The energy of an electron in Bohr's orbit of hydrogen atom is `-13.6eV`. The total electronic energy of a hypothetical He atom in which there are no electron - electron repulsions or interactions is

Options

A. 27.2 eV
B. `-27.2 eV`
C. `-108.8 eV`
D. 108. eV

Answer

To solve the problem, we need to determine the total electronic energy of a hypothetical helium atom (He) where there are no electron-electron repulsions or interactions.

### Step-by-Step Solution:

1. **Understanding the Energy of Electrons in Hydrogen Atom**:
The energy of an electron in the first orbit (n=1) of a hydrogen atom is given as:
\[ E_H = -13.6 \, \text{eV} \]

2. **Applying the Formula for Helium**:
For a hydrogen-like atom, the energy of an electron in the nth orbit can be expressed as:
\[ E_n = -\frac{13.6 \, \text{eV} \times Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

3. **Substituting Values for Helium**:
Helium has an atomic number \( Z = 2 \). For the ground state (n=1):
\[ E_{He} = -\frac{13.6 \, \text{eV} \times 2^2}{1^2} = -\frac{13.6 \, \text{eV} \times 4}{1} = -54.4 \, \text{eV} \]

4. **Calculating Total Energy for Two Electrons**:
Since there are two electrons in helium and we are assuming no electron-electron repulsions, the total energy for the two electrons will be:
\[ E_{total} = 2 \times E_{He} = 2 \times (-54.4 \, \text{eV}) = -108.8 \, \text{eV} \]

5. **Final Answer**:
Therefore, the total electronic energy of the hypothetical helium atom is:
\[ \boxed{-108.8 \, \text{eV}} \]