Assertion : Photoelectric saturation current increases with the increase in frequency of incident light.
Reason : Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.

Energy conversion in a photoelectric cell takes place from

Photoelectric effect experiments are performed using three different metal plates $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ having work functions $\phi_p=2.0 \mathrm{eV}, \phi_q=2.5 \mathrm{eV}$ and $\phi_r=3.0 \mathrm{eV}$, respectively. A light beam containing wavelengths of $550 \mathrm{~nm}, 450 \mathrm{~nm}$ and $350 \mathrm{~nm}$ with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is [Take $h c=1240 \mathrm{eV} \mathrm{nm}$ ]

The energy of a photon is $E = hv$ and the momentum of photon $p = \frac{h}{\lambda }$, then the velocity of photon will be

The work functions of Caesium $(Cs)$,Potassium $(K)$ and Sodium $(Na)$ are $2.14\,eV,2.30\,eV$ and $2.75\,eV$ respectively. If incident electromagnetic radiation has an incident energy of $2.20\, eV$, which of these photosensitive surfaces may emit photoelectrons?