If the frequency of the incident radiation is equal to the threshold frequency, what will be the value of the stopping potential?

1. Break Down the Problem

We need to analyze the relationship between the frequency of incident radiation, threshold frequency, and stopping potential in the context of the photoelectric effect.

2. Relevant Concepts

• The threshold frequency ($f_0$) is the minimum frequency of incident light necessary to eject electrons from a material.
• The energy of the incident photons is given by the equation: $E = h f$ where $E$ is the energy, $h$ is Planck's constant ($h \approx 6.626 \times 10^{-34} \, \text{J s}$), and $f$ is the frequency.
• The stopping potential ($V_s$) is defined as the potential difference needed to stop the ejected electrons, which can be related to energy as: $E = e V_s$ where $e$ is the elementary charge ($e \approx 1.602 \times 10^{-19} \, \text{C}$).

3. Analysis and Detail

• When the frequency of the incident radiation is equal to the threshold frequency ($f = f_0$): $E = h f_0$
• The energy at this frequency is exactly equal to the work function ($\phi$) of the material, which means: $E = \phi$ This indicates that there is no excess energy to do work on the ejected electrons, which leads to: $e V_s = \phi \rightarrow V_s = \frac{\phi}{e}$
• At this point, since $E = \phi$, there is no kinetic energy imparted to the electrons, meaning they have zero velocity when they leave the material.

4. Verify and Summarize

Since at the threshold frequency, photons provide just enough energy to overcome the work function and no further energy for kinetic energy, the stopping potential ($V_s$) is effectively leading to only one conclusion:

• If $f = f_0$: $V_s = 0 \, \text{V}$

Final Answer

The stopping potential will be $V_s = 0 \, \text{V}$.