Which of the following graphs shows the correct variation of maximum kinetic energy $(E)$ of photoelectrons with the intensity of incident radiation $(I)$?

When electromagnetic radiation is incident on a metallic surface,the maximum kinetic energy of the photoelectrons depends on .......

Light of incident frequency $3$ times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{4}\right)^{\text{th}}$ and intensity is tripled,then the photoelectric current will

$A$ beam of light of wavelength $\lambda$ falls on a metal having work function $\phi$ placed in a magnetic field $B$. The most energetic electrons,moving perpendicular to the field,are bent in circular arcs of radius $R$. If the experiment is performed for different values of $\lambda$,then the $B^2$ vs. $\frac{1}{\lambda}$ graph will look like (keeping all other quantities constant):

Assuming photoemission takes place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10} \, m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10} \, m$ is (Given $h = 4.14 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):