Light of wavelength 3320 Å incidents on metal surface (work function = 1.07 eV ). To stop emission of photo electron, retarding potential required to be -

Options

A. 3.74 V
B. 2.67 V
C. 1.07 V
D. 4.81 V

Answer

To solve the problem of determining the retarding potential required to stop the emission of photoelectrons when light of wavelength 3320 Å is incident on a metal surface with a work function of 1.07 eV, we can follow these steps:

### Step 1: Calculate the Energy of Incident Photons
The energy of the incident photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant (\( 4.14 \times 10^{-15} \) eV·s)
- \( c \) is the speed of light (\( 3 \times 10^8 \) m/s)
- \( \lambda \) is the wavelength in meters (3320 Å = \( 3320 \times 10^{-10} \) m)

First, convert the wavelength from angstroms to meters:
\[ \lambda = 3320 \, \text{Å} = 3320 \times 10^{-10} \, \text{m} = 3.32 \times 10^{-7} \, \text{m} \]

Now, substituting the values into the energy formula:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{3.32 \times 10^{-7} \, \text{m}} \approx 3.73 \, \text{eV} \]

### Step 2: Use the Photoelectric Equation
The maximum kinetic energy (\( KE_{max} \)) of the emitted photoelectrons can be calculated using the photoelectric equation:
\[ KE_{max} = E - \phi \]
Where:
- \( E \) is the energy of the incident photons (calculated above as 3.73 eV)
- \( \phi \) is the work function of the metal (given as 1.07 eV)

Substituting the values:
\[ KE_{max} = 3.73 \, \text{eV} - 1.07 \, \text{eV} = 2.66 \, \text{eV} \]

### Step 3: Relate Maximum Kinetic Energy to Retarding Potential
The maximum kinetic energy is also related to the retarding potential (\( V_s \)) by the equation:
\[ KE_{max} = eV_s \]
Where \( e \) is the charge of the electron (1 eV = 1 V·C).

Thus, we can express the retarding potential as:
\[ V_s = \frac{KE_{max}}{e} \]
Since \( KE_{max} \) is already in eV, we can directly write:
\[ V_s = KE_{max} = 2.66 \, \text{V} \]

### Final Answer
The retarding potential required to stop the emission of photoelectrons is approximately:
\[ \boxed{2.66 \, \text{V}} \]