The maximum velocity of photoelectrons emitted from a metal surface is $1.2 \times 10^6 \ m/s$. If the specific charge of the electron is $1.8 \times 10^{11} \ C/kg$,then the value of the stopping potential in volts is:

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electric potential to the photoemissive substance.
Reason $(R) :$ $A$ negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation.
In the light of the above statements, choose the most appropriate answer from the options given below:

When light of wavelength $\lambda$ incidents on a photosensitive material,photoelectrons are emitted. If the wavelength of the incident light is reduced by $50 \%$,the maximum kinetic energy of the emitted photoelectrons becomes $3$ times the initial maximum kinetic energy. The work function of the material is ($h$ - Planck's constant,$c$ - Speed of light in vacuum).

How does the maximum kinetic energy of a photoelectron depend on the frequency of incident radiation? On which factors does it not depend?

What is the threshold frequency of a metal with a work function of $1.65 \ eV$?

When a photon of energy $6 \ eV$ is incident on a metal surface with a work function of $2.1 \ eV$,what is the stopping potential of the emitted electrons in $V$?