One mole of `He^(o+)` ions is excited. An anaylsis showed that `50%` of ions are in the third energy level `25%` are in the second energy level and the remaining are in the first energy level. Calculate the energy emitted in kilojoules when all the ions return to the ground state.

Explanation

To solve the problem of calculating the energy emitted when one mole of `He^(o+)` ions return to the ground state, we will follow these steps:

### Step 1: Identify the Energy Levels
We know that:
- 50% of the ions are in the third energy level (n=3).
- 25% of the ions are in the second energy level (n=2).
- The remaining 25% of the ions are in the first energy level (n=1).

### Step 2: Calculate the Ionization Potential
The ionization potential (IP) for `He^(o+)` can be calculated using the formula:
\[ \text{IP} = -13.6 \times Z^2 \]
where \( Z \) is the atomic number. For helium, \( Z = 2 \):
\[ \text{IP} = -13.6 \times 2^2 = -13.6 \times 4 = -54.4 \text{ eV} \]

### Step 3: Calculate the Energy Change for Each Transition
Using the formula for energy change between two levels:
\[ \Delta E = -\text{IP} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

#### Transition from n=3 to n=1:
For the transition from n=3 to n=1:
\[ \Delta E_{3 \to 1} = 54.4 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ = 54.4 \left( 1 - \frac{1}{9} \right) \]
\[ = 54.4 \left( \frac{8}{9} \right) \]
\[ = 48.64 \text{ eV} \]

#### Transition from n=2 to n=1:
For the transition from n=2 to n=1:
\[ \Delta E_{2 \to 1} = 54.4 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ = 54.4 \left( 1 - \frac{1}{4} \right) \]
\[ = 54.4 \left( \frac{3}{4} \right) \]
\[ = 40.8 \text{ eV} \]

### Step 4: Calculate Total Energy Emitted
Now, we need to calculate the total energy emitted when all ions return to the ground state.

- For the ions in the third energy level (50% of 1 mole):
\[ \text{Number of ions} = 0.5 \times 6.022 \times 10^{23} \]
\[ \Delta E_{total, 3 \to 1} = 0.5 \times 6.022 \times 10^{23} \times 48.64 \text{ eV} \]

- For the ions in the second energy level (25% of 1 mole):
\[ \text{Number of ions} = 0.25 \times 6.022 \times 10^{23} \]
\[ \Delta E_{total, 2 \to 1} = 0.25 \times 6.022 \times 10^{23} \times 40.8 \text{ eV} \]

### Step 5: Combine the Energies
Now, we can combine the energies:
\[ \Delta E_{total} = \Delta E_{total, 3 \to 1} + \Delta E_{total, 2 \to 1} \]

### Step 6: Convert to Kilojoules
To convert the total energy from electron volts to kilojoules:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules} \]
\[ \Delta E_{total} \text{ (in Joules)} = \Delta E_{total} \text{ (in eV)} \times 1.6 \times 10^{-19} \]
\[ \Delta E_{total} \text{ (in kJ)} = \frac{\Delta E_{total} \text{ (in Joules)}}{1000} \]

### Final Calculation
After performing all the calculations, we find:
\[ \Delta E_{total} \approx 331.13 \text{ kJ} \]