Question

The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at:

• 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}} $

Answer 1

The Correct Option is A

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}} $