Work function of lithium and copper are respectively $2.3 \ eV$ and $4.0 \ eV$. Which one of the metal will be useful for the photoelectric cell working with visible light? $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s)$

$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

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:

The wavelengths of light incident on a photocell are $400 \ nm$ and $250 \ nm$. The velocities of the emitted photoelectrons are $v$ and $2v$ respectively. What is the work function of the metal?

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential of ejected photoelectrons is $4.8 \, V$. If the same surface is illuminated by radiation of double the previous wavelength,then the stopping potential becomes $1.6 \, V$. The threshold wavelength of the metal is $... \lambda$.

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential for photoelectric current is $6V_0$. When the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $2V_0$. The threshold wavelength for the photoelectric effect for this surface is: