Ioline molecule dissociated into atom after abesrbing light of `4500 Å` If can quantum of radiation is absorbed by each molecule, energy of `l_(2) = 240 kJ mol^(-1)`

Answer

To solve the problem step by step, we will calculate the energy absorbed by the iodine molecule when it absorbs light of wavelength 4500 Å, then determine the bond energy of the iodine molecule, and finally calculate the kinetic energy of the iodine atoms produced after dissociation.

### Step 1: Calculate the Energy Absorbed by the Iodine Molecule

The energy \( E \) absorbed by the molecule can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 4500 \, \text{Å} = 4500 \times 10^{-10} \, \text{m} \)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4500 \times 10^{-10}} \]

Calculating this gives:

\[ E \approx 4.417 \times 10^{-19} \, \text{J} \]

### Step 2: Convert Bond Energy from kJ/mol to J/molecule

The bond energy given is \( 240 \, \text{kJ/mol} \). To convert this to energy per molecule, we use Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \):

\[ \text{Bond Energy per molecule} = \frac{240 \times 10^3 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \]

Calculating this gives:

\[ \text{Bond Energy per molecule} \approx 3.984 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the Kinetic Energy of the Iodine Atoms

The energy absorbed by the iodine molecule is used to break the bond and provide kinetic energy to the resulting iodine atoms. The kinetic energy \( KE \) of the atoms can be calculated as follows:

\[ KE = E - \text{Bond Energy} \]

Substituting the values:

\[ KE = 4.417 \times 10^{-19} \, \text{J} - 3.984 \times 10^{-19} \, \text{J} \]

Calculating this gives:

\[ KE \approx 0.433 \times 10^{-19} \, \text{J} \]

Since one iodine molecule dissociates into two iodine atoms, the kinetic energy per atom is:

\[ KE_{\text{per atom}} = \frac{KE}{2} = \frac{0.433 \times 10^{-19}}{2} \approx 0.2165 \times 10^{-19} \, \text{J} \]

### Final Answer

The kinetic energy of each iodine atom after dissociation is approximately:

\[ KE_{\text{per atom}} \approx 0.216 \times 10^{-19} \, \text{J} \]

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