A) \[\frac{5R}{36}c{{m}^{-1}}\]
B) \[\frac{3R}{4}c{{m}^{-1}}\]
C) \[\frac{7R}{144}c{{m}^{-1}}\]
D) \[\frac{9R}{400}c{{m}^{-1}}\]
Correct Answer: A
Solution :
For Balmer series in the atomic spectrum of hydrogen \[{{n}_{1}}=2\]and \[{{n}_{2}}=3,\]the \[v={{R}_{H}}\left[ \frac{1}{\eta _{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[={{R}_{H}}\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right]\] \[={{R}_{H}}\left[ \frac{1}{4}-\frac{1}{9} \right]\] \[={{R}_{H}}\left[ \frac{9-4}{36} \right]\] \[={{R}_{H}}\frac{5}{36}\] \[=\frac{5R}{36}c{{m}^{-1}}\]You need to login to perform this action.
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