Photoelectric Effect

[Shark 774]

Will doubling the intensity of the incident light double the photocurrent or is it not directly proportoinal? 

[b^3]

As long as everything else stays the same then doubling the intensity of incident light will double the photocurrent. More photons are being delivered per second but since the wavelenght of photons stays the same, then the amount of energy the electrons absorb remain the same so more electrons will be emitted with the same amount of energy.

[cranberry]

* most energetic electrons have same energy yeah?

[Lasercookie]

* most energetic electrons have same energy yeah?

I am not entirely sure what you mean, but I'll try and explain as I understand it.

What gives the ejected electron it's energy is the photon. The electron absorbs the energy from a single photon (conservation of energy principle). Increasing the intensity means that there are more of these photons. These photons each have the same amount of energy (E=hf). So if there's more of these photons, and one photon gives it's energy up completely to one electron and each has the same amount of energy - then each electron will receive the same amount of energy.

BUT! We know that with our energy level diagrams and etc., the first electron to be emitted will be the "most energetic electron" (i.e. the least bound electron or the outer-most electron - whatever you want to call it). The other electrons require a bit more energy to get to, as they may be in different states. This uses up some of the energy delivered by the photons, so these electrons may have less excess energy to convert to Ek when they are emitted.

This allows us to find the the maximum kinetic energy of the photoelectron. This is . Note that concept of work function and stopping voltage are clearly linked to "most energetic electron". The work function is the energy required to eject the least-bound electron from the metal and the stopping voltage is the energy required to stop the most energetic electron. Different words that mean the same thing really.

Notice that these is pretty much the only calculation we refer to (other than ) when we do the photoelectric stuff. For the sake of simplicity, it looks like we only really discuss these "most energetic electrons" and don't have to refer to the fact that these other electrons will have less energy (except when we're talking about energy in the case of emission and absorption spectra, we get asked questions like that then). This might be a limitation set by the study design (I haven't really checked thoroughly for anything that implies this, other than the fact that our equations are limited to being about the "most energetic electron") - my teacher told us earlier that we pretty much only consider the most energetic electrons in our calculations.

So to recap, the photons deliver the same energy but the electrons may not be emitted with the same energy. But this is VCE Physics and I'm pretty sure we ignore the other electrons.

This is another one of my posts where I feel like I've explained something in a dodgy way, that implies something incorrect. (I'm not that great at explaining stuff). It might be that my understanding is flawed. If that is the case, I apologise in advance and hopefully somebody who knows a bit more/can explain better will post. 

[b^3]

* most energetic electrons have same energy yeah?

This is another one of my posts where I feel like I've explained something in a dodgy way, that implies something incorrect. (I'm not that great at explaining stuff). It might be that my understanding is flawed. If that is the case, I apologise in advance and hopefully somebody who knows a bit more/can explain better will post. 

laserred, you can explain physics concepts much better than preety much everyone one here, and you put your time and effort into typing all that out and explaining it. Good job and we know that you are going to do well this year in physics.

Thanks for cleary that up, I'd been thinking about cranberr's post for the last 40 minutes.

[Shark 774]

As long as everything else stays the same then doubling the intensity of incident light will double the photocurrent. More photons are being delivered per second but since the wavelenght of photons stays the same, then the amount of energy the electrons absorb remain the same so more electrons will be emitted with the same amount of energy.

I asked my teacher and apparently that's not true. He isn't the best at explaining things, but he usually does know his facts... But he said that's beyond the scope of the course anyway.

[b^3]

As long as everything else stays the same then doubling the intensity of incident light will double the photocurrent. More photons are being delivered per second but since the wavelenght of photons stays the same, then the amount of energy the electrons absorb remain the same so more electrons will be emitted with the same amount of energy.

Really? I'm just using the theory I learnt out of the Heinemann book, but isn't this got to do with the photoelectric effect "partly" dispproving the wave model why "partly" proving the particle model?
I asked my teacher and apparently that's not true. He isn't the best at explaining things, but he usually does know his facts... But he said that's beyond the scope of the course anyway.

[xZero]

In vce physics, b^3 is right but in an experiment we did in uni, stopping voltage (max kinetic energy of electrons) does depend on intensity but they never told us why :S

[cranberry]

yeah i thought so....it's just knowing that if we get another question asking to explain what "Ek" means in relation to the experiment, you have to mention that its the energy of the "most energetic" electrons, hence max ek yeah

[Lasercookie]

In vce physics, b^3 is right but in an experiment we did in uni, stopping voltage (max kinetic energy of electrons) does depend on intensity but they never told us why :S

Hey xZero (or anyone else), did I end up saying anything incorrect in my post? I've been looking up and trying to find references for what I said and I can't figure out if I've said something completely wrong or if what I said is correct. Feeling mega-confused at the moment .

Oh yeah, with the stopping voltage and intensity thing - is it because intensity increases the amount of electrons ejected. Because there are more electrons, you'd need more energy to stop them?

[xZero]

seems good to me, as for your explanation, i dont think so because as you said in your post, we only look at the most energetic electrons and what repulsive force i need to dissipate all its kinetic energy. Its like setting up a unbreakable barrier, if you throw 1 ball at it, it bounces back, you throw a million balls at it, they all bounce back.

[Lasercookie]

seems good to me,

Thanks.

as for your explanation, i dont think so because as you said in your post, we only look at the most energetic electrons and what repulsive force i need to dissipate all its kinetic energy. Its like setting up a unbreakable barrier, if you throw 1 ball at it, it bounces back, you throw a million balls at it, they all bounce back.

Hmm yeah. I'll try to find out more about the reason (hopefully it's not way beyond my level) after the physics exam - I want to know the answer lol.

[Shark 774]

Just want to know if doubling the number of photons hitting the metal will always double the number of electrons coming out of the metal, or not necessarily. Anyone know for sure, or not exactly? I wish my teacher could explain himself better...

[jimbothebest]

My understanding of it is that photocurrent is directly proportional to light intensity, so doubling the intensity will double the photocurrent. I'm pretty sure in the 'real world' this is not the case, as not every electron is the 'most loosely bound' or won't have the 'maximum kinetic energy' or whatever. But in VCE physics I've been told that a double in light intensity will double the amount of incident photons, and therefore double the amount of ejected electrons and hence photocurrent. Hope this is of any assistance.

[Lasercookie]

Yeah, that's my understanding of it as well. However, I tend to avoid explicitly using the word 'double' and just go for increase when answering theory questions though. I don't know why I do that, I guess because I'm not 100% convinced that it doubles perfectly.

There was that question in VCAA 2009 (Q7ii) about electrons emitted per second when intensity is doubled. VCAA used the term 'increase' (though my teacher used the term 'double'). It's probably safer to go with VCAA's wording on the matter.

If I was asked to draw a graph of photo current after intensity is doubled, I would draw it to be doubled. I can't immediately recall any VCAA questions like that (I think they tend opt for the multiple choice of graphs).

I'll try to find more stuff relating to it when I read through the assessor reports again tomorrow.

About the 'real world' explanation, I think I went into that in that post I made earlier. I did some checking and I'm pretty sure most of what I said there is correct.