# Some questions on the photoelectric effect

I was reading about the photoelectric effect and a few questions come to my mind.

For the context, I read that if you shine some light (of high energy) on a metal, this latter will emit electrons.

So I was wondering:

• how can a metal lose an electron? Would electrons 'fly' in the air? Or are we just talking about the metal having more "free" electrons?
• if electrons are freed 'in the air', would that mean the metal would become electropositive? I recall in my chemistry classes at high-school when making batteries, we needed a salt bridge between the oxydation and reduction half cells to maintain their electrical neutrality. Do metals need electrical neutrality?
• again if the metal "loses" electrons in the air: if we radiate enough light would the metal end up being short of electrons? as in, would the photo electric effect stop working on a given piece of metal after e.g an hour?

(note: I have read many questions on this website, including Photoelectric effect vs. electronegativity? and the wikipedia pages)

• "Without the current of electrons flowing from anode to cathode, a cloud of electrons will established around the cathode" ─ not quite. Electrons will fly away, but as they do so, the cathode will start to develop a positive charge, and, depending on the geometry (specifically, the cathode's self capacitance), this will produce an additional voltage $V_{\rm c\infty}$ compared to infinity, which acts in the same way as the bias voltage used in normal experiments. (cont.) – Emilio Pisanty Nov 11 '19 at 11:06
• The kinetic energy gained by electrons at infinity will then be $K = h\nu - \Phi - V_{rm c\infty}$, with $\Phi$ the work function of the metal. Once $V_{rm c\infty}$ becomes greater than $\Phi-h\nu$, electrons can no longer be emitted by the cathode: they will attempt to fly away, but they'll be trapped in the charged cathode's Coulombic electrostatic potential well, and they'll fall back to the cathode. – Emilio Pisanty Nov 11 '19 at 11:08