Which one of the following graphs correctly represents the variation of maximum kinetic energy $(E_{k})$ of the emitted electrons with frequency $(\nu)$ of incident light in the photoelectric effect?

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:

The work function of a metal surface is $1.6 \times 10^{-19} \text{ J}$. When light of wavelength $6400 \ \mathring{A}$ is incident on this surface,the maximum kinetic energy of the emitted photoelectrons is ....... $\text{J}$.

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

Energy of the incident photons on the photosensitive metal surface is $3W$ and then $5W$ where '$W$' is the work function of that metal. The ratio of velocities of the emitted photoelectrons is

Light of wavelength $488 \, nm$ produced by an Argon laser is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping potential of the photoelectrons is $0.38 \, V$. The work function of the cathode material is (in $ \, eV$)