The work function for a metal is `40 eV`. To emit photo electrons of zero velocity from the surface of the metal the wavelength of incident light should be `x nm`.

Explanation

To solve the problem, we need to find the wavelength of the incident light that will cause photoelectrons to be emitted from a metal surface with a work function of 40 eV, such that these photoelectrons have zero velocity.

### Step-by-Step Solution:

1. **Understanding the Work Function**:
The work function (W₀) is the minimum energy required to remove an electron from the surface of a metal. In this case, W₀ = 40 eV.

2. **Relating Energy and Wavelength**:
The energy of a photon can be expressed in terms of its wavelength (λ) using the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

3. **Setting Kinetic Energy to Zero**:
When the photoelectrons are emitted with zero velocity, their kinetic energy (KE) is zero. According to the photoelectric equation:
\[ KE = E - W₀ \]
Since KE = 0, we have:
\[ E = W₀ \]

4. **Converting Work Function to Joules**:
We need to convert the work function from electron volts to joules. The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore:
\[ W₀ = 40 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.4 \times 10^{-18} \, \text{J} \]

5. **Substituting into the Energy-Wavelength Equation**:
Now, substituting \(E = W₀\) into the energy-wavelength equation:
\[ 6.4 \times 10^{-18} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{\lambda} \]

6. **Solving for Wavelength (λ)**:
Rearranging the equation to solve for λ gives:
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{6.4 \times 10^{-18}} \]
Calculating the right-hand side:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{6.4 \times 10^{-18}} \approx 3.105 \times 10^{-8} \, \text{m} \]

7. **Converting to Nanometers**:
To convert meters to nanometers:
\[ 1 \, \text{m} = 10^9 \, \text{nm} \]
Therefore:
\[ \lambda \approx 3.105 \times 10^{-8} \, \text{m} \times 10^9 \, \text{nm/m} \approx 31.05 \, \text{nm} \]

### Final Answer:
The wavelength of the incident light should be approximately **31.05 nm**.