The work functions of three metals $A, B$ and $C$ are $W_A, W_B$ and $W_C$ respectively. They are in decreasing order $(W_A > W_B > W_C)$. The correct graph between the maximum kinetic energy $E_k$ of the emitted electron and the frequency $v$ of the incident radiation is:

Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

$A$ photon of energy $8 eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} Hz$. The maximum kinetic energy of the emitted photoelectrons is ............... $eV$ (Given $h = 6.6 \times 10^{-34} Js$).

The threshold frequency of a metal is $f_0$. When light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of the photoelectrons is $v_1$. When the frequency of the incident radiation is increased to $5f_0$, the maximum velocity of the photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is

What is the slope of the $V_0 - \nu$ graph? Write its equation. Also,mention whether it depends on the type of material or not.