In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of photoelectrons will become

Light of frequency $7.21 \times 10^{14} \; Hz$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5} \; m/s$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

In the photoelectric effect,when photons of energy $h \nu$ fall on a photosensitive surface (work function $h \nu_0$),electrons are emitted from the metallic surface. It is possible to say that:

From a metallic surface,photoelectric emission is observed for frequencies $v_1$ and $v_2$ $(v_1 > v_2)$ of the incident light. The maximum kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence,the threshold frequency of the metallic surface is

$A$ photosensitive metallic surface is illuminated alternately with lights of wavelength $3100 \mathring A$ and $6200 \mathring A$. It is observed that the maximum speeds of the photoelectrons in the two cases are in the ratio $2:1$. The work function of the metal is $(hc = 12400 \, eV \mathring A)$.

Light of frequency $4v_0$ is incident on a metal surface with a threshold frequency $v_0$. The maximum kinetic energy of the emitted photoelectrons is: