A photon of 3000 `A^o` is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of `I_2` ​ molecule is 255.5 kJ/mol is:

Options

A. 17%
B. 35.6%
C. 40.5%
D. 18.6%

Explanation

To solve the problem step by step, we need to calculate the energy of a photon, the energy supplied by one mole of photons, and then find the percentage of energy converted to the kinetic energy of iodine atoms.

### Step 1: Calculate the energy of one photon.

The energy of a photon can be calculated using the formula:

\[ E = \frac{12400}{\lambda} \]

where \(E\) is the energy in electron volts (eV) and \(\lambda\) is the wavelength in angstroms (Å).

Given:
\(\lambda = 3000 \, \text{Å}\)

Substituting the value:

\[ E = \frac{12400}{3000} = 4.1333 \, \text{eV} \]

### Step 2: Convert the energy of one photon from eV to Joules.

To convert eV to Joules, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\).

\[ E = 4.1333 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.6133 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the energy supplied by one mole of photons.

The energy supplied by one mole of photons can be calculated using Avogadro's number (\(N_A = 6.02 \times 10^{23}\)):

\[ E_{\text{mole}} = E \times N_A \]

Substituting the values:

\[ E_{\text{mole}} = 6.6133 \times 10^{-19} \, \text{J} \times 6.02 \times 10^{23} \, \text{mol}^{-1} \]

Calculating this gives:

\[ E_{\text{mole}} = 398.2 \, \text{kJ/mol} \]

### Step 4: Calculate the percentage of energy converted to kinetic energy.

The bond dissociation energy of \(I_2\) is given as \(255.5 \, \text{kJ/mol}\). The energy converted to kinetic energy is the difference between the energy supplied by one mole of photons and the bond dissociation energy.

\[ \text{K.E.} = E_{\text{mole}} - \text{Bond Dissociation Energy} \]

Substituting the values:

\[ \text{K.E.} = 398.2 \, \text{kJ/mol} - 255.5 \, \text{kJ/mol} = 142.7 \, \text{kJ/mol} \]

Now, we calculate the percentage of energy converted to kinetic energy:

\[ \text{Percentage} = \left( \frac{\text{K.E.}}{E_{\text{mole}}} \right) \times 100 \]

Substituting the values:

\[ \text{Percentage} = \left( \frac{142.7}{398.2} \right) \times 100 \approx 35.8\% \]

### Final Result:

The percentage of energy converted to the kinetic energy of iodine atoms is approximately **35.8%**.

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