Figure shows the plot of the stopping potentialversus the frequency of the light used in anexperiment on photoelectric effect. The ratio h/e is:

To determine the ratio $\frac{h}{e}$ from the stopping potential versus frequency plot in the context of the photoelectric effect, we use the following analysis:

1. Break Down the Problem

• We need to find the slope of the line in the plot of stopping potential $V_s$ versus frequency $\nu$.
• According to the photoelectric equation, the stopping potential can be related to frequency as follows: $eV_s = h\nu - \phi$
• Rearranging this gives: $V_s = \frac{h}{e} \nu - \frac{\phi}{e}$
• From this equation, the slope of the line will be $\frac{h}{e}$ and the y-intercept will be $-\frac{\phi}{e}$.

2. Relevant Concepts

• The photoelectric effect equation relates the stopping potential and the frequency of incident light.
• The slope of the graph $V_s$ vs. $\nu$ gives us $\frac{h}{e}$.
• $h$ is Planck's constant, and $e$ is the elementary charge.

3. Analysis and Detail

• Plot the data points from the experiment to determine the linear relationship.
• Calculate the slope by selecting two points on the line: $\text{Slope} = \frac{\Delta V_s}{\Delta \nu}$
• Ensure that the points used for calculation are accurately reflecting the linear region of the plot.

4. Verify and Summarize

• After calculating the slope, verify if the linearity holds true across the selected points.
• This slope $\frac{\Delta V_s}{\Delta \nu}$ will represent $\frac{h}{e}$.

Final Answer

To find the final value of $\frac{h}{e}$, simply insert the values obtained from the slope into the relationship and verify against known values or conditions of the experiment. The exact numerical solution will depend on the slope derived from the specific data points from the plot.