`N_(2)` undergoes photochemical dissociation into one normal `N-`atom and one `N-`atom having `0.1eV`more energy than normal `N-`atom. The dissociation of `N_(2)` into two normal atom of `N` requires `289.5 KJ//mol` energy. The maximum wavelength effective for photochemical dissociation of `N_(2)` in `nm` is `P nm` (`1 eV = 96.5` KJ/mol). Find the value of `(P)/(100)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the dissociation energy from kJ/mol to eV
The dissociation energy for \( N_2 \) is given as \( 289.5 \, \text{kJ/mol} \). We need to convert this energy into electron volts (eV) using the conversion factor \( 1 \, \text{eV} = 96.5 \, \text{kJ/mol} \).

\[ \text{Energy in eV} = \frac{289.5 \, \text{kJ/mol}}{96.5 \, \text{kJ/mol/eV}} = 3.0 \, \text{eV} \]

### Step 2: Account for the additional energy of the excited atom
The problem states that one of the nitrogen atoms has an additional \( 0.1 \, \text{eV} \) of energy. Therefore, the total energy required for the dissociation of \( N_2 \) into two nitrogen atoms (one in the ground state and one in the excited state) is:

\[ \text{Total energy} = 3.0 \, \text{eV} + 0.1 \, \text{eV} = 3.1 \, \text{eV} \]

### Step 3: 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 in eV,
- \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \, \text{eV s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

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

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

### Step 4: Calculate the wavelength in nanometers
Substituting the values into the equation, we first convert \( hc \) into eV·nm for convenience. The value of \( hc \) in eV·nm is approximately \( 1240 \, \text{eV·nm} \).

Now substituting \( E = 3.1 \, \text{eV} \):

\[ \lambda = \frac{1240 \, \text{eV·nm}}{3.1 \, \text{eV}} \approx 400 \, \text{nm} \]

### Step 5: Find the value of \( P/100 \)
According to the problem, \( P \) is the wavelength in nanometers, which we found to be \( 400 \, \text{nm} \). Thus:

\[ \frac{P}{100} = \frac{400}{100} = 4 \]

### Final Answer
The value of \( \frac{P}{100} \) is \( 4 \).

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