A) \[\frac{9R}{400}c{{m}^{-1}}\]
B) \[\frac{7R}{144}c{{m}^{-1}}\]
C) \[\frac{3R}{4}c{{m}^{-1}}\]
D) \[\frac{5R}{36}c{{m}^{-1}}\]
Correct Answer: D
Solution :
For Balmer series,\[{{n}_{1}}=2\]and\[{{n}_{2}}=3,\]thus\[\overline{v}={{R}_{H}}\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]=\frac{5R}{36}\]
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