Electromagnetic radiations of wavelength `2000Å` are incident on a metal surface which has a work function `5.01 eV` The stopping potential for the given setup is

Options

A. 1.19 eV
B. 6.19 eV
C. 3.19 eV
D. 4.19 eV

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Wavelength of the incident electromagnetic radiation, \( \lambda = 2000 \, \text{Å} \)
- Work function of the metal, \( \phi = 5.01 \, \text{eV} \)

### Step 2: Convert the wavelength to meters (if necessary)
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):
\[ \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} = 2 \times 10^{-7} \, \text{m} \]

### Step 3: Calculate the energy of the incident photons
The energy of the incident photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 4.14 \times 10^{-15} \, \text{eV s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

Substituting the values:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{2 \times 10^{-7} \, \text{m}} \]
Calculating this gives:
\[ E = \frac{1.242 \times 10^{-6} \, \text{eV m}}{2 \times 10^{-7} \, \text{m}} = 6.21 \, \text{eV} \]

### Step 4: Apply the photoelectric equation
The maximum kinetic energy of the emitted electrons is given by:
\[ K.E. = E - \phi \]
Where:
- \( K.E. \) is the maximum kinetic energy of the electrons
- \( E \) is the energy of the incident photons
- \( \phi \) is the work function of the metal

Substituting the values:
\[ K.E. = 6.21 \, \text{eV} - 5.01 \, \text{eV} = 1.20 \, \text{eV} \]

### Step 5: Relate kinetic energy to stopping potential
The maximum kinetic energy can also be expressed in terms of stopping potential \( V_s \):
\[ K.E. = e \cdot V_s \]
Where \( e \) is the charge of the electron (approximately 1 eV when in electron volts).

Thus, we have:
\[ 1.20 \, \text{eV} = V_s \]

### Final Answer
The stopping potential \( V_s \) is:
\[ V_s = 1.20 \, \text{V} \]