Question: 90: If radius of second Bohr orbit of the \mathrm{He}^{+} ion is 105.8 pm , what is the radius of third Bohr orbit of \mathrm{Li}^{2+} ion?
(1) 1.587 pm
(2) 158.7 \AA
(3) 158.7 pm
(4) 15.87 pm
Answer: Option (3)
Explanation:
The radius of the n^{\text{th}} Bohr orbit for a hydrogen-like ion is given by
r_n=\frac{a_0 n^2}{Z}where a_0 is the Bohr radius, n is the principal quantum number,
and Z is the atomic number.
Since a_0 is constant, the radius is proportional to
r \propto \frac{n^2}{Z}For \mathrm{He}^{+}, n=2 and Z=2,
and the radius is given as 105.8 pm.
For \mathrm{Li}^{2+}, n=3 and Z=3.
Using the ratio,
\frac{r_{\mathrm{Li}^{2+}}}{r_{\mathrm{He}^{+}}}=\frac{(3^2/3)}{(2^2/2)} \frac{r_{\mathrm{Li}^{2+}}}{105.8}=\frac{3}{2} r_{\mathrm{Li}^{2+}}=105.8 \times \frac{3}{2} r_{\mathrm{Li}^{2+}}=158.7 \, \mathrm{pm}Therefore, the radius of the third Bohr orbit of \mathrm{Li}^{2+} ion is 158.7 \, \mathrm{pm}.