A) \[\frac{16}{25}{{\lambda }_{0}}\]
B) \[\frac{20}{27}{{\lambda }_{0}}\]
C) \[\frac{27}{20}{{\lambda }_{0}}\]
D) \[\frac{25}{16}{{\lambda }_{0}}\]
Correct Answer: B
Solution :
\[\frac{1}{\lambda }=R\,\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] Þ \[\frac{1}{{{\lambda }_{3\to 2}}}=R\,\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]=\frac{5R}{36}\] and \[\frac{1}{{{\lambda }_{4\to 2}}}=R\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(4)}^{2}}} \right]=\frac{3R}{16}\] \ \[\frac{{{\lambda }_{4\to 2}}}{{{\lambda }_{3\to 2}}}=\frac{20}{27}\] Þ \[{{\lambda }_{4\to 2}}=\frac{20}{27}{{\lambda }_{0}}\]
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