An electron is revolving in a Bohr orbit of a doubly ionized...

The angular momentum of an electron in the $n^{\text{th}}$ Bohr orbit is given by $L = \dfrac{nh}{2\pi}$.

Given $L = \dfrac{h}{\pi}$, we have:
$\dfrac{nh}{2\pi} = \dfrac{h}{\pi} \Rightarrow n = 2$

The total energy of an electron in a hydrogen-like ion is given by $E = -13.6 \dfrac{Z^2}{n^2} \text{ eV}$.

For a doubly ionized lithium atom ($Li^{2+}$), the atomic number $Z = 3$.

Substituting $Z = 3$ and $n = 2$, we get:
$E = -13.6 \times \dfrac{3^2}{2^2}$
$E = -13.6 \times \dfrac{9}{4} = -30.6 \text{ eV}$

Failing to account for the atomic number $Z$ leads to the incorrect value of $-3.4 \text{ eV}$.

Answer: $-30.6 \text{ eV}$