What happens to the incident photon involved in the photoelectric effect experiment?

If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

Two monochromatic light beams $A$ and $B$ of equal intensity are incident normally on a metallic surface per unit area. Their wavelengths are $\lambda_A$ and $\lambda_B$ respectively. The incident light is effective in ejecting photoelectrons. The ratio of the number of photoelectrons emitted by beam $A$ to that by beam $B$ is:

When a point source of light is at a distance of $1 \ m$ from a photocell,the cut-off voltage is found to be $V$. If the same source is placed at $2 \ m$ distance from the photocell,the cut-off voltage will be

$A$ metal is exposed to light of wavelength $800 \, nm$ and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of the photoelectrons doubles when light of wavelength $500 \, nm$ is used. The work function of the metal is $... \, eV$ (Take $hc = 1230 \, eV \cdot nm$).

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $1:k$, then the threshold frequency of the metallic surface is: