A) \[\frac{27}{32}\lambda \]
B) \[\frac{32}{27}\lambda \]
C) \[\frac{2}{3}\lambda \]
D) \[\frac{3}{2}\lambda \]
Correct Answer: A
Solution :
\[\frac{1}{\lambda }=R\,\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] First condition \[\frac{1}{\lambda }=R\,\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]\Rightarrow R=\frac{4}{3\lambda }\] Second condition \[\frac{1}{\lambda '}=R\,\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right]\] \[\Rightarrow \lambda '=\frac{9}{8R}\Rightarrow \lambda '=\frac{9}{8\times \frac{4}{3\lambda }}=\frac{27\lambda }{32}\]
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