A) \[\frac{3}{4}\times {{10}^{15}}\]
B) \[\frac{3}{16}\times {{10}^{15}}\]
C) \[\frac{3}{16}\times {{10}^{15}}\]
D) \[\frac{9}{16}\times {{10}^{15}}\]
Correct Answer: D
Solution :
Using the relation, \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[={{10}^{-5}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[={{10}^{-5}}\times \left( \frac{1}{4}-\frac{1}{16} \right)\] \[\lambda =\frac{16}{3}\times {{10}^{-5}}\,cm\] \[\therefore \] Frequency, \[\therefore \eta =\frac{c}{\lambda }\] \[=\frac{3\times {{10}^{10}}}{\frac{16}{3}\times {{10}^{-5}}}\] \[=\frac{9}{16}\times {{10}^{15}}Hz\]
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