`IE_(1)` for H and He are 13.6 eV and 24.6 eV respectively. Thus, energy required in eV during the formation of `He^(2+)` by `HerarrHe^(2+)+2e^(-)` is

Answer

To solve the problem of finding the energy required for the formation of \( \text{He}^{2+} \) from neutral helium by removing two electrons, we can follow these steps:

### Step 1: Understand Ionization Energies
The first ionization energy (IE) of an atom is the energy required to remove one electron from a neutral atom. We are given:
- \( \text{IE}_1 \) for Hydrogen (H) = 13.6 eV
- \( \text{IE}_1 \) for Helium (He) = 24.6 eV

### Step 2: Calculate the Energy Required to Remove the First Electron from Helium
The first ionization energy of helium (to form \( \text{He}^+ \)) is given as 24.6 eV. This is the energy required to remove the first electron from a neutral helium atom.

### Step 3: Calculate the Energy Required to Remove the Second Electron from Helium
To find the energy required to remove the second electron from \( \text{He}^+ \) to form \( \text{He}^{2+} \), we can use the formula for hydrogen-like atoms:
\[ \text{IE}_1(\text{He}^+) = Z^2 \times \text{IE}_1(\text{H}) \]
where \( Z \) is the atomic number of helium, which is 2. Thus:
\[ \text{IE}_1(\text{He}^+) = 2^2 \times 13.6 \, \text{eV} = 4 \times 13.6 \, \text{eV} = 54.4 \, \text{eV} \]

### Step 4: Total Energy Required to Remove Two Electrons
Now, we need to sum the energies required to remove both electrons:
1. Energy to remove the first electron from He: 24.6 eV
2. Energy to remove the second electron from \( \text{He}^+ \): 54.4 eV

Thus, the total energy required is:
\[ \text{Total Energy} = 24.6 \, \text{eV} + 54.4 \, \text{eV} = 79 \, \text{eV} \]

### Final Answer
The energy required during the formation of \( \text{He}^{2+} \) by \( \text{He} + 2e^- \) is **79 eV**.

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