1
$\begingroup$

I'm a super amateur wannabe physicist, and I'm trying to learn the fundamental workings of the photovoltaic effect.

I haven't been able to understand "how" or "what" is displaced/transmitted in the current generated by the electron absorbing the photon's energy.

So, in order to make my question particularly simple, let's imagine a solar cell connected to a 10 cm copper cable, and the cable is connected, on the other end, to an electricity-consuming apparatus.

When a photon, propagating with high enough frequency, hits an electron in the solar cell, and this electron absorbes the photon's energy, what is it exactly that's coming out the other end? Is it the excess energy from "that" electron that's moving from electron-to-electron until it reaches the other end of the copper cable, or is it the electron itself that is moving? And if the second is true.. does that electron get replaced?

I hope I'm not to annoyingly basic and thanks a lot!

$\endgroup$
0
$\begingroup$

If you are interested in the physics of solar cells this series of lectures is great. It may at times be over your head but you should be able to get a general idea.

But to answer your question: Yes, the electron is excited by the photon and will then travel through the circuit, retaining some of the extra energy that was given to it by the photon.

To go into more detail I will steal a slide from the linked lecture series(lecture 2):

Solar cell under illumination

This diagram shows information about the energy states in the semiconductor absorber of the solar cell. The vertical axis is energy and the horizontal axis is position through the solar cell. For more information google search 'band diagrams'.

When a photon is absorbed an electron(red dot) is excited to a higher energy state. This process leaves behind a 'missing electron' known as a hole(blue dot). You can see that this electron travels to the left and exits the cell, goes through the copper wire, and re-enters the cell at the other end. It then will finally recombine with the hole that was left behind.

$\endgroup$
0
$\begingroup$

So let's ignore electrons and in the honor of the World Cup think of footballs (soccer balls) instead. And let's imagine them on the Moon instead. Let’s think of some big crater full of footballs.

Imagine I am in a spaceship and I fire some projectile into this crater: what happens? Ripples in the sea of football. If I fire harder: a splash. If I fire hard enough?

If I can hit the sea hard enough that some footballs get launched a few km per second, they might go so high that they escape the Moon and fall towards Earth instead. We would see a ”projectileofootball current” from the Moon to the Earth. But only if we can overcome the Moon’s “work function,” the energy cost that gravity taxes us to overcome the Moon’s gravity with a football.

Now here's where the analysis breaks down, it turns out footballs are destructible where electrons aren’t, and it turns out that the entire energy of the projectile needs to be absorbed by just one football where real projectiles hit many footballs at once. It’s not a perfect analogy. But that energy becomes kinetic energy of the electron flying out from the “pull” of the photoelectric material it was “bound” in, and some “work function” of energy is stolen by the process of breaking that “pull” before the electron can escape.

These differences are important because they are how we discovered that the energy of light comes in these lumps that are only absorbed by one electron at a time. Here's how: count the number of electrons hitting the other surface, or the number of indestructible footballs hitting the Earth. For projectiles and footballs, you will see that a greater energy of projectile will cause more footballs to hit the Earth. But that's not what we see for light. When we increase the frequency of the light, but keep a measure of intensity that ultimately counts the number of photons constant, we do not see more current. The current is a constant. That's because each photon is being absorbed by one electron, so the number of electrons that are emitted, is equal to the number of photons that get absorbed, minus the electrons that happened to go in the wrong direction.

$\endgroup$
0
$\begingroup$

First, the photovoltaic effect is the creation of voltage and electric current in a material upon exposure to light.

You are talking about the electron being excited to a higher energy level, so I will write about the photoelectric effect too.

In the photoelectric effect, the electron is ejected out of the material, because the photon's energy will be enough to match the electron's work function.

Now in the photovoltaic effect, the electron is excited too, but remains inside the material.

These electrons in the material, in your case the solar cell, are not free. They are loosely bound to the nuclei of the atoms of the material. These electrons, when they interact with a photon, and absorb the photon, gain kinetic energy.

Because these electrons are loosely bound, they can travel from nucleus to nucleus, and when they do, they travel very slowly. This is called drift velocity.

The drift velocity is very slow, but because the electrons are densely packed inside the material, the speed of electricity is close to the speed of light.

You are correct, that the energy of the photon will be transformed into the kinetic energy of the electron, and so the electron moves to the next nucleus. Now that nucleus already has a loosely bound electron, so that electron will be kicked off (will gain kinetic energy), and will travel to the next nucleus.

You are asking whether it is the kinetic energy of the electron that is moving, or is it the electron itself. In your case, the electron that absorbs the photon, will move to the next nucleus. That next nucleus already has a loosely bound electron, that will be kicked off, will gain kinetic energy and will move to the next nucleus. So the answer is that electrons are moving along the material to the next nucleus. The kinetic energy of the electron (that absorbed the photon) will be transferred this way along the material. The charge carriers are the electrons, because each electron in the way will move to the next nucleus. This is called drift velocity.

So it is the electron that is moving to the next nucleus, but then the next electron (that was loosely bound to the next nucleus) moves to the next nucleus too. So the kinetic energy is transferred too. But the kinetic energy is transferred step by step, nucleus to nucleus. It is not the original electron. Each electron moves just to the next nucleus, and kicks off the electron that was there to the next nucleus.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.