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Asked by Sonia Last Modified
Answer To determine which of the two photosensitive metals, A or B, has a higher value of the work function, we need to analyze their respective graphs showing the variation of stopping potential (VsVs) with frequency (ff) of incident radiation.
The work function (WW) of a metal is the minimum energy required to remove an electron from the surface of the metal and is directly related to the threshold frequency (f0f0) by the equation:
W=hf0W=hf0
where:
From the graph, the stopping potential (VsVs) is directly proportional to the frequency (ff) of the incident radiation and is given by the equation:
Vs=hfe−WeVs=ehf−eW
where:
By analyzing the graph, we can observe the following:
Since the work function is proportional to the intercept, the metal with the higher stopping potential (i.e., the metal with the higher intercept on the VsVs axis) has a higher work function.
Therefore, in the given graph, the metal associated with the higher stopping potential (further above the x-axis) has the higher value of the work function.
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