Photon having wavelength `310nm` is used to break the bond of `A_(2)` molecule having bond energy `28kg mol^(-1)` then `%` of energy of photon converted to the `K.E.` is `[hc=12400 evÅ,1ev=96kJ //mol]`

Options

A. 25
B. 50
C. 75
D. 80

Explanation

To solve the problem, we need to find out what percentage of the energy of a photon is converted into kinetic energy when it breaks the bond of an \( A_2 \) molecule. Here are the steps to find the solution:

### Step 1: Convert the wavelength from nanometers to angstroms
Given the wavelength \( \lambda = 310 \, \text{nm} \):
\[ \lambda = 310 \, \text{nm} = 3100 \, \text{Å} \]
(Hint: Remember that \( 1 \, \text{nm} = 10 \, \text{Å} \))

### Step 2: Calculate the energy of one photon
Using the formula for the energy of a photon:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant in eV·Å (given as \( 12400 \, \text{eV·Å} \))
- \( \lambda \) is the wavelength in Å (calculated in Step 1)

Substituting the values:
\[ E = \frac{12400 \, \text{eV·Å}}{3100 \, \text{Å}} = 4 \, \text{eV} \]
(Hint: Make sure to substitute the correct values for \( h \) and \( \lambda \))

### Step 3: Convert the energy from eV to kJ/mol
We know that \( 1 \, \text{eV} = 96 \, \text{kJ/mol} \):
\[ E = 4 \, \text{eV} \times 96 \, \text{kJ/mol} = 384 \, \text{kJ/mol} \]
(Hint: Use the conversion factor provided in the question)

### Step 4: Calculate the bond energy of the \( A_2 \) molecule
The bond energy given is \( 28 \, \text{kJ/mol} \).

### Step 5: Calculate the kinetic energy of the photon after breaking the bond
The kinetic energy (K.E.) of the photon after breaking the bond is given by:
\[ \text{K.E.} = E - \text{Bond Energy} \]
Substituting the values:
\[ \text{K.E.} = 384 \, \text{kJ/mol} - 28 \, \text{kJ/mol} = 356 \, \text{kJ/mol} \]
(Hint: Make sure to subtract the bond energy from the total energy of the photon)

### Step 6: Calculate the percentage of energy converted to kinetic energy
The percentage of energy converted to kinetic energy is calculated as follows:
\[ \text{Percentage} = \left( \frac{\text{K.E.}}{E} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage} = \left( \frac{356 \, \text{kJ/mol}}{384 \, \text{kJ/mol}} \right) \times 100 \approx 92.71\% \]
(Hint: Divide the kinetic energy by the total energy and multiply by 100 to get the percentage)

### Final Answer
The percentage of energy of the photon converted to kinetic energy is approximately \( 92.71\% \).