The energy released during conversion of million atoms of iodine in gaseous state to iodide ions in gaseous state is `4.9xx210^(-13)J`. What is the electron gain enthalpy in eV/atom.
[Report your answer by rounding it up to nearest whole number ]

Answer

To find the electron gain enthalpy (ΔH_e_g) in eV/atom for the conversion of iodine atoms to iodide ions, we can follow these steps:

### Step 1: Understand the given data
We are given that the energy released during the conversion of 1 million (10^6) iodine atoms in the gaseous state to iodide ions is \(4.9 \times 10^{-13} \, \text{J}\).

### Step 2: Calculate the energy released per atom
To find the energy released per atom, we divide the total energy by the number of atoms:

\[ \text{Energy per atom} = \frac{4.9 \times 10^{-13} \, \text{J}}{10^6} \]

Calculating this gives:

\[ \text{Energy per atom} = 4.9 \times 10^{-19} \, \text{J} \]

### Step 3: Convert energy from joules to electron volts
We know that \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\). To convert the energy from joules to electron volts, we use the following formula:

\[ \text{Energy in eV} = \frac{\text{Energy in J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]

Substituting the energy per atom:

\[ \text{Energy in eV} = \frac{4.9 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]

Calculating this gives:

\[ \text{Energy in eV} \approx 3.0625 \, \text{eV} \]

### Step 4: Round to the nearest whole number
Rounding \(3.0625\) to the nearest whole number gives us:

\[ \text{Electron gain enthalpy} \approx 3 \, \text{eV/atom} \]

### Final Answer
The electron gain enthalpy for the conversion of iodine atoms to iodide ions is approximately **3 eV/atom**.

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