Planck-Einstein relation and the Photoelectric Effect

[Abu]

Hi everyone, I just have some confusion regarding Planck's and Einstein's equation.
The following is an explanation of the photoelectric effect using Einsteins theory:
Light is composed of photons. Each photon has energy hf and mass hf/c^2. When ultraviolet photons are brought to rest by zinc, the mass of the photon changes to energy. This energy is used to break the binding energy between the outermost electron and the nucleus. The excess energy is carried by the electrons as kinetic energy.

Here is my confusion though:

When we ask for the energy of a photon, we say it is equal to E = hf. When a photon is brought to rest in the explanation of the photoelectric effect, it says the mass of the photon during its motion changes to energy. So shouldn't the energy of the photon be hf/c^2, not hf?

For example, we say that the kinetic energy of the electrons in the photoelectric effect is:
KE = hf - w where w is the work function.
How come it isn't KE = hf/c^2 minus w

Maybe I am misunderstanding the photoelectric effect when I say that the mass of the photon changes to energy? Perhaps the mass does not completely change into energy, and thus hf/c^2 is not applicable, only for E = mc^2?

If my question is unclear, let me know. Thank you very much.

 

[mfb]

Photons have a mass of zero. Forget the concept of relativistic mass, it is not used any more. Photons cannot be brought to rest (at least not in the way you try to do that here). If the photons hit the metal their energy can be transferred to something else (typically an electron), they stop existing then.

This energy is used to break the binding energy between the outermost electron and the nucleus.

While that is possible, most of the time the electron that absorbs the photon is not bound in a specific atom.

So shouldn't the energy of the photon be hf/c^2, not hf?

That doesn't even have the right units to be an energy.

 

[Abu]

Photons have a mass of zero. Forget the concept of relativistic mass, it is not used any more. Photons cannot be brought to rest (at least not in the way you try to do that here). If the photons hit the metal their energy can be transferred to something else (typically an electron), they stop existing then.While that is possible, most of the time the electron that absorbs the photon is not bound in a specific atom.That doesn't even have the right units to be an energy.

Thank you so much for your input. Unfortunately, I cannot forget about relativistic mass because it is the way I am being taught currently, so I have to try my best to understand it. But I thought about my question as well as your response:
So from what I know now and what I am currently being taught, the photon has a relativistic mass of hf/c^2 (not used anymore, but I am still being taught it). So we can apply that to E = mc^2 where m is the relativistic mass, correct? So e = hf/c^2(c/^2), and now we have E = hf. So from that, in the explanation provided above that was said by my teacher, we can say that the energy provided by the relativistic mass of the photon is what is allowing the electrons to gain kinetic energy.

I'm sorry I had to disregard the comment on relativistic mass, but its crucial I understand it, I think, even though from what I've read it can run into a whole load of problems.

Thank you very much.

 

[mfb]

and now we have E = hf.

Right, and the detour via the mass didn't help. The photon has energy, this energy can be transferred to an electron. No masses involved.