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You are increasing the energy of each photon, but holding the intensity constant. If you do this, fewer and fewer photons are leaving the lamp, so fewer and fewer are hitting the metal. This is probably the effect you are seeing.

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The minimum energy required to eject an electron out of the metal surface is called the work function of the metal. For simplicity let us denote it by $w$ instead of $\phi$. When a photon of energy $hv$ (where $h$ is plancks constant and $v$ is frequency) is absorbed by an electron, an amount of energy atleast equal to $w$ (provided $hv$>$w$) is used up ...

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Answering your questions one by one: This equation assumes a monochromatic source of light. So $f$ is same for all photons. 'Why can't the photons release many electrons which only just exceed hf=ϕ?' this is because not all the energy of photons are transferred to electrons, or some of the electrons may lose kinetic energy via other collisions, etc... ...

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In particle interactions the total number of particles is not conserved. For example in a collision in the LHC two photons collide and many hundreds of particles are created in the collision. There are still some conserved quantities, for example lepton number is still conserved so you cannot just create an electron. You need to create an electron and ...

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First: momentum is conserved, energy is conserved, electric charge is conserved, and a few other things. Particles and particle number can, in some instances, flit into and out of existence. The photon is absorbed by the electron. I.e. the photon ceases to exist, its energy and momentum have to go somewhere (the electron and the bulk material). In the ...

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Have you read the wiki link electrons are only dislodged by the photoelectric effect if light reaches or exceeds a threshold frequency, below which no electrons can be emitted from the metal regardless of the amplitude and temporal length of exposure of light. To make sense of the fact that light can eject electrons even if its intensity is low, Albert ...

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The work function, $\phi$ is the amount of energy required to free the electron from the pull of the nuclei of the atoms of the photosurface. Here $\phi=3\text{ eV}$. Since the kinetic energy of the electron is given by, $$E_k=hf-\phi$$ it becomes evident that the condition of electron emission is when $hf>\phi$. Clearly this is not the case, which is ...

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