A) \[\frac{25}{16}{{\lambda }_{0}}\]
B) \[\frac{27}{20}{{\lambda }_{0}}\]
C) \[\frac{20}{27}{{\lambda }_{0}}\]
D) \[\frac{16}{25}{{\lambda }_{0}}\]
Correct Answer: C
Solution :
Form the relation \[\frac{1}{{{\lambda }_{0}}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] ?(i) also \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}\] ?(ii) From equations (i) and (ii) we get \[\frac{\lambda }{{{\lambda }_{0}}}=\frac{5R/36}{3R/16}=\frac{20}{27}\] or \[\lambda =\frac{20}{27}{{\lambda }_{0}}\]You need to login to perform this action.
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