A) \[L{{i}^{2+}}\]
B) \[H{{e}^{+}}\]
C) H
D) \[{{H}^{+}}\]
Correct Answer: A
Solution :
\[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right){{Z}^{2}}\] \[=R\left( \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(2)}^{2}}} \right){{Z}^{2}}\] \[\frac{1}{\lambda }=\frac{3}{4}R{{Z}^{2}}\] \[\therefore \] \[\lambda \propto \frac{1}{{{Z}^{2}}}\] \[\therefore \] For shortest \[\lambda ,Z\]must be maximum, which is for \[L{{i}^{2+}}.\]
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