$A$ light source of wavelength $\lambda$ illuminates a metal surface and electrons are ejected with maximum kinetic energy of $2 \ \text{eV}$. If the same surface is illuminated by a light source of wavelength $\frac{\lambda}{2}$, then the maximum kinetic energy of ejected electrons will be (The work function of the metal is $1 \ \text{eV}$). (in $\text{eV}$)

By the photoelectric effect,Einstein proved:

Given below are two statements:
Statement-$I$: The figure shows the variation of stopping potential $(V_0)$ with frequency $(v)$ for two photosensitive materials $M_1$ and $M_2$. The slope gives the value of $\frac{h}{e}$,where $h$ is Planck's constant and $e$ is the charge of an electron.
Statement-$II$: $M_2$ will emit photoelectrons of greater kinetic energy for incident radiation having the same frequency.
In the light of the above statements,choose the most appropriate answer from the options given below.

$A$ photocell is illuminated by a $100 \ W$ mercury source. It emits ultraviolet light of wavelength $2271 \ \mathring A$. If the stopping potential is $1.3 \ V$,find the work function of the metal in $eV$.

Photoemission from a metal surface takes place for frequencies $v_1$ and $v_2$ of incident rays. If the ratio of the maximum kinetic energy of photoelectrons is $1:K$,then the threshold frequency of the metallic surface is

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be