A) 40
B) 30
C) 20
D) 10
Correct Answer: A
Solution :
[a] \[\frac{1}{\lambda }=R{{Z}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3}{4}R{{Z}^{2}}\] \[or\text{ }{{Z}^{2}}=\frac{4}{3R\lambda }\] \[=\frac{4}{3\times \left( 1.097\times {{10}^{7}} \right)\times \left( 0.76\times {{10}^{-10}} \right)}=1599.25\] \[or\text{ }{{Z}^{2}}=1600\text{ }\therefore Z=40.\]
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