What is the binding energy (in J/mol or kJ/mol) of an electron in a metal, whose threshold frequency for photoelectron is `3.5 xx 10^(13) s^(-1)` ?

Options

A. `66.38 kJ//mol`
B. `6.95 kJ//mol`
C. `13.91 kJ//mol`
D. `27.82 kJ//mol`

Answer

To find the binding energy of an electron in a metal given the threshold frequency, we can use the formula:

\[ E = h \nu_0 \]

where:
- \(E\) is the binding energy,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(\nu_0\) is the threshold frequency.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Threshold frequency (\(\nu_0\)) = \(3.5 \times 10^{13} \, \text{s}^{-1}\)
- Planck's constant (\(h\)) = \(6.626 \times 10^{-34} \, \text{J s}\)

2. **Substitute the Values into the Formula**:
\[ E = h \nu_0 = (6.626 \times 10^{-34} \, \text{J s}) \times (3.5 \times 10^{13} \, \text{s}^{-1}) \]

3. **Perform the Multiplication**:
\[ E = 6.626 \times 3.5 \times 10^{-34} \times 10^{13} \, \text{J} \]
\[ E = 23.191 \times 10^{-21} \, \text{J} \]

4. **Convert to kJ/mol**:
- To convert joules to kilojoules, divide by 1000.
- To convert joules to kJ/mol, use Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)):
\[ E = \frac{23.191 \times 10^{-21} \, \text{J}}{1000} \times (6.022 \times 10^{23} \, \text{mol}^{-1}) \]
\[ E = 27.82 \, \text{kJ/mol} \]

### Final Answer:
The binding energy of the electron in the metal is approximately **27.82 kJ/mol**.