In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is (Where $h =$ Planck's constant,$m =$ mass of electron,and $c =$ speed of light).

This question has Statement $1$ and Statement $2.$ Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1:$ $A$ metallic surface is irradiated by a monochromatic light of frequency $v > v_0$ (the threshold frequency). If the incident frequency is now doubled,the photocurrent and the maximum kinetic energy are also doubled.
Statement $2:$ The maximum kinetic energy of photoelectrons emitted from a surface is linearly dependent on the frequency of the incident light. The photocurrent depends only on the intensity of the incident light.

In the photoelectric effect experiment, if the frequency of incident radiation $(v)$ is increased, keeping all other factors constant, what happens to the stopping potential $(V_s)$, given $(v > v_0)$ where $(v_0)$ is the threshold frequency?

When the energy of incident radiation is increased by $20 \%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5 \ eV$ to $0.8 \ eV$. The work function of the metal is (in $eV$)

The maximum velocity of electrons emitted from a metal surface is $V$,when the frequency of light falling on it is $f$. What is the maximum velocity when the frequency becomes $4f$?