In the experiment of the photoelectric effect, if the frequency of incident light is $v_1$, the maximum kinetic energy $(K.E.)$ of photoelectrons is $K_0$. If the frequency of light is $v_2$, the maximum $K.E.$ of photoelectrons is $2K_0$. Which of the following relations is correct?

The stopping potential for photoelectrons from a metal surface is $V_{1}$ when monochromatic light of frequency $v_{1}$ is incident on it. The stopping potential becomes $V_{2}$ when monochromatic light of another frequency is incident on the same metal surface. If $h$ is the Planck's constant and $e$ is the charge of an electron,then the frequency of light in the second case 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 $(hc = 12400 \, eV \mathring A)$.

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

When the wavelength of radiation falling on a metal is changed from $500 \, nm$ to $200 \, nm$,the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to $..... \, eV$.

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?