Considering no electronic repulsion in Helium atom what will be Bohr's energy for both the electrons ? Bohr's energy of ground state of `H-` atom is `- 13.6 eV`

Options

A. `- 13.6 eV`
B. `- 27.2 eV`
C. `- 108.8 eV`
D. `- 54.4 eV`

Explanation

To find the Bohr's energy for both electrons in a helium atom considering no electronic repulsion, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Bohr's Energy Formula**:
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \(E_n\) is the energy at the nth level, \(Z\) is the atomic number, and \(n\) is the principal quantum number (energy level).

2. **Identify the Atomic Number for Helium**:
Helium has an atomic number \(Z = 2\).

3. **Determine the Energy for the First Electron**:
For the first electron in the ground state (\(n = 1\)):
\[ E_1 = -\frac{13.6 \, \text{eV} \cdot 2^2}{1^2} = -\frac{13.6 \, \text{eV} \cdot 4}{1} = -54.4 \, \text{eV} \]

4. **Calculate the Energy for the Second Electron**:
Since we are considering no electronic repulsion, the second electron will also have the same energy as the first electron in the ground state:
\[ E_2 = -54.4 \, \text{eV} \]

5. **Total Energy for Both Electrons**:
To find the total energy of the system, we sum the energies of both electrons:
\[ E_{\text{total}} = E_1 + E_2 = -54.4 \, \text{eV} + (-54.4 \, \text{eV}) = -108.8 \, \text{eV} \]

### Final Answer:
The total Bohr's energy for both electrons in a helium atom, considering no electronic repulsion, is:
\[ \boxed{-108.8 \, \text{eV}} \]