Energy required to ionise 1 model of gaseous`He^+ion` present in its ground state is

Options

A. `108.8 N_AeV`
B. `13.6 `eV
C. `54.4` eV
D. `54.4 N_A eV`

Answer

To find the energy required to ionize one mole of gaseous He⁺ ions in their ground state, we can use the formula derived from the Bohr model of the hydrogen-like atoms. Helium ion (He⁺) has a nuclear charge (Z) of +2 since it has two protons.

### Step-by-Step Solution:

1. **Identify the Formula**:
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{Z^2 \cdot R_H}{n^2} \]
where:
- \( E_n \) = energy of the electron at level n,
- \( Z \) = atomic number (for He⁺, Z = 2),
- \( R_H \) = Rydberg constant for hydrogen, approximately \( 13.6 \, \text{eV} \),
- \( n \) = principal quantum number (for ground state, \( n = 1 \)).

2. **Substitute Values**:
For He⁺ in the ground state (n=1):
\[ E_1 = -\frac{(2)^2 \cdot 13.6 \, \text{eV}}{(1)^2} \]
\[ E_1 = -\frac{4 \cdot 13.6 \, \text{eV}}{1} \]
\[ E_1 = -54.4 \, \text{eV} \]

3. **Calculate Ionization Energy**:
The ionization energy is the energy required to remove the electron from the ground state to infinity (where the energy is 0). Therefore, the ionization energy \( I \) is:
\[ I = -E_1 = 54.4 \, \text{eV} \]

4. **Convert to Energy per Mole**:
To find the energy required to ionize one mole of He⁺ ions, we multiply by Avogadro's number (\( N_A \approx 6.022 \times 10^{23} \)):
\[ I_{\text{mole}} = 54.4 \, \text{eV} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \]
To convert eV to joules (1 eV = \( 1.602 \times 10^{-19} \, \text{J} \)):
\[ I_{\text{mole}} = 54.4 \times 1.602 \times 10^{-19} \, \text{J} \times 6.022 \times 10^{23} \]
\[ I_{\text{mole}} \approx 5.22 \times 10^3 \, \text{J/mol} \]

### Final Answer:
The energy required to ionize one mole of gaseous He⁺ ions in their ground state is approximately **5220 J/mol**.