The electric field of light wave is given as `vec(E)=10^(-3)cos((2 pi x)/(5xx10^(-7))-2pi xx 6xx10^(14)t) hat(x) (N)/(C)`. This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is:
Given, E (in eV) `= (12375)/(lambda ( "in" Å ))`

Options

A. 2.0 V
B. 0.72 V
C. 0.48 V
D. 2.48 V

Explanation

To find the stopping potential of the photoelectrons when light with a given electric field falls on a metal plate, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electric Field Expression**:
The electric field of the light wave is given as:
\[ \vec{E} = 10^{-3} \cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t\right) \hat{x} \, \text{N/C} \]

2. **Determine the Wavelength**:
From the expression of the electric field, we can identify the wave number \( k \):
\[ k = \frac{2\pi}{\lambda} \]
Given that:
\[ k = \frac{2\pi}{5 \times 10^{-7}} \implies \lambda = 5 \times 10^{-7} \, \text{m} \]
Converting this to angstroms (1 m = \( 10^{10} \) Å):
\[ \lambda = 5 \times 10^{-7} \, \text{m} = 5000 \, \text{Å} \]

3. **Calculate the Energy of the Photon**:
The energy of a photon can be calculated using the formula:
\[ E = \frac{12375}{\lambda \, (\text{in Å})} \]
Substituting the value of \( \lambda \):
\[ E = \frac{12375}{5000} = 2.475 \, \text{eV} \]

4. **Determine the Stopping Potential**:
The stopping potential \( V_s \) is given by:
\[ V_s = E - \text{Work Function} \]
Given the work function \( \phi = 2 \, \text{eV} \):
\[ V_s = 2.475 \, \text{eV} - 2 \, \text{eV} = 0.475 \, \text{eV} \]

5. **Final Result**:
The stopping potential of the photoelectrons is approximately:
\[ V_s \approx 0.48 \, \text{eV} \]