Question

The potential energy of the electron present in the ground state of $L{i}^{2+}$ ion is represented by:

• A. $\frac{-3e^2}{4\pi {\epsilon }_{0}r}$
• B. $\frac{3e}{4\pi {\epsilon }_{0}r}$
• C. $\frac{-2e^2}{4\pi {\epsilon }_0{r}^2}$
• D. $-\frac{-e^2}{4\pi {\epsilon }_{0}r}$

Answer 1

The correct option is A $\frac{-3e^2}{4\pi {\epsilon }_{0}r}$

As we know,
the potential energy of an electron in Bohr atomic model is

$PE=\frac{-Z{e}^2}{4\pi {\epsilon }_{0}r}$

for $L{i}^{2+}$, $Z=3$ (atomic no.)

so for $L{i}^{2+}$,

$PE=\frac{-3e^2}{4\pi {\epsilon }_{0}r}$

Hence, the correct option is $A$