When light of frequency twice the threshold frequency is incident on a metal plate,the maximum velocity of the emitted electron is $v_{1}$. When the frequency of incident radiation is increased to five times the threshold value,the maximum velocity of the emitted electron becomes $v_{2}$. If $v_{2} = x v_{1}$,the value of $x$ will be...........

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.

The electric field associated with a light wave is given by $E = E_0 \sin [1.57 \times 10^7 (x - ct)]$,where $x$ is in meters and $t$ is in seconds. If this light is used to produce photoelectric emission from a metal surface with a work function of $1.9 \ eV$,what will be the stopping potential in $V$?

The photoelectric threshold wavelength of Tungsten is $2300\; \mathring{A}$. The kinetic energy of the electrons ejected from the surface by ultraviolet light of wavelength $1800\; \mathring{A}$ is $.............\,eV$.

Initially,a photon of wavelength $\lambda_1$ falls on a photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to $\lambda_2$,the maximum energy of the electron emitted becomes $E_2$. Then the value of $hc$ ($h=$ Planck's constant,$c=$ velocity of light) is

Explain Einstein's photoelectric effect and derive Einstein's photoelectric equation.