A) \[\frac{16}{R}\]
B) \[\frac{16}{3R}\]
C) \[\frac{16}{5R}\]
D) \[\frac{16}{7R}\]
Correct Answer: B
Solution :
\[\frac{1}{\lambda }=R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]=R\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(4)}^{2}}} \right]\] Þ \[\lambda =\frac{16}{3R}\]You need to login to perform this action.
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