A) 7/16 R
B) 3R/16
C) 16/3 R
D) 9R/16
Correct Answer: C
Solution :
\[\frac{1}{\lambda }=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]=\frac{3R}{16}\] \[\therefore \] \[\lambda =\left( \frac{16}{3R} \right)\]You need to login to perform this action.
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