Photoelectric effect experiments are performed using three...

$\begin{gathered}\text{ 1. }K_p=E_p-\phi_p=\frac{1240}{550}-20=0.2545 \mathrm{eV}\\ K_q=E_q-\phi_q=\frac{1240}{450}-25=0.255 \mathrm{eV}\\ K_r=E_r-\phi_r=\frac{1240}{350}-3.0=0.543 \mathrm{eV} \end{gathered}$In the above equation $K$ represents maximum kinetic energy of photoelectrons and $E$, the energy of incident right. From the above values we can see that stopping potential, $\left|V_r\right| \gt \left|V_q\right| \gt \left|V_p\right|$ Further, their intensities are equal, but energy of individual photon of $r$ is maximum. Hence, number of photons incident (per unit area per unit time)$of r$ can be assumed to be least. Hence, saturation current of $r$ should be minimum. Keeping these points in mind no option seems to be correct. The correct graph is shown below, $<img src="/api/proxy-image?url=https$ \\ \therefore $\text{ No choice is correct. }$