Ozone in the upper atmoshphere absorbs ultraviolet radiation which induces the following chemical reaction
`O_(3)(g)rightarrowO_(2)(g)+O(g)`
`O_(2)` produced in the above photochemical dissociation undergoes further dissociation into one normal oxygen atom (O) and more energetic oxygen atom `O**`.
`O_(2)(g) rightarrowO+O**`
`(O**)` has 1 eV more energy than(O) and normal dissociation energy of `O_(2)` is 480 kJ `"mol"^(-1)`.
[1 eV/Photon =96 kJ `"mol"^(-1)`]
What is the maximum wavelength effective for the photochemical dissociation of `O_(2)` molecule

Options

A. 2440 Å
B. 2066.67 Å
C. 1000 Å
D. 155 Å

Answer

To solve the problem, we need to determine the maximum wavelength effective for the photochemical dissociation of the O2 molecule. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Energy Requirements
The dissociation of O2 involves two steps:
1. The dissociation of O3 to O2 and O.
2. The further dissociation of O2 into O and O* (where O* has 1 eV more energy than O).

The normal dissociation energy of O2 is given as 480 kJ/mol. Since O* has 1 eV more energy than O, we need to convert this additional energy into kJ/mol.

### Step 2: Convert 1 eV to kJ/mol
1 eV corresponds to 96 kJ/mol. Therefore, the total energy required for the dissociation of O2 into O and O* is:
\[ \text{Total Energy} = \text{Dissociation Energy of O2} + \text{Energy from O*} \]
\[ \text{Total Energy} = 480 \text{ kJ/mol} + 96 \text{ kJ/mol} = 576 \text{ kJ/mol} \]

### Step 3: Convert Total Energy to Electron Volts
Now, we need to convert this total energy from kJ/mol to eV. We know that:
\[ 1 \text{ kJ/mol} = \frac{1}{96} \text{ eV} \approx 0.0104 \text{ eV} \]
Thus, to convert 576 kJ/mol to eV:
\[ E_{\text{min}} = \frac{576 \text{ kJ/mol}}{96 \text{ kJ/eV}} = 6 \text{ eV} \]

### Step 4: Calculate Maximum Wavelength
The relationship between energy (E) and wavelength (λ) is given by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{ J s}\))
- \(c\) is the speed of light (\(3 \times 10^8 \text{ m/s}\))

To find the maximum wavelength (λ), we rearrange the equation:
\[ \lambda = \frac{hc}{E} \]

Since we want the wavelength in angstroms (1 angstrom = \(10^{-10} \text{ m}\)), we can use the formula:
\[ \lambda_{\text{max}} \text{ (in angstroms)} = \frac{12400}{E_{\text{min}} \text{ (in eV)}} \]
Substituting \(E_{\text{min}} = 6 \text{ eV}\):
\[ \lambda_{\text{max}} = \frac{12400}{6} \approx 2066.67 \text{ angstroms} \]

### Final Answer
The maximum wavelength effective for the photochemical dissociation of the O2 molecule is approximately **2066.67 angstroms**.

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