Assume that there is no repulsive force between the electrons in an atom,but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances,calculate the ground state energy of a $He$ atom.

(N/A) As per the assumption made in the statement,since the $He$ atom behaves like a hydrogen-like atom (where electron-electron repulsion is ignored),we can apply Bohr's atomic model to the $He$ atom.
The energy of an electron in an orbit is given by:
$E_{n} = -\frac{m Z^{2} e^{4}}{8 \epsilon_{0}^{2} n^{2} h^{2}}$
This can be simplified as:
$E_{n} = -13.6 \frac{Z^{2}}{n^{2}} \text{ eV}$
For a $He$ atom,the atomic number $Z = 2$. For the ground state,the principal quantum number $n = 1$.
Substituting these values:
$E_{1} = -13.6 \times \frac{(2)^{2}}{(1)^{2}} \text{ eV}$
$E_{1} = -13.6 \times 4 \text{ eV}$
$E_{1} = -54.4 \text{ eV}$
Since there are two electrons in the $He$ atom and we assume no repulsion between them,the total ground state energy of the $He$ atom is the sum of the energies of both electrons:
$E_{\text{total}} = 2 \times E_{1} = 2 \times (-54.4 \text{ eV}) = -108.8 \text{ eV}$.