Up to what potential V can a zinc ball (work function 3.74 eV) removed from other bodies be charged by irradiating it with light of `lamda=200 nm`?

Options

A. 2.5 V
B. 1.8 V
C. 2.2 V
D. 3 V

Explanation

To solve the problem of determining the maximum potential \( V \) that a zinc ball can be charged to when irradiated with light of wavelength \( \lambda = 200 \, \text{nm} \), we will use the photoelectric effect principles. The work function \( \phi \) of zinc is given as \( 3.74 \, \text{eV} \).

### Step-by-Step Solution:

1. **Convert the wavelength to meters**:
\[ \lambda = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m} \]

2. **Calculate the energy of the incident photons**:
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{200 \times 10^{-9}} \]

3. **Calculate the numerical value of \( E \)**:
\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{200 \times 10^{-9}} = \frac{1.9878 \times 10^{-25}}{200 \times 10^{-9}} = 9.939 \times 10^{-19} \, \text{J} \]

4. **Convert the energy from Joules to electron volts**:
Since \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \text{ (in eV)} = \frac{9.939 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 6.21 \, \text{eV} \]

5. **Apply the photoelectric effect equation**:
The stopping potential \( V_0 \) can be found using the equation:
\[ eV_0 = E - \phi \]
where \( \phi = 3.74 \, \text{eV} \).

Substituting the values:
\[ eV_0 = 6.21 \, \text{eV} - 3.74 \, \text{eV} = 2.47 \, \text{eV} \]

6. **Final result for the stopping potential**:
Therefore, the maximum potential \( V \) that the zinc ball can be charged to is approximately:
\[ V_0 \approx 2.47 \, \text{V} \]

### Conclusion:
The maximum potential \( V \) that the zinc ball can be charged to is approximately **2.47 V**.