The work function of a metal is $2.1 \text{ eV}$. Which of the following wavelengths will be able to emit photoelectrons from its surface?

$A$ metal surface of work function $1.13 eV$ is irradiated with light of wavelength $310 nm$. The retarding potential required to stop the escape of photoelectrons is [Take $hc = 1240 eV \cdot nm$] (in $V$)

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 a photoelectric experiment,a graph of maximum kinetic energy $(KE_{\max})$ against the frequency of incident radiation $(\nu)$ is plotted. If $A$ and $B$ are the intercepts on the $X$ and $Y$ axis respectively,the Planck's constant is given by:

In a photoelectric effect experiment,the maximum kinetic energy of emitted photoelectrons is $K_0$. If the frequency of the incident radiation is increased by a factor of $n_1$,the new maximum kinetic energy becomes $n_2K_0$. Find the work function of the metal.

The wave equation of an electric field at a point is $E = 100 \frac{V}{m} [\sin(7 \omega t) + \cos(10 \omega t) + \cos(15 \omega t)]$ at instant $t$. If the work function of the photocell is $\phi$,then the stopping potential is: