A) \[\frac{3R}{4}\]
B) \[\frac{27R}{4}\]
C) \[\frac{4}{3R}\]
D) \[\frac{4}{27R}\]
Correct Answer: D
Solution :
\[\frac{1}{\lambda }=R{{Z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] where, \[Z=3\] (\[\because \] for \[\text{L}{{\text{i}}^{\text{2+}}}\]atomic no. =3) \[{{n}_{2}}=2,\,\,{{n}_{1}}=1\] \[\frac{1}{\lambda }=R{{(3)}^{2}}\left( \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right)\] \[\lambda =\frac{R\times 9\times 3}{4}\] \[\lambda =\frac{4}{27R}\]
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