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

Question

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

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Solution

I'm sorry, but you didn't provide a specific question or a figure to analyze. However, I can tell you that in a photoelectric effect experiment, the slope of the graph of stopping potential versus frequency gives the ratio h/e, where h is Planck's constant and e is the charge of an electron. This is derived from the equation of the photoelectric effect:

E = hν - φ

where E is the energy of the ejected electron, h is Planck's constant, ν is the frequency of the incident light, and φ is the work function of the material. When the electron is just about to escape the surface of the material, its energy E is zero and the stopping potential V is given by:

eV = hν - φ

Rearranging this gives:

V = (h/e)ν - φ/e

This is the equation of a straight line y = mx + c, where the slope m is h/e. So, by measuring the slope of the graph, you can determine the ratio h/e.

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