Which one of the following four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident light $(I)$ is correct?

When a photon of energy $4.25\, eV$ strikes the surface of a metal $A$,the ejected photoelectrons have a maximum kinetic energy $T_A\, eV$ and de-Broglie wavelength ${\lambda _A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.70\, eV$ is ${T_B} = ({T_A} - 1.50)\, eV$. If the de-Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$,then:

In the graph of maximum kinetic energy of photoelectrons emitted from a metal surface versus the frequency of incident radiation based on Einstein's photoelectric equation,the slope is .......

The work function of a surface of a photosensitive material is $6.2\, eV$. The wavelength of the incident radiation for which the stopping potential is $5\, V$ lies in the

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$ : The photoelectric effect does not take place,if the energy of the incident radiation is less than the work function of a metal.
Reason $R$ : Kinetic energy of the photoelectrons is zero,if the energy of the incident radiation is equal to the work function of a metal.
In the light of the above statements,choose the most appropriate answer from the options given below.

When photons of energy $5 eV$ are incident on a metal surface with a work function of $4.36 eV$,photoelectrons are emitted. What is the maximum momentum of the emitted photoelectrons?