The energy released during conversion of million atoms of iodine in gaseous state to iodide ions in gaseous state is `4.9xx10^(-13)J`. What is the electron gain enthalpy in eV/atom.
[Report your answer by rounding it up to nearest whole number ]

Answer

To solve the problem, we need to find the electron gain enthalpy of iodine in electron volts per atom, given the energy released during the conversion of a million iodine atoms to iodide ions.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Energy released for 1 million atoms of iodine: \( 4.9 \times 10^{-13} \, \text{J} \)
- Number of atoms: \( 1 \, \text{million} = 10^6 \, \text{atoms} \)

2. **Calculate the Energy Released per Atom**:
- To find the energy released per atom, divide the total energy by the number of atoms:
\[ \text{Energy per atom} = \frac{4.9 \times 10^{-13} \, \text{J}}{10^6} = 4.9 \times 10^{-19} \, \text{J} \]

3. **Convert Joules to Electron Volts**:
- We know that \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).
- To convert the energy per atom from Joules to electron volts, we use the formula:
\[ \text{Energy in eV} = \frac{\text{Energy in J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]
- Substituting the values:
\[ \text{Energy in eV} = \frac{4.9 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 3.0625 \, \text{eV} \]

4. **Round to the Nearest Whole Number**:
- Rounding \( 3.0625 \) to the nearest whole number gives us \( 3 \).

### Final Answer:
The electron gain enthalpy of iodine in eV/atom is approximately **3 eV**.