question_answer 36)

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer:

\[\overset{\_}{\mathop{v}}\,=\frac{1}{\lambda }=R{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] (i) For hydrogen, \[z=1,\,\,\,\,\,\,\,R=109676\,c{{m}^{-1}}\] \[{{n}_{1}}=2,\,\,\,\,{{n}_{2}}=3\](for longest wavelength line) \[\overset{\_}{\mathop{v}}\,=109676\times {{1}^{2}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]\] \[=1.523\times {{10}^{4}}c{{m}^{-1}}\] \[=1.523\times {{10}^{6}}{{m}^{-1}}\]