A) \[5.099\ c{{m}^{-1}}\]
B) \[20.497\ c{{m}^{-1}}\]
C) \[40.994\ c{{m}^{-1}}\]
D) \[81.988\ c{{m}^{-1}}\]
Correct Answer: A
Solution :
| [a] \[E\left( =\frac{hc}{\lambda } \right)\propto \frac{{{Z}^{2}}}{{{n}^{2}}}\] Þ \[\lambda \propto \frac{1}{{{Z}^{2}}}\] |
| Hence \[{{\lambda }_{H{{e}^{+}}}}=\frac{20.397}{4}=5.099\,cm\] |
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