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++
In Bohr's model the radius of the nth orbit of a one-electron system of atomic number Z is \( r_n = r_1\,\frac{n^{2}}{Z} \), where \( r_1 \) is the Bohr radius (the radius of the first orbit of hydrogen). The shortest orbit is the first, n = 1.
\( \displaystyle r = \frac{r_1}{Z} \Rightarrow Z = \frac{r_1}{r} = \frac{0.54\,\mathrm{\mathring{A}}}{0.18\,\mathrm{\mathring{A}}} = 3 \)
The one-electron system with \( Z = 3 \) is the doubly ionised lithium ion, Li++.