• question_answer The wavelength of a light emitting from second orbit to the first orbit in a hydrogen atom will be: A)  $1.215\times {{10}^{-7}}m$        B)  $1.215\times {{10}^{-3}}m$C)  $1.215\times {{10}^{-5}}m$        D)  none of these

Initial orbit ${{n}_{1}}=1,$ final orbit Tie ${{n}_{2}}=2$ The relation for wavelength is given by $\frac{1}{\lambda }=R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]=R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]=\frac{3R}{4}$ $\Rightarrow$ $\lambda =\frac{4}{3R}=\frac{4}{3\times 1.098\times {{10}^{7}}}=1.215\times {{10}^{-7}}$ (Where Rydberg constant                         $R=1.098\times {{10}^{7}}/m$).