10}^{-19}\mathrm{joule}=6.625\times {10}^{-34}\mathrm{Js}\times \mathrm{\nu } \\ \mathrm{\nu }=\frac{3.01\times {10}^{-19}\mathrm{joule}}{6.625\times {10}^{-34}\mathrm{Js}}=4.5\times {10}^{14}{\mathrm{s}}^{-1} \end{gathered}$$

The frequency of light emitted when an electron in a hydrogen atom jumps from the 3^rd Orbit to the 2^nd Orbit is $4.5\times {10}^{14}{\mathrm{s}}^{-1}$.