Does the Photoelectric Effect Require a Battery or a Circuit?

Removing each electron from the metal requires energy measured in electron volts. The number of volts is the work function, W. The illuminating photons have an energy related to the frequency (E = hf, where E is energy, h is Planck's constant and f is the frequency of the light).

When the energy of the photons exceeds the energy needed to free an electron, the photon will cause an electron to be ejected with kinetic energy equal to the photon energy minus the work function. E_k = hf - W. If the photon energy is less than the work function, there will be no current.

When the plate is illuminated and electrons are ejected, it acquires a positive charge.

If there is no connection between the illuminated plate and the other plate, electrons will be ejected and land on the other plate until the opposing electric field is strong enough to counter the electrons kinetic energy, E_k. There will be a voltage between the plates.

If the plates are connected with a wire, both plates will be the same potential, there will be no electric field between the plates, and a current will flow through the wire. The amount of current will be proportional to the intensity of the light (number of photons).

If there is a battery in the circuit, it depends on the battery's polarity, whether it opposes or aids the photoelectric effect. If it aids the photoelectric effect, the result will be that lower energy photons (lower frequency) would be required for electron emission (current flow). If it opposes the photoelectric effect, higher energy (higher frequency) photons would be required. The amount of current would be solely determined by the intensity of the light (number of photons).

http://photonicswiki.org/index.php?title=Work_Function_of_Metals

http://physicsnet.co.uk/a-level-physics-as-a2/electromagnetic-radiation-quantum-phenomena/photoelectric-effect/

"If there is no connection between the illuminated plate and the other plate, electrons will be ejected and land on the other plate until the opposing electric field is strong enough to counter the electrons kinetic energy, E_k. There will be a voltage between the plates."

How to find the voltage between the plates
in this case the energy of an X-ray photon is 1000kev

since "E_k = hf - W "

Work function of iron is 4.47 eV

Ek=1000Kev-4.47ev

= 999995.53 let us take it as 1Mev

so if the plate gets charged more than 1 Mev then the electrons coming with a charge of 1 Mev gets repealed.

so what will be the potential difference between the plates if one plate gets charged to 1 Mev?

1MeV is pretty high for an x-ray, unless your naming convention is source based. Beyond that, some other effects are going to come into play that will lead to electrons being not nearly that energetic.

Anyway, an electron gains or loses 1eV traversing a 1 volt difference in potential. So 1MeV would require of or provide to an electron 10⁶ volts.

As a practical consideration, if the plates are 1mm apart you would have a hard time illuminating one of them. If the plates are far apart, then some of the ejected electrons will land elsewhere and not on the other plate. Putting a battery in the circuit provides an electric field in the vacuum, attracting the electrons to the other plate so that the total current can be measured.

It can be shown that increasing the intensity of the light does not increase the electrons' energy but does increase the number of electrons (current). The electron energy is proportional to the frequency (color) of the light, so below a certain light frequency, the electrons do not have sufficient energy to escape the illuminated plate. This is the opposite of what you would expect from the wave theory of light.

This site provides an excellent description of the photoelectric effect.

https://physics.info/photoelectric/