Question

What is the photoelectric effect? For any photosensitive surface threshold frequency is \( 3.3 \times 10^{14} \, \text{Hz} \). If frequency of incident light becomes \( 8.2 \times 10^{14} \, \text{Hz} \), then calculate the stopping potential and give the value of 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 emission of electrons from a metal surface when it is illuminated by light of sufficient frequency. The energy of the incident photon is absorbed by the electron, and if the energy is enough, the electron is ejected.
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 light. For \( f = 8.2 \times 10^{14} \, \text{Hz} \), we get: \[ E_{\text{photon}} = (6.626 \times 10^{-34}) \times (8.2 \times 10^{14}) = 5.43 \times 10^{-19} \, \text{J} \]
Step 3: Work Function and Stopping Potential.
The work function \( \phi \) is the minimum energy required to remove an electron from the metal surface. It is related to the threshold frequency \( f_0 \) by: \[ \phi = h f_0 \] Substituting the threshold frequency \( f_0 = 3.3 \times 10^{14} \, \text{Hz} \): \[ \phi = (6.626 \times 10^{-34}) \times (3.3 \times 10^{14}) = 2.19 \times 10^{-19} \, \text{J} \] The maximum kinetic energy of the emitted photoelectron is: \[ K_{\text{max}} = E_{\text{photon}} - \phi = 5.43 \times 10^{-19} - 2.19 \times 10^{-19} = 3.24 \times 10^{-19} \, \text{J} \] 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 \( V_s \): \[ V_s = \frac{K_{\text{max}}}{e} = \frac{3.24 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.025 \, \text{V} \]