The binding energy of electron in a metal is `193 kJ mol^(-1)` .Find the threshold frequency of the metal

Explanation

To find the threshold frequency of the metal given the binding energy of the electron, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept**: The threshold frequency (\( \nu_0 \)) is the minimum frequency of light required to remove an electron from the metal surface. The binding energy (BE) of the electron is the energy needed to remove it.

2. **Convert Binding Energy to Joules**:
The binding energy is given as \( 193 \, \text{kJ/mol} \). We need to convert this to joules per electron (photon).
\[ \text{Binding Energy in Joules} = 193 \, \text{kJ/mol} \times \frac{10^3 \, \text{J}}{1 \, \text{kJ}} \times \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{electrons}} \]
\[ = \frac{193 \times 10^3}{6.022 \times 10^{23}} \, \text{J/electron} \]

3. **Calculate the Binding Energy per Electron**:
\[ = \frac{193000}{6.022 \times 10^{23}} \approx 3.20 \times 10^{-19} \, \text{J/electron} \]

4. **Use the Equation for Binding Energy**:
The binding energy is related to the threshold frequency by the equation:
\[ \text{Binding Energy} = h \nu_0 \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{J s} \)).

5. **Rearrange to Find Threshold Frequency**:
\[ \nu_0 = \frac{\text{Binding Energy}}{h} \]
\[ \nu_0 = \frac{3.20 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J s}} \]

6. **Calculate the Threshold Frequency**:
\[ \nu_0 \approx 4.83 \times 10^{14} \, \text{s}^{-1} \]

### Final Answer:
The threshold frequency of the metal is approximately \( 4.83 \times 10^{14} \, \text{s}^{-1} \).

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