Question

What is the photoelectric effect? For any photosensitive surface, the threshold frequency is \( 3.3 \times 10^{14} \, \text{Hz} \). If the frequency of incident light becomes \( 8.2 \times 10^{14} \, \text{Hz} \), then calculate the stopping potential and give the value of the work function of the surface also.

Hint:

The stopping potential is the potential required to stop the most energetic photoelectrons, and it is directly related to the maximum kinetic energy of the emitted electrons.

Answer 1

Step 1: Photoelectric Effect.
The photoelectric effect refers to the phenomenon where electrons are ejected from a material (usually a metal) when it is exposed to light or other electromagnetic radiation of sufficient frequency. The energy of the incident photons is transferred to the electrons, causing them to be ejected from the surface of the material.
Step 2: Energy of Incident Photon.
The energy of a photon is given by: \[ E_{\text{photon}} = h f \] where: - \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) is Planck’s constant, - \( f \) is the frequency of the incident light. For the incident light frequency \( f = 8.2 \times 10^{14} \, \text{Hz} \), the energy is: \[ E_{\text{photon}} = (6.626 \times 10^{-34}) \times (8.2 \times 10^{14}) = 5.43 \times 10^{-19} \, \text{J} \]
Step 3: Work Function of the Surface.
The work function \( \phi \) is the minimum energy required to release an electron from the metal surface. The threshold frequency \( f_0 \) is the minimum frequency required to eject an electron. The work function is related to the threshold frequency by: \[ \phi = h f_0 \] Given \( f_0 = 3.3 \times 10^{14} \, \text{Hz} \), the work function is: \[ \phi = (6.626 \times 10^{-34}) \times (3.3 \times 10^{14}) = 2.19 \times 10^{-19} \, \text{J} \]
Step 4: Maximum Kinetic Energy of the Emitted Electrons.
The maximum kinetic energy \( K_{\text{max}} \) of the emitted photoelectrons is given by: \[ K_{\text{max}} = E_{\text{photon}} - \phi \] Substituting the values: \[ K_{\text{max}} = 5.43 \times 10^{-19} \, \text{J} - 2.19 \times 10^{-19} \, \text{J} = 3.24 \times 10^{-19} \, \text{J} \]
Step 5: Stopping Potential.
The stopping potential \( V_s \) is related to the maximum kinetic energy by: \[ K_{\text{max}} = e V_s \] where \( e = 1.6 \times 10^{-19} \, \text{C} \) is the charge of an electron. Solving for \( V_s \): \[ V_s = \frac{K_{\text{max}}}{e} = \frac{3.24 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.025 \, \text{V} \]