HSC Question 15 2005

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mwatson
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HSC Question 15 2005

Postby mwatson » Fri Oct 31, 2008 9:39 pm

Hi

I was hoping for an explanation to HSC question 15 2005.
My understanding is that electrons in the rod, moving under the influence of both the applied electric and magnetic fields, will experience a force downwards to the bottom of the rod (and out of the page). As these electrons move in this direction they may fill ‘holes’ in the lattice creating a net movement of holes in the opposite direction i.e. into the page and upwards i.e. ans D
The answer is however, C. What is the flaw in my reasoning?

Thanks MW

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joe
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Postby joe » Mon Nov 03, 2008 7:55 am

I was hoping for an explanation to HSC question 15 2005.


What is the question?
Joe

mwatson
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Postby mwatson » Mon Nov 03, 2008 8:30 am

joe wrote:
I was hoping for an explanation to HSC question 15 2005.


What is the question?
Joe


Image

answer is C

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joe
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Postby joe » Mon Nov 03, 2008 8:39 am

This question concerns the Hall effect. If you look it up in a text or the net you will get a nice, three-dimensional diagram, which you may need to analyse the magnetic forces.

When particle with velocity v and charge q moves in a magnetic field B , the magnetic force on the particle is

F = q v X B

In this case, because they are driven by an electric field, the holes and electrons have opposite v. However, they also have opposite q, so the force is in the same direction.

Joe

mwatson
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Postby mwatson » Mon Nov 03, 2008 10:01 am

I’ve read up on the Hall effect and think I understand it. I think my problem is with interpreting holes as simply an absence of an electron that ‘appears’ to move as electrons move the other way. Would I be correct to say that this is an oversimplification and that I need to consider holes in a semiconductor as being real positive charges that can experience forces as do electrons?

Thanks again for your help. :)

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joe
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Postby joe » Mon Nov 03, 2008 10:08 am

I think my problem is with interpreting holes as simply an absence of an electron that ‘appears’ to move as electrons move the other way.


That's not a problem, that is correct, as far as it goes. However, the electron motion is very limited: most electrons cannot move: only an electron that is next to a hole can move, and it can only move into the hole. If you see a picture of this, you'll see that it is very much like a hole moving.

A useful analogy is the apparent motion of a bubble in a liquid. Of course we know that, as the bubble moves up, what is really happening is that fluid is flowing downwards to fill the hole. However, it does look rather like a bubble moving up.

Joe

mwatson
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Postby mwatson » Mon Nov 03, 2008 12:30 pm

Hi Joe,

I think I have a good understanding of what the conduction electrons are doing - the conduction band electrons (while drifting due to the externally applied voltage) experience a downwards force due to the applied magnetic field. This causes a build up of negative electrons on the lower face of the conductor and hence sets up an electric field which will in turn exert an upwards force on the electron. As charge on the lower surface builds so does this secondary electric field. Eventually the force due to the electric and magnetic fields are balanced and there is no net force acting vertically on the electrons. This new potential difference is called the Hall voltage.

When you say that most electrons cannot move, I assume that conventional current will still flow as before, directed into the page.

I still have some difficulty with hole movement though. I understand a hole to be an absence of an electron in the lattice structure of the semiconductor.

Before the magnetic field is applied electron movement in the diagram is out of the page while hole movement is into the page i.e. electrons and holes move in opposite directions under the influence of an externally applied voltage.

The external magnetic field is then applied. I still imagine that as the conduction electrons experience a force downwards that some may move into vacant holes (the fluid filling the holes in your analogy?). Then as before, holes would appear to move up. With so many XS electrons on the lower face how can there be unfilled holes?

Thankyou for your patience!

mw

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joe
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Postby joe » Tue Nov 11, 2008 2:55 pm

Even though I have done so (implicitly) above, it's rather dangerous to treat conduction electrons and holes as classical particles. However, I don't suppose that a high school physics exam was looking for a quantum answer, so I'll have a go, but warn you not to take this too seriously.

Let's take a specific case: suppose E is to the right, so electrons go left and holes to the right. Now put B into the page.

An electron immediately to the right of a hole moves into the hole. Yes, it has a magnetic force acting up, but there is nowhere for it to go: it has filled the hole.

Now imagine a 3x3 square (like noughts and crosses) with a hole in the middle and electrons everywhere else.
e-e-e B X
e-0-e E ->
e-e-e
We've just argued that the one on the right can fill the hole. And the electrons on the left won't (on average*) move into the hole against the field.

But consider an electron to the right of the hole, but above it. It has an electric force to the left and, once it moves, a magnetic force with a component to the left as well. So this electron could move to fill the hole. The result would be that the hole has moved to the right and also upwards. On the other hand, an electron to the right of the hole, but below it would have a magnetic force to the right.

* caveat included because of the thermal excitation.

Joe


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