The reaction between `H_(2)` and `Br_(2)` to form `HBr` in presence of light is initiated by the photo decomposition of `Br_(2)` into free Br atoms (free radicals) by absorption of light. The bond dissociation energy of `Br_(2)` is `192kJ//"mole"`.What is the longest wavelength of the photon that would initiate the reaction?

Explanation

To determine the longest wavelength of the photon that would initiate the reaction between \( H_2 \) and \( Br_2 \) to form \( HBr \), we need to follow these steps:

### Step 1: Understand the Energy Requirement
The bond dissociation energy of \( Br_2 \) is given as \( 192 \, \text{kJ/mol} \). This is the minimum energy required to break the \( Br-Br \) bond and produce free radicals.

### Step 2: Convert Energy to Joules
Since the bond dissociation energy is provided in kilojoules per mole, we need to convert it to joules. The conversion factor is:
\[ 1 \, \text{kJ} = 1000 \, \text{J} \]
Thus,
\[ 192 \, \text{kJ/mol} = 192 \times 10^3 \, \text{J/mol} \]

### Step 3: Use the Energy-Wavelength Relation
The energy of a photon can be related to its wavelength using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( E \) is the energy in joules,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.0 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 4: Rearranging the Formula
To find the wavelength \( \lambda \), we rearrange the formula:
\[ \lambda = \frac{hc}{E} \]

### Step 5: Substitute the Values
Now, we can substitute the values into the formula. First, we need to convert the energy from per mole to per photon. Since \( 1 \, \text{mol} \) contains \( 6.022 \times 10^{23} \) photons (Avogadro's number), we divide the energy by this number:
\[ E_{\text{photon}} = \frac{192 \times 10^3 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{photons/mol}} \approx 3.19 \times 10^{-19} \, \text{J} \]

Now substituting into the wavelength formula:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.0 \times 10^8 \, \text{m/s})}{3.19 \times 10^{-19} \, \text{J}} \]

### Step 6: Calculate the Wavelength
Calculating the above expression:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{3.19 \times 10^{-19}} \approx 6.22 \times 10^{-7} \, \text{m} \]

### Step 7: Convert to Nanometers
To express the wavelength in nanometers (1 nm = \( 10^{-9} \, \text{m} \)):
\[ \lambda \approx 622 \, \text{nm} \]

### Final Answer
The longest wavelength of the photon that would initiate the reaction is approximately \( 622 \, \text{nm} \).

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