Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: n$,then the threshold frequency of the metallic surface is

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.

The threshold wavelength for photoelectric emission from a material is $5200 \, \mathring{A}$. Photo-electrons will be emitted when this material is illuminated with monochromatic radiation from a

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other,the maximum velocities of the photoelectrons emitted in the two cases are $V_1$ and $V_2$ respectively. The ratio $V_1: V_2$ is

$A$ metal is exposed to light of wavelength $800 \, nm$ and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of the photoelectrons doubles when light of wavelength $500 \, nm$ is used. The work function of the metal is $... \, eV$ (Take $hc = 1230 \, eV \cdot nm$).

When radiation of the wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then,the threshold wavelength for the surface is :