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 :
\[\bar{v}=R\left( \frac{1}{{{n}_{1}}^{2}}-\frac{1}{{{n}_{2}}^{2}} \right)\] For Balmer series: \[{{n}_{1}}=2,\,{{n}_{2}}=3,\,4,\,5,..\infty \] For first emission line \[{{n}_{2}}=3\] \[\therefore \] \[\bar{v}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\] \[=R\left( \frac{1}{4}-\frac{1}{9} \right)=R\left( \frac{5}{36} \right)\] \[\bar{v}=\frac{5R}{36}c{{m}^{-1}}\]
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