Question

Hydrogen atom in ground state is excited by a monochromatic radiation of wavelenth975 ampere. Number of spectral lines in the resulting spectrum emitted will be?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Ionisation}\mathrm{energy}\mathrm{for}\mathrm{hydrogen}\mathrm{atom}=13.6\mathrm{eV} \\ {\mathrm{E}}_{\mathrm{n}}=-\frac{\mathrm{Rhc}}{{\mathrm{n}}^2}=-\frac{13.6}{{\mathrm{n}}^2}\mathrm{eV} \\ \mathrm{For},\mathrm{n}=1;{\mathrm{E}}_1=-13.6\mathrm{eV} \\ \mathrm{For},\mathrm{n}=2;{\mathrm{E}}_2=-3.4\mathrm{eV} \\ \mathrm{For},\mathrm{n}=3;{\mathrm{E}}_3=-1.51\mathrm{eV} \\ \mathrm{For},\mathrm{n}=4;{\mathrm{E}}_4=-0.85\mathrm{eV}\mathrm{and}\mathrm{so}\mathrm{on}. \\ \mathrm{The}\mathrm{enrgy}\mathrm{of}\mathrm{incident}\mathrm{photon}\mathrm{is}, \\ \mathrm{E}=\mathrm{h\nu }=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ \mathrm{E}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}}\mathrm{J} \\ \mathrm{or},\mathrm{E}=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}\times 1.6\times {10}^{-19}}\mathrm{eV} \\ \mathrm{or},\mathrm{E}=12.75\mathrm{eV} \\ \mathrm{When}\mathrm{hydrogen}\mathrm{atom}\mathrm{absorbs}\mathrm{this}\mathrm{incident}\mathrm{photon},\mathrm{let}\mathrm{the}\mathrm{electron}\mathrm{of}\mathrm{ground}\mathrm{state} \\ \mathrm{occupies}\mathrm{the}{\mathrm{n}}^{\mathrm{th}}\mathrm{excited}\mathrm{state}.\mathrm{Then}, \\ \mathrm{E}=-\frac{\mathrm{Rhc}}{{\mathrm{n}}^2}-\left(-\frac{\mathrm{Rhc}}{1^2}\right) \\ \mathrm{E}=\mathrm{Rhc}\left(1-\frac{1}{{\mathrm{n}}^2}\right)=12.75\mathrm{eV} \\ 13.6\left(1-\frac{1}{{\mathrm{n}}^2}\right)=12.75 \\ \left(1-\frac{1}{{\mathrm{n}}^2}\right)=\frac{12.75}{13.6} \\ \frac{1}{{\mathrm{n}}^2}=1-\frac{12.75}{13.6} \\ \frac{1}{{\mathrm{n}}^2}=0.0625 \\ {\mathrm{n}}^2=16 \\ \mathrm{or},\mathrm{n}=4 \end{gathered}$$