Bond dissociation on energy of `Cl_(2)` is 240 kJ/mol. The longest wavelength of photon that can break this bond would be `[N_(A)=6xx10^(23), h=6.6xx10^(-34)J/s]`

Options

A. `4.95xx10^(-7)m`
B. `9.9xx10^(-7)m`
C. `4.95xx10^(-6)m`
D. `9.9xx10^(-6)m`

Answer

To solve the problem of finding the longest wavelength of a photon that can break the Cl-Cl bond, we will use the relationship between energy, wavelength, and the constants provided.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Bond dissociation energy of Cl₂: \( E = 240 \, \text{kJ/mol} \)
- Avogadro's number: \( N_A = 6 \times 10^{23} \, \text{molecules/mol} \)
- Planck's constant: \( h = 6.6 \times 10^{-34} \, \text{J s} \)
- Speed of light: \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Convert the Bond Dissociation Energy to Joules:**
\[ E = 240 \, \text{kJ/mol} = 240 \times 10^3 \, \text{J/mol} \]

3. **Use the Energy-Wavelength Relationship:**
The energy of a photon can be expressed as:
\[ E = \frac{N_A \cdot h \cdot c}{\lambda} \]
Rearranging this equation to solve for wavelength \( \lambda \):
\[ \lambda = \frac{N_A \cdot h \cdot c}{E} \]

4. **Substitute the Values into the Equation:**
\[ \lambda = \frac{(6 \times 10^{23}) \cdot (6.6 \times 10^{-34}) \cdot (3 \times 10^8)}{240 \times 10^3} \]

5. **Calculate the Numerator:**
\[ \text{Numerator} = (6 \times 10^{23}) \cdot (6.6 \times 10^{-34}) \cdot (3 \times 10^8) = 1.188 \times 10^{-2} \]

6. **Calculate the Wavelength:**
\[ \lambda = \frac{1.188 \times 10^{-2}}{240 \times 10^3} = 4.95 \times 10^{-7} \, \text{m} \]

7. **Convert Wavelength to Nanometers (if needed):**
\[ \lambda = 4.95 \times 10^{-7} \, \text{m} = 495 \, \text{nm} \]

### Final Answer:
The longest wavelength of a photon that can break the Cl-Cl bond is \( 4.95 \times 10^{-7} \, \text{m} \) or \( 495 \, \text{nm} \).

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