`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 during the formation of \( \text{He}^{2+} \) by \( \text{He} + 2e^{-} \), we can break it down into a few steps:

### Step 1: Understand the Ionization Energies
We are given the ionization energies (IE) for Hydrogen (H) and Helium (He):
- \( IE_1 \) for H = 13.6 eV
- \( IE_1 \) for He = 24.6 eV

### Step 2: Formation of \( \text{He}^{2+} \)
The formation of \( \text{He}^{2+} \) involves removing two electrons from a neutral Helium atom. The process can be represented as:
\[ \text{He} \rightarrow \text{He}^+ + e^- \]
\[ \text{He}^+ \rightarrow \text{He}^{2+} + e^- \]

### Step 3: Calculate the Energy Required for Each Step
1. The energy required to remove the first electron from Helium (to form \( \text{He}^+ \)) is equal to the ionization energy of Helium, which is 24.6 eV.
2. The energy required to remove the second electron from \( \text{He}^+ \) (to form \( \text{He}^{2+} \)) can be calculated using the formula for hydrogen-like atoms:
\[ E = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
For \( \text{He}^{+} \), \( Z = 2 \) and \( n = 1 \):
\[ E = -\frac{2^2 \cdot 13.6 \, \text{eV}}{1^2} = -54.4 \, \text{eV} \]
The energy required to remove the second electron (ionization energy of \( \text{He}^+ \)) is 54.4 eV.

### Step 4: Total Energy Required
The total energy required to form \( \text{He}^{2+} \) from neutral Helium is the sum of the energies required for both ionization steps:
\[ \text{Total Energy} = 24.6 \, \text{eV} + 54.4 \, \text{eV} = 79.0 \, \text{eV} \]

### Final Answer
The energy required during the formation of \( \text{He}^{2+} \) is **79.0 eV**.

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