A photon having `lambda = 854 Å` cause the ionization of a nitrogen atom. Give the I.E. per mole of nitrogen in KJ

Explanation

To find the ionization energy per mole of nitrogen when a photon with a wavelength of 854 Å (angstroms) is involved, we will follow these steps:

### Step 1: Convert the wavelength from angstroms to meters
The given wavelength is \( \lambda = 854 \, \text{Å} \).

1 Å = \( 10^{-10} \) meters, so:
\[ \lambda = 854 \, \text{Å} = 854 \times 10^{-10} \, \text{m} = 8.54 \times 10^{-8} \, \text{m} \]

### Step 2: Calculate the energy of the photon using the formula
The energy \( E \) of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) (speed of light) = \( 3.00 \times 10^{8} \, \text{m/s} \)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^{8} \, \text{m/s})}{8.54 \times 10^{-8} \, \text{m}} \]

Calculating the energy:
\[ E = \frac{1.9878 \times 10^{-25} \, \text{J m}}{8.54 \times 10^{-8} \, \text{m}} \approx 2.33 \times 10^{-18} \, \text{J} \]

### Step 3: Convert the energy from joules to kilojoules per mole
To find the ionization energy per mole, we multiply the energy of one photon by Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)):
\[ \text{Ionization Energy per mole} = E \times N_A \]
\[ \text{Ionization Energy per mole} = (2.33 \times 10^{-18} \, \text{J}) \times (6.022 \times 10^{23} \, \text{mol}^{-1}) \]
Calculating this gives:
\[ \text{Ionization Energy per mole} \approx 1401.68 \, \text{kJ/mol} \]

### Final Answer
The ionization energy per mole of nitrogen is approximately \( 1401.68 \, \text{kJ/mol} \).

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