Photons of minimum energy `496k,J. mol^(-1)` are needed to an atoms. Calculate the lowest frequency of light that will ionize a sodium atom.

Options

A. `12.4xx10^(14)s^(-1)`
B. `1.24xx10^(15)s^(-1)`
C. `2.48xx10^(15)s^(-1)`
D. `2.48xx10^(14)s^(-1)`

Explanation

To solve the problem of calculating the lowest frequency of light that will ionize a sodium atom given the minimum energy of photons required (496 kJ/mol), we will follow these steps:

### Step 1: Convert the energy from kJ/mol to J/atom
We know that 1 mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) atoms.

1. Convert the energy from kilojoules to joules:
\[ 496 \, \text{kJ/mol} = 496 \times 10^3 \, \text{J/mol} = 496000 \, \text{J/mol} \]

2. Now, calculate the energy per atom:
\[ E_{\text{atom}} = \frac{496000 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \approx 8.23 \times 10^{-19} \, \text{J/atom} \]

### Step 2: Use the energy to find the frequency
The relationship between energy (E) and frequency (ν) is given by the equation:
\[ E = h \nu \]
Where:
- \(E\) is the energy in joules,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(\nu\) is the frequency in Hz (s\(^{-1}\)).

3. Rearranging the equation to solve for frequency:
\[ \nu = \frac{E}{h} \]

4. Substitute the values:
\[ \nu = \frac{8.23 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J s}} \approx 1.24 \times 10^{15} \, \text{Hz} \]

### Final Answer
The lowest frequency of light that will ionize a sodium atom is approximately \(1.24 \times 10^{15} \, \text{Hz}\).

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