question_answer 52)

What transition in the hydrogen atom spectrum would have the same wavelength as the Balmer transition \[n=4\] to \[n=2\] of \[H{{e}^{+}}\] spectrum?

Answer:

\[\frac{1}{{{\lambda }_{H{{e}^{+}}}}}=R{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]=R\times 4\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]\] \[\frac{1}{{{\lambda }_{H{{e}^{+}}}}}=\frac{3R}{4}\] .. (i) \[\frac{1}{{{\lambda }_{H}}}=R{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]=\frac{1}{{{\lambda }_{H{{e}^{+}}}}}\] \[\therefore \] \[\frac{3R}{4}=R\times {{1}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[\therefore \] \[{{n}_{1}}=1,{{n}_{2}}=2\] \[\therefore \] The transition will be from \[n=2\] to \[n=1\].