Explanation

(a) $K_{p} = E_{p} - \phi_{p} = \frac{1240}{550} - 2.0 = 0.2545\ eV$
$K_{q} = E_{q} - \phi_{q} = \frac{1240}{450} - 2.5 = 0.255\ eV$
$K_{r} = E_{r} - \phi_{r}\ = \frac{1240}{350} - 3.0 = 0.543$eV
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,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left| V_{r} \right| > \left| V_{q} \right| > \left| V_{p} \right|$
Further, their intensities are equal, but energy of individual photon $r$ 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
[image:11_image39.emf]
$\therefore$ No choice is correct.