Energy of a photoelectron depends on which factors?

An electron and a proton are separated by a large distance. The electron starts approaching the proton with an initial kinetic energy of $3 \, eV$. The proton captures the electron and forms a hydrogen atom in the second excited state. The resulting photon is incident on a photosensitive metal with a threshold wavelength of $4000 \, Å$. What is the maximum kinetic energy of the emitted photoelectron? (In $eV$)

The difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work functions $\phi_A = 9 \, eV$ and $\phi_B = 4.5 \, eV$ in $nm$ is: (Given,$hc = 1242 \, eV \, nm$)

Photons with energy $5\, eV$ are incident on a cathode $C$ in a photoelectric cell. The maximum energy of emitted photoelectrons is $2\, eV.$ When photons of energy $6\, eV$ are incident on $C,$ no photoelectrons will reach the anode $A,$ if the stopping potential of $A$ relative to $C$ is ............ $V$.

If the frequency of light in a photoelectric experiment is doubled,the stopping potential will

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: