The radius of the shortest orbit in a one-electron system is 0.18 Å for:

(Given Bohr's orbit radius = 0.54 Å)

$Li^{++}$

The correct answer is Option (2) → $Li^{++}$

Using Bohr radius formula,

$r_n=n^2.r_1×\frac{1}{Z}$

The radius of the shortest orbit (n = 1) is,

$r_1=\frac{0.54}{Z}$

$⇒\frac{0.54}{Z}=0.18$

$⇒Z=3$ which correspond to $Li^{++}$