`O_(2)` undergoes photochemical dissociation into one normal oxygen atom and one excited oxygen atom. Excited oxygen atom is `1.967 eV` more energetic than normal . The dissociation of `O_(2)` into two normal atoms of oxygen required `498 kJ mol^(-1)`, what is the maximum wavelength effective for photochemical dissociation of `O_(2)`?

Options

A. `1.01 nm`
B. `1.64 nm`
C. `1.74 nm`
D. `2.74 nm`

Explanation

To solve the problem, we need to find the maximum wavelength effective for the photochemical dissociation of \( O_2 \) into one normal oxygen atom and one excited oxygen atom. We will follow these steps:

### Step 1: Calculate the energy required for the dissociation of \( O_2 \) into two normal oxygen atoms.

The energy required for the dissociation of \( O_2 \) into two normal oxygen atoms is given as \( 498 \, \text{kJ/mol} \). To find the energy required for one molecule, we will divide this value by Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)).

\[ E_{\text{normal}} = \frac{498 \, \text{kJ/mol}}{N_A} = \frac{498 \times 10^3 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 82.72 \times 10^{-20} \, \text{J} \]

### Step 2: Convert the energy of the excited oxygen atom from eV to Joules.

The energy of the excited oxygen atom is given as \( 1.967 \, \text{eV} \). To convert this to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \).

\[ E_{\text{excited}} = 1.967 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \approx 3.1552 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the total energy required for the photochemical dissociation.

The total energy required for the photochemical dissociation into one normal and one excited oxygen atom is the sum of the energy of the normal oxygen atom and the energy of the excited oxygen atom.

\[ E_{\text{total}} = E_{\text{normal}} + E_{\text{excited}} = 82.72 \times 10^{-20} \, \text{J} + 3.1552 \times 10^{-19} \, \text{J} \]

Converting \( 82.72 \times 10^{-20} \, \text{J} \) to the same power of ten as \( 3.1552 \times 10^{-19} \, \text{J} \):

\[ E_{\text{total}} = 8.272 \times 10^{-19} \, \text{J} + 3.1552 \times 10^{-19} \, \text{J} = 4.9822 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the maximum wavelength using the energy-wavelength relationship.

We use the formula \( E = \frac{hc}{\lambda} \) to find the wavelength \( \lambda \). Rearranging gives:

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

Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3.00 \times 10^{8} \, \text{m/s} \) (speed of light)

Substituting the values:

\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^{8} \, \text{m/s})}{4.9822 \times 10^{-19} \, \text{J}} \approx 3.98 \times 10^{-7} \, \text{m} = 398 \, \text{nm} \]

### Step 5: Convert to Angstroms.

Since \( 1 \, \text{nm} = 10 \, \text{Å} \):

\[ \lambda = 398 \, \text{nm} = 3980 \, \text{Å} \]

### Conclusion

The maximum wavelength effective for the photochemical dissociation of \( O_2 \) is approximately \( 3980 \, \text{Å} \).