Light in incident on a metal plate whose work function is 2 eV. Electric field associated with light is given by `E=E_(0)sin(omegat-(2pi)/(5xx10^(7))x)`{ S.I unit} if energy of photom is given by `(12375)/(lamda("in"A))` eV then stopping potential is

Options

A. `2.48eV`
B. `0.48eV`
C. `0.78eV`
D. `1.24eV`

Answer

To find the stopping potential for the given problem, we will follow these steps:

### Step 1: Identify the work function and the equation for energy of the photon
The work function of the metal plate is given as \( \phi = 2 \, \text{eV} \). The energy of the photon is given by the formula:
\[ E = \frac{12375}{\lambda} \, \text{eV} \]
where \( \lambda \) is in Angstroms.

### Step 2: Determine the wave number \( k \) from the electric field equation
The electric field associated with the light is given by:
\[ E = E_0 \sin(\omega t - kx) \]
From the equation, we can see that:
\[ k = \frac{2\pi}{5 \times 10^7} \, \text{m}^{-1} \]

### Step 3: Calculate the wavelength \( \lambda \)
Using the relationship between \( k \) and \( \lambda \):
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{\frac{2\pi}{5 \times 10^7}} = 5 \times 10^7 \, \text{m} \]
Converting this 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 4: Calculate the energy of the incident photon
Now substituting \( \lambda = 5000 \, \text{Å} \) into the energy formula:
\[ E = \frac{12375}{5000} = 2.475 \, \text{eV} \]

### Step 5: Apply the photoelectric effect equation
The energy of the incident photon is related to the work function and the stopping potential \( V_s \) by the equation:
\[ E = \phi + eV_s \]
Rearranging gives:
\[ V_s = \frac{E - \phi}{e} \]
Substituting the values:
\[ V_s = 2.475 \, \text{eV} - 2 \, \text{eV} = 0.475 \, \text{eV} \]

### Step 6: Final answer
Thus, the stopping potential \( V_s \) is approximately:
\[ V_s \approx 0.48 \, \text{eV} \]