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Given below are two statements :
Statement I :
Figure shows the variation of stopping potential $$(V_0)$$ with frequency $$(\nu)$$ for the two photosensitive materials $$M_1$$ and $$M_2$$. The slope gives value of $$\frac{h}{e}$$, where $$h$$ is Planck's constant, $$e$$ is the charge of electron.
Statement II : $$M_2$$ will emit photoelectrons of greater kinetic energy for the incident radiation having same frequency.
In the light of the above statements, choose the most appropriate answer from the options given below.
$$eV_0 = h\nu - \phi \implies V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e}$$
$$\text{Slope} = \frac{h}{e} \implies \text{Statement I is correct}$$
$$\text{From graph, threshold frequency: } \nu_{02} > \nu_{01} \implies \phi_2 > \phi_1$$
$$K_{\text{max}} = h\nu - \phi \implies K_{\text{max1}} > K_{\text{max2}}\ \text{for same } \nu \implies \text{Statement II is incorrect}$$
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