Bond energy of `F_2` is 150 kJ `mol^(-1)`. Calculate the minimum frequency of photon to break this bond.

Answer

To calculate the minimum frequency of a photon required to break the bond in the `F_2` molecule, we can follow these steps:

### Step 1: Convert Bond Energy from kJ/mol to J/molecule
The bond energy of `F_2` is given as 150 kJ/mol. We need to convert this energy into joules per molecule.

1. **Convert kJ to J**:
\[ 150 \text{ kJ/mol} = 150 \times 10^3 \text{ J/mol} = 150000 \text{ J/mol} \]

2. **Calculate energy per molecule**:
To find the energy for one molecule, we divide the total energy by Avogadro's number (\(N_A \approx 6.022 \times 10^{23} \text{ molecules/mol}\)):
\[ \text{Energy per molecule} = \frac{150000 \text{ J/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 2.49 \times 10^{-19} \text{ J} \]

### Step 2: Use Planck's Equation to Find Frequency
Planck's equation relates the energy of a photon to its frequency:
\[ E = h \nu \]
Where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{ J s}\)),
- \(\nu\) is the frequency of the photon.

Rearranging the equation to solve for frequency:
\[ \nu = \frac{E}{h} \]

### Step 3: Substitute the Values
Now we can substitute the energy we calculated and Planck's constant into the equation:
\[ \nu = \frac{2.49 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} \approx 3.76 \times 10^{14} \text{ Hz} \]

### Final Answer
The minimum frequency of a photon required to break the bond in `F_2` is approximately:
\[ \nu \approx 3.76 \times 10^{14} \text{ Hz} \]