The maximum wavelength of radiation that can ionize a sodium atom is 2414 A the ionization energy of sodium per mole shall be

Options

A. `241.4 j mol^(-1)`
B. `479.7 j mol^(-1)`
C. `241.4 k j mol^(-1)`
D. `497.11k j mol^(-1)`

Answer

To find the ionization energy of a sodium atom given the maximum wavelength of radiation that can ionize it, we can follow these steps:

### Step 1: Convert Wavelength to Meters
The wavelength is given as 2414 Å (angstroms). We need to convert this to meters for our calculations.

1 Å = 10^-10 meters, so:
\[ 2414 \, \text{Å} = 2414 \times 10^{-10} \, \text{m} = 2.414 \times 10^{-7} \, \text{m} \]

### Step 2: Use the Energy-Wavelength Relationship
The energy (E) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) = Planck's constant = \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) = speed of light = \( 3 \times 10^{8} \, \text{m/s} \)
- \( \lambda \) = wavelength in meters

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^{8} \, \text{m/s})}{2.414 \times 10^{-7} \, \text{m}} \]

### Step 3: Calculate the Energy
Now we can calculate the energy:
\[ E = \frac{1.9878 \times 10^{-25} \, \text{J m}}{2.414 \times 10^{-7} \, \text{m}} \]
\[ E \approx 8.23 \times 10^{-18} \, \text{J} \]

### Step 4: Convert Energy to Per Mole
To find the ionization energy per mole, we multiply the energy of one photon by Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)):
\[ \text{Ionization Energy per mole} = E \times N_A \]
\[ \text{Ionization Energy per mole} = (8.23 \times 10^{-18} \, \text{J}) \times (6.022 \times 10^{23} \, \text{mol}^{-1}) \]

### Step 5: Calculate the Final Result
Calculating this gives:
\[ \text{Ionization Energy per mole} \approx 4.96 \times 10^{6} \, \text{J/mol} \]
Converting to kilojoules:
\[ \text{Ionization Energy per mole} \approx 496 \, \text{kJ/mol} \]

### Final Answer
The ionization energy of sodium per mole is approximately **496 kJ/mol**.

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