Question

What is the photoelectric effect? Which nature of light is shown by it? Threshold frequency of a metal is \(2.3 \times 10^{14}\) Hz. If the frequency of incident light is \(8.2 \times 10^{14}\) Hz, what will be the stopping potential? Calculate work function of the metal also.

Hint:

The photoelectric effect shows the particle nature of light. The stopping potential is related to the energy of the photons and the work function of the metal. Use the equation \(e V_s = h \nu - \Phi\) to calculate the stopping potential.

Answer 1

The photoelectric effect is the phenomenon in which electrons are ejected from the surface of a metal when light (or any electromagnetic radiation) of a certain frequency falls on it. This effect provides strong evidence for the particle nature of light. According to Einstein's theory, light consists of photons, and when a photon with energy \(E = h \nu\) strikes an electron, it can transfer its energy to the electron. If the photon's energy is greater than the work function (\(\Phi\)) of the metal, the electron is ejected.
The stopping potential is the minimum voltage required to stop the emitted electrons. The relationship between the energy of the photon, the work function of the metal, and the stopping potential is given by the equation:
\[ K_{\text{max}} = h \nu - \Phi \] Where:
- \(K_{\text{max}}\) is the maximum kinetic energy of the emitted electron (which is related to the stopping potential \(V_s\) by \(K_{\text{max}} = e V_s\)),
- \(h\) is Planck's constant (\(6.626 \times 10^{-34}~\text{J} \cdot \text{s}\)),
- \(\nu\) is the frequency of the incident light,
- \(\Phi\) is the work function of the metal.
Given:
- \(\nu = 8.2 \times 10^{14}~\text{Hz}\),
- \(\Phi = h \nu_0 = h \times 2.3 \times 10^{14}~\text{Hz}\) (where \(\nu_0\) is the threshold frequency).
First, calculate the energy of the incident photon:
\[ E_{\text{photon}} = h \nu = (6.626 \times 10^{-34}) \times (8.2 \times 10^{14}) = 5.428 \times 10^{-19}~\text{J} \] Next, calculate the work function:
\[ \Phi = h \nu_0 = (6.626 \times 10^{-34}) \times (2.3 \times 10^{14}) = 1.524 \times 10^{-19}~\text{J} \] Now, using the photoelectric equation, find the stopping potential:
\[ e V_s = h \nu - \Phi \] \[ V_s = \frac{h \nu - \Phi}{e} = \frac{5.428 \times 10^{-19} - 1.524 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{3.904 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.44~\text{V} \] Thus, the stopping potential is approximately \(2.44~\text{V}\).
Finally, the work function of the metal is \(\Phi = 1.524 \times 10^{-19}~\text{J}\), which is the energy required to release an electron from the metal surface.