The radius of the shortest orbit in a one-electron system is 0.18 Å for:
(Given Bohr's orbit radius = 0.54 Å)
Answer & explanation
Correct answer: option 2
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 $Li^{++}$