Calculate the wavelength of light required to break the bond between two Cl atoms in `Cl_(2)` molecules `((BE)_(Cl-Cl)= 243 kJ mol^(-1))`

Options

A. 8180 `Å`
B. `9860Å`
C. `4930Å`
D. 500nm

Answer

To calculate the wavelength of light required to break the bond between two chlorine atoms in a Cl₂ molecule, we can follow these steps:

### Step 1: Convert the bond energy from kJ/mol to J/molecule
The bond energy given is \( BE_{Cl-Cl} = 243 \, \text{kJ/mol} \). We need to convert this energy into joules per molecule.

\[ E_{\text{one molecule}} = \frac{243 \times 10^3 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} \]

Calculating this gives:

\[ E_{\text{one molecule}} = \frac{243000}{6.022 \times 10^{23}} \approx 4.03 \times 10^{-19} \, \text{J} \]

### Step 2: Use the energy-wavelength relationship
The energy of a photon is related to its wavelength by the equation:

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

Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

Rearranging this equation to solve for \( \lambda \):

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

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{4.03 \times 10^{-19} \, \text{J}} \]

Calculating this gives:

\[ \lambda \approx \frac{1.9878 \times 10^{-25}}{4.03 \times 10^{-19}} \approx 4.93 \times 10^{-7} \, \text{m} \]

### Step 4: Convert the wavelength from meters to nanometers
To convert the wavelength from meters to nanometers, we use the conversion factor \( 1 \, \text{m} = 10^9 \, \text{nm} \):

\[ \lambda \approx 4.93 \times 10^{-7} \, \text{m} \times 10^9 \, \text{nm/m} \approx 493 \, \text{nm} \]

### Step 5: Convert to angstroms if needed
To convert to angstroms, where \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):

\[ \lambda \approx 4.93 \times 10^{-7} \, \text{m} \times 10^{10} \, \text{Å/m} \approx 4930 \, \text{Å} \]

### Final Answer
The wavelength of light required to break the bond between two Cl atoms in Cl₂ is approximately **493 nm** or **4930 Å**.