A) \[\frac{16}{3R}\]
B) \[\frac{16}{5R}\]
C) \[\frac{5R}{16}\]
D) \[\frac{3R}{16}\]
Correct Answer: A
Solution :
[a] \[\frac{1}{\lambda }=R\,\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[{{n}_{1}}=2,{{n}_{2}}=4\] \[\frac{1}{\lambda }=R\,\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(4)}^{2}}} \right]\] \[\frac{1}{\lambda }=R\,\left[ \frac{1}{4}-\frac{1}{16} \right]\] \[\lambda =\frac{16}{3R}\] Note: use \[\frac{1}{R}=911\overset{o}{\mathop{A}}\,\] If \[\lambda \] asked
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