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?

When light is incident on a surface, photoelectrons are emitted. For photoelectrons:

Statement $1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of photoelectrons is $K_{max}$. When $X$-rays are used instead of ultraviolet light, both $V_0$ and $K_{max}$ increase.
Statement $2$: Photoelectrons are emitted with a range of speeds from $0$ to a maximum value because the incident light contains a range of frequencies.

$A$ metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value,then the maximum $KE$ of the emitted photoelectrons would be

The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

In a photoelectric effect experiment,$f$ is the frequency of radiations incident on the metal surface and $I$ is the intensity of the incident radiations. Consider the following statements. Which of the following statements are correct?
$(A)$ If $f$ is increased keeping $I$ and work function constant,then the maximum kinetic energy of the photoelectron increases.
$(B)$ If the distance between the cathode and anode is increased,the stopping potential increases.
$(C)$ If $I$ is increased keeping $f$ and work function constant,then the stopping potential remains the same and the saturation current increases.
$(D)$ If the work function is decreased keeping $f$ and $I$ constant,then the stopping potential increases.