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(D) Given that,work function $\phi_0 = 1.5 \ eV$. The stopping potential $V_s$ is given by Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambd

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$III, II, IV$

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(D) Given that,work function $\phi_0 = 1.5 \ eV$. The stopping potential $V_s$ is given by Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \phi_0$. Since $V_s \propto \frac{1}{\lambda}$,smaller wavelength corresponds to a larger magnitude of stopping potential (more negative). Comparing wavelengths: $\lambda_A = 200 \ nm  |V_B| > |V_C|$. From the graph,the stopping potentials are $V_1, V_2, V_3, V_4$ where $|V_1| > |V_2| > |V_3| > |V_4|$. Therefore,$A \rightarrow V_2$,$B \rightarrow V_3$,and $C \rightarrow V_4$. Saturation photocurrent is directly proportional to the intensity of incident light. Given intensities: $I_A = 1.8 \ W/m^2$,$I_B = 1 \ W/m^2$,$I_C = 0.5 \ W/m^2$. Thus,the saturation current order is $I_A > I_B > I_C$. Matching these with the curves: Curve $III$ has stopping potential $V_2$ and high saturation current,corresponding to light $A$. Curve $II$ has stopping potential $V_3$ and medium saturation current,corresponding to light $B$. Curve $IV$ has stopping potential $V_4$ and low saturation current,corresponding to light $C$. Thus,the correct correspondence is $A \rightarrow III, B \rightarrow II, C \rightarrow IV$. Option $(d)$ is correct.

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