The radius of shortest orbit in one electron system is $18$ pm.It may be.

_{1}^{1} H

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_{2}^{1} H

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H e^{+}

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L i^{+}

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Solution

The correct option is D $L i^{+}$
Given : $r_{1} = 18$ pm $= 0.18$ $A^{o}$
Radius of shortest orbit $r_{1} = \frac{0.529}{Z}$ $A^{o}$ $(n = 1)$
$∴$ $\frac{0.529}{Z} = 0.18$ $⟹ Z = 3$
Thus the ion must be $L i^{+}$ .

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