The number of photoelectrons emitted for light of frequency $v$ (higher than the threshold frequency $v_0$) is proportional to

What is the intensity of radiation (light)?

For photoelectric emission, tungsten requires light of $2300 \mathring{A}$. If light of $1800 \mathring{A}$ wavelength is incident, then emission:

Two metallic plates $A$ and $B$,each of area $5 \times 10^{-4} \, m^2$,are placed parallel to each other at a separation of $1 \, cm$. Plate $B$ carries a positive charge of $33.7 \, pC$. $A$ monochromatic beam of light,with photons of energy $5 \, eV$ each,starts falling on plate $A$ at $t = 0$,so that $10^{16}$ photons fall on it per square meter per second. Assume that one photoelectron is emitted for every $10^6$ incident photons. Also assume that all the emitted photoelectrons are collected by plate $B$ and the work function of plate $A$ remains constant at $2 \, eV$. The electric field between the plates at the end of $10 \, seconds$ is:

For alkali metals, threshold frequencies lie in which region?

When monochromatic radiation of intensity $I$ falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are $N$ and $K$ respectively. If the intensity of radiation is $2I$, the number of emitted electrons and their maximum kinetic energy are respectively