Calculate the energy in calories required to produce, from neutral He atoms, 1 mole of (a) `He^(+)` ions (b ) `'He^(+ +)` ions using Bohr's equations

Explanation

To calculate the energy required to produce 1 mole of \( \text{He}^+ \) ions and 1 mole of \( \text{He}^{2+} \) ions from neutral helium atoms using Bohr's equations, we will follow these steps:

### Step 1: Understand the Ionization Process
- **Neutral Helium Atom**: The neutral helium atom has an atomic number \( Z = 2 \).
- **Ionization**: To form \( \text{He}^+ \), one electron is removed from the neutral helium atom. To form \( \text{He}^{2+} \), two electrons are removed.

### Step 2: Calculate the Ionization Energy for \( \text{He}^+ \)
- The formula for the ionization energy (IE) in the nth orbit is given by:
\[ \text{IE} = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
- For \( \text{He}^+ \):
- \( Z = 2 \)
- \( n = 1 \)

Substituting these values:
\[ \text{IE}_{\text{He}^+} = -\frac{13.6 \cdot 2^2}{1^2} = -\frac{13.6 \cdot 4}{1} = -54.4 \, \text{eV} \]

### Step 3: Convert Ionization Energy to Calories
- To convert electron volts to calories, we use the conversion factor:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
\[ 1 \, \text{cal} = 4.184 \, \text{J} \]

Therefore, the conversion from eV to calories is:
\[ 1 \, \text{eV} = \frac{1.6 \times 10^{-19}}{4.184} \approx 3.83 \times 10^{-20} \, \text{cal} \]

- Now, converting \( -54.4 \, \text{eV} \) to calories:
\[ \text{Energy in calories} = -54.4 \times 3.83 \times 10^{-20} \approx -2.08 \times 10^{-19} \, \text{cal} \]

### Step 4: Calculate the Energy for 1 Mole of \( \text{He}^+ \)
- Since we need the energy for 1 mole (Avogadro's number \( N_A = 6.022 \times 10^{23} \)):
\[ \text{Energy for 1 mole} = -2.08 \times 10^{-19} \times 6.022 \times 10^{23} \approx -125.1 \, \text{cal} \]

### Step 5: Calculate the Ionization Energy for \( \text{He}^{2+} \)
- For \( \text{He}^{2+} \):
- \( Z = 2 \)
- \( n = 1 \)

Using the same formula:
\[ \text{IE}_{\text{He}^{2+}} = -\frac{13.6 \cdot 2^2}{1^2} = -54.4 \, \text{eV} \]

### Step 6: Convert Ionization Energy for \( \text{He}^{2+} \) to Calories
- The energy in calories for \( \text{He}^{2+} \) is the same as for \( \text{He}^+ \) since the calculation is identical:
\[ \text{Energy for 1 mole of } \text{He}^{2+} = -125.1 \, \text{cal} \]

### Final Results
- Energy required to produce 1 mole of \( \text{He}^+ \): **-125.1 cal**
- Energy required to produce 1 mole of \( \text{He}^{2+} \): **-125.1 cal**