Light of wavelength `330nm` falling on a piece of metal ejects electrons with sufficient energy with required voltage `V_0` to prevent them reaching a collector. In the same set up, light of wavelength `220 nm` ejects electrons which require twice the voltage `V_0` to stop them in reaching a collector. the numerical value of voltage `V_0` is

Answer

To solve the problem, we will use the photoelectric effect equation given by Einstein, which relates the energy of the incident photons to the work function of the metal and the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two cases with different wavelengths of light:
- Case 1: Wavelength \( \lambda_1 = 330 \, \text{nm} \)
- Case 2: Wavelength \( \lambda_2 = 220 \, \text{nm} \)

For Case 1, the stopping voltage is \( V_0 \).
For Case 2, the stopping voltage is \( 2V_0 \).

2. **Using the Photoelectric Equation**:
The energy of the photons can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

3. **Case 1**:
For the first case:
\[ \frac{hc}{\lambda_1} = \phi + eV_0 \]
where \( \phi \) is the work function and \( e \) is the charge of an electron.

Substituting \( \lambda_1 = 330 \, \text{nm} = 330 \times 10^{-9} \, \text{m} \):
\[ \frac{hc}{330 \times 10^{-9}} = \phi + eV_0 \quad \text{(1)} \]

4. **Case 2**:
For the second case:
\[ \frac{hc}{\lambda_2} = \phi + 2eV_0 \]
Substituting \( \lambda_2 = 220 \, \text{nm} = 220 \times 10^{-9} \, \text{m} \):
\[ \frac{hc}{220 \times 10^{-9}} = \phi + 2eV_0 \quad \text{(2)} \]

5. **Subtracting Equations**:
Now, we will subtract equation (1) from equation (2):
\[ \frac{hc}{220 \times 10^{-9}} - \frac{hc}{330 \times 10^{-9}} = (2eV_0 - eV_0) \]
Simplifying gives:
\[ \frac{hc}{220 \times 10^{-9}} - \frac{hc}{330 \times 10^{-9}} = eV_0 \]

6. **Finding a Common Denominator**:
The common denominator for the left side is \( 660 \times 10^{-9} \):
\[ \frac{3hc - 2hc}{660 \times 10^{-9}} = eV_0 \]
This simplifies to:
\[ \frac{hc}{660 \times 10^{-9}} = eV_0 \]

7. **Substituting Values**:
Now, substituting \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{660 \times 10^{-9}} = eV_0 \]
Calculating \( hc \):
\[ hc = 1.9878 \times 10^{-25} \, \text{Jm} \]
Thus:
\[ V_0 = \frac{1.9878 \times 10^{-25}}{660 \times 10^{-9} \cdot 1.6 \times 10^{-19}} \]

8. **Final Calculation**:
After performing the calculation:
\[ V_0 \approx 15/8 \, \text{V} \approx 1.875 \, \text{V} \]

### Final Answer:
The numerical value of voltage \( V_0 \) is \( \frac{15}{8} \, \text{V} \).