Question

The ratio of the difference between 1st and 2nd Bohr's orbits Energy to that between 2nd and 3rd orbits energy orbits energy per H atom is ?

Answer 1

$$\begin{gathered} \text{Energy of an electron is nth orbit of H atom=}\frac{-13.6}{{\mathrm{n}}^2}\mathrm{eV}\mathrm{per}\mathrm{atom} \\ \mathrm{So},\mathrm{energy}\mathrm{of}\mathrm{an}\mathrm{electron}\mathrm{in}1\mathrm{st}\mathrm{orbit},{\mathrm{E}}_1=\frac{-13.6}{(1{)}^2}=-13.6\mathrm{eV} \\ \mathrm{Moreover},{\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{E}}_1}{{\mathrm{n}}^2} \\ \mathrm{Therefore},{\mathrm{E}}_2=\frac{{\mathrm{E}}_1}{(2{)}^2}\mathrm{and}{\mathrm{E}}_3=\frac{{\mathrm{E}}_1}{(3{)}^2} \\ \mathrm{So},{\mathrm{E}}_2=\frac{-13.6}{4}=-3.4\mathrm{eV} \\ \mathrm{And},{\mathrm{E}}_3=\frac{-13.6}{9}=-1.51\mathrm{eV} \\ \mathrm{Difference}\mathrm{between}\mathrm{energy}\mathrm{of}\mathrm{electron}\mathrm{in}1\mathrm{st}\mathrm{and}2\mathrm{nd}\mathrm{orbit}, \\ ∆{\mathrm{E}}_1={\mathrm{E}}_2-{\mathrm{E}}_1 \\ =-3.4-(-13.6)=10.2\mathrm{eV} \\ \mathrm{Difference}\mathrm{between}\mathrm{energy}\mathrm{of}\mathrm{electron}\mathrm{in}2\mathrm{nd}\mathrm{and}3\mathrm{rd}\mathrm{orbit}, \\ ∆{\mathrm{E}}_2={\mathrm{E}}_3-{\mathrm{E}}_2 \\ =-1.51-(-3.4)=1.89\mathrm{eV} \\ \mathrm{So},\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{the}\mathrm{difference}\mathrm{between}1\mathrm{st}\mathrm{and}2\mathrm{nd}\mathrm{orbit}\mathrm{to}\mathrm{that}\mathrm{between}2\mathrm{md}\mathrm{and}3\mathrm{rd}\mathrm{orbit}\mathrm{per}\mathrm{H}\mathrm{atom}\mathrm{is} \\ \frac{∆{\mathrm{E}}_1}{∆{\mathrm{E}}_2}=\frac{10.2}{1.89}=\frac{340}{63} \\ \mathrm{So},∆{\mathrm{E}}_1:∆{\mathrm{E}}_2=340:63 \end{gathered}$$