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 the light that causes photoemission from a metallic surface when a stopping potential of 3V is applied. We know that no photoelectrons are emitted above a wavelength of 5000 Å, which gives us the work function of the material.

### Step-by-Step Solution:

1. **Identify the Work Function (Φ):**
The work function can be calculated using the threshold wavelength (λ₀ = 5000 Å). The formula to find the work function in electron volts (eV) is:
\[ \Phi = \frac{hc}{\lambda_0} \]
where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda_0 = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)

Plugging in the values:
\[ \Phi = \frac{(4.1357 \times 10^{-15} \, \text{eV s}) \times (3 \times 10^8 \, \text{m/s})}{5000 \times 10^{-10} \, \text{m}} \]
\[ \Phi = \frac{1.241 \times 10^{-6} \, \text{eV m}}{5000 \times 10^{-10} \, \text{m}} = 2.475 \, \text{eV} \]

2. **Determine the Maximum Kinetic Energy (K.E.):**
The maximum kinetic energy of the emitted photoelectrons when a stopping potential (V₀) of 3V is applied is given by:
\[ K.E. = eV_0 = 3 \, \text{eV} \]

3. **Apply the Photoelectric Equation:**
According to Einstein's photoelectric equation:
\[ K.E. = E - \Phi \]
where \( E \) is the energy of the incoming photon. We can express the energy in terms of wavelength:
\[ E = \frac{hc}{\lambda} \]
Therefore, we can rewrite the equation as:
\[ 3 = \frac{hc}{\lambda} - 2.475 \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{hc}{\lambda} = 3 + 2.475 = 5.475 \]
Now substituting \( hc \):
\[ \frac{1.241 \times 10^{-6} \, \text{eV m}}{\lambda} = 5.475 \]

5. **Solving for Wavelength (λ):**
\[ \lambda = \frac{1.241 \times 10^{-6}}{5.475} \]
\[ \lambda \approx 2.26 \times 10^{-7} \, \text{m} = 2260 \, \text{Å} \]

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