A metallic surface is irradiated with monochromatic light of variable wavelength. Above a wavelength of `5000 Å` no photoelectrons are emitted from the surface. With an unknown wavelength, stopping potential fo 3V is neceddsry ot eliminate the photocurrent. Find the unknown wavelenght.

Explanation

To solve the problem, we need to find the unknown wavelength of light that causes photoemission from a metallic surface, given that the stopping potential required to eliminate the photocurrent is 3V. We also know that no photoelectrons are emitted for wavelengths longer than 5000 Å, which indicates that 5000 Å is the threshold wavelength.

### Step-by-Step Solution:

1. **Identify the Threshold Wavelength:**
The threshold wavelength (\( \lambda_0 \)) is given as 5000 Å. This means that any light with a wavelength longer than this will not cause photoemission.

2. **Calculate the Work Function:**
The work function (\( \phi \)) can be calculated using the formula:
\[ \phi = \frac{12375}{\lambda_0} \]
where \( \phi \) is in electron volts (eV) and \( \lambda_0 \) is in angstroms (Å). Substituting \( \lambda_0 = 5000 \) Å:
\[ \phi = \frac{12375}{5000} = 2.475 \, \text{eV} \]

3. **Use the Stopping Potential:**
The stopping potential (\( V_0 \)) is given as 3V. The maximum kinetic energy (\( K_{max} \)) of the emitted photoelectrons can be expressed as:
\[ K_{max} = e \cdot V_0 = 3 \, \text{eV} \]

4. **Relate Photon Energy to Work Function and Kinetic Energy:**
The energy of the incoming photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light. According to the photoelectric effect, we have:
\[ K_{max} = E - \phi \]
Substituting the expressions for \( K_{max} \) and \( E \):
\[ 3 = \frac{12375}{\lambda} - 2.475 \]

5. **Rearranging the Equation:**
Rearranging the equation to solve for \( \lambda \):
\[ \frac{12375}{\lambda} = 3 + 2.475 \]
\[ \frac{12375}{\lambda} = 5.475 \]

6. **Solve for the Unknown Wavelength:**
Now, we can solve for \( \lambda \):
\[ \lambda = \frac{12375}{5.475} \approx 2260 \, \text{Å} \]

### Final Answer:
The unknown wavelength is approximately **2260 Å**.