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

When a point source of monochromatic light is at a distance of $0.2 \ m$ from a photoelectric cell,the cut-off voltage and the saturation current are $0.6 \ V$ and $18 \ mA$ respectively. If the same source is placed $0.6 \ m$ away from the photoelectric cell,then:

Light of energy $E$ falls normally on a metal of work function $\frac{E}{3}$. The kinetic energies $K$ of the photoelectrons are

Light is incident on a metallic plate having a work function of $ 110 \times 10^{-20} \ J $. If the produced photoelectrons have zero kinetic energy, then the angular frequency of the incident light is . . . . . . $ rad/s $. $( h = 6.63 \times 10^{-34} \ J \cdot s )$

The work functions of Silver and Sodium are $4.6 \ eV$ and $2.3 \ eV$,respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is :

$A$ 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 photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)