The electric field of light wave is given as `vec(E )=10^(-3)cos((2pi x)/(5xx10^(-7))-2pixx6xx10^(14)t)hat(x)(N)/(C )`
This light falls on a metal plate of work function 2eV. 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`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the wavelength (λ) from the electric field equation
The electric field of the light wave is given as:
\[ \vec{E} = 10^{-3} \cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t\right) \hat{x} \text{ (N/C)} \]
From the term \(\frac{2\pi x}{5 \times 10^{-7}}\), we can identify that:
\[ k = \frac{2\pi}{\lambda} \]
Thus, we have:
\[ \lambda = 5 \times 10^{-7} \text{ m} \]

### Step 2: Convert the wavelength from meters to angstroms
To convert meters to angstroms (1 m = \(10^{10}\) Å):
\[ \lambda = 5 \times 10^{-7} \text{ m} = 5 \times 10^{-7} \times 10^{10} \text{ Å} = 5000 \text{ Å} \]

### Step 3: Calculate the energy (E) of the photons using the wavelength
Using the formula for energy:
\[ E = \frac{12375}{\lambda \text{ (in Å)}} \]
Substituting \(\lambda = 5000 \text{ Å}\):
\[ E = \frac{12375}{5000} = 2.475 \text{ eV} \]

### Step 4: Calculate the stopping potential (V_s)
The stopping potential is given by:
\[ V_s = E - \Phi \]
where \(\Phi\) (the work function) is given as 2 eV. Thus:
\[ V_s = 2.475 \text{ eV} - 2 \text{ eV} = 0.475 \text{ eV} \]

### Step 5: Final answer
The stopping potential of the photo-electrons is approximately:
\[ V_s \approx 0.48 \text{ V} \]

### Summary of Steps:
1. Identify the wavelength from the electric field equation.
2. Convert the wavelength from meters to angstroms.
3. Calculate the energy of the photons using the wavelength.
4. Calculate the stopping potential using the energy and work function.
5. Present the final answer.