The variation of the stopping potential (Vo) with the frequency (v) of the light incident on two different photosensitive surfaces M1 and M2 is shown in the figure. Identify the surface which has greater value of the work function.

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The variation of the stopping potential (V_o) with the frequency (v) of the light incident on two different photosensitive surfaces M₁ and M₂ is shown in the figure. Identify the surface which has greater value of the work function.

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Detailed Solution

${eV}_{0} = hv - {hh}_{0}$
Here, e is the charge of electron, h=planck's constant, v₀ is the threshold frequency and v is the frequency of the incident light
on comparing with the Y = mX + C we can observe that more the intercept on the Y-axis or V0 more
is the frequency(v₀).
Hence from the graph, the threshold frequency of Metal A is greater than the Metal B, therefore the
work function of Metal A is more than Metal B.
M2
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