Question

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

• A. Hydrogen
• B. Deuterium
• C. $H{e}^{+}$
• D. $L{i}^{++}$

Answer 1

The correct option is D $L{i}^{++}$

Radius $r=a_o\times \frac{n^2}{Z}$
where $a_o$ is the Bohr's radius for hydrogen atom at $n=1$ level
which is $53pm$. ($pm$ for picometer)

Also for shortest orbit $n=1$
Putting the values given in the question
$18=53\times \frac{1}{Z}$

This gives $Z=3$

For $L{i}^{++}$, we have $Z=3$.