# What if all the electrons leave a metal?

I was studying photoelectric effect. Then I thought that what will happen if all the electrons from a metal piece come out as photoelectrons by using a light source of particular frequency? Will the metal still exist?

## 3 Answers

In a small metal bar (let's say 1cm of length, a few grams of weight) there are about $10^{24}$ protons, amounting to a total charge of $10^{14}$ e.s.u. The interaction energy between these protons is $Q^2/r$, which amounts to $10^{28}$ erg, or $10^{21}$ J. This is repulsive energy, of course, so that the value we just estimated is the work we would need to perform in order to assemble this configuration.

In order to get a feeling for the size of this amount of energy, you should compare it to the values contained in this Wikipedia article: it roughly equals the total energy consumptions in the world over a year, and it is less than an order of magnitude below the total estimated energy reserves in the world of gas or petroleum.

Given the size of the energy required to perform the experiment you suggest, the correct answer is: we could never even get close to this configuration, because at some point the excess positive charge left on the metal will start tearing electrons off nearby objects, preventing any further increase in the bar net charge.

You can get an order of magnitude estimate for the moment when this happens as follows. Extracting electrons from nearby objects requires work; the typical work function is of order $1\; keV$. This amount of work must be performed by a net charge $Q$ over a distance of roughly 1 skin depth, i.e. the distance over which the effect of an outside charge is felt inside a conductor. For gold (see this Wikipedia link) a typical skin depth is about $1\; cm$. Thus the bar must generate a field of roughly $1 kV/1 cm$. This is generated, at a distance of $1\; cm$, by a total charge of $10^{10}\; e$, about $14$ (fourteen) orders of magnitude less than required in your question.

Alternatively, you may ask whether this experiment could be performed in space, as follows: place the metal bar far from any object, so that the field necessary to tear electrons off them (still $1\; kV/1\; cm$) cannot be produced even when the metal bar is totally stripped of electrons ($Q = 10^{14}$ esu) because it is too far, at distance $D$. In other words, $$\frac{Q}{D^2} < 1\; kV/1\; cm\;.$$ This requires $D= 10^7\; cm = 100\; km$. On these scales, the solar wind (see here) provides about $10^{23}$ particles in this same volume, and never mind the Earth's atmosphere.

It still looks like a desperate enterprise.

The configuration that you describe (all electrons leaving the metal) would be impossible to generate by using a light source.

The electrostatic repulsion of two point charges is given by Coulomb's law:

$F=\frac{k_e q_1 q_2}{r^2}$

If the positive (protons) and negative (electrons) charges occur in roughly the same quantity, then the material is electrically neutral and the net electrostatic forces cancel. This is the case for most materials you encounter in everyday life. If all of the electrons were to "leave" the metal, the resulting electrostatic repulsive forces between the nuclei would be immense, and the metal would fly apart. The resulting configuration would be unstable.

This can't happen. As you remove electrons the metal will get more and more positively charged. Either it becomes so imbalanced it takes too much energy to remove the electrons anymore, or, the metal tears itself apart from the repulsive force of too much charge in one place.