The photoelectric work function for a metal surface is `4.125 eV`. The cut - off wavelength for this surface is

Options

A. 1. `4125 Å`
B. 2. `3000 Å`
C. 3. `6000 Å`
D. 4. `2062.5 Å`

Answer

To find the cut-off wavelength for a metal surface with a given photoelectric work function, we can use the relationship between energy and wavelength. The work function (Φ) is the minimum energy required to eject an electron from the metal surface, and it is given in electron volts (eV).

### Step-by-Step Solution:

1. **Understand the relationship between energy and wavelength**:
The energy (E) of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{12400}{\lambda} \]
where E is in electron volts (eV) and λ is in angstroms (Å).

2. **Set the energy equal to the work function**:
Since the cut-off wavelength corresponds to the energy equal to the work function, we can set:
\[ E = \Phi = 4.125 \text{ eV} \]

3. **Substitute the work function into the equation**:
We can now substitute the work function into the energy equation:
\[ 4.125 = \frac{12400}{\lambda} \]

4. **Rearrange the equation to solve for λ**:
To find λ, we rearrange the equation:
\[ \lambda = \frac{12400}{4.125} \]

5. **Calculate λ**:
Performing the calculation:
\[ \lambda = \frac{12400}{4.125} \approx 3000 \text{ Å} \]

6. **Final Result**:
The cut-off wavelength for the metal surface is approximately:
\[ \lambda \approx 3000 \text{ Å} \]