Gap energy and changing diameters

In summary, David found a way to solve the homework equation using a radius of 4 and found that the result was close to what was graphed in the answer.
  • #1
dscot
32
0

Homework Statement


Hi,

I'm trying to show how the band gap energy changes with the radius as Part 2 of this questions asks us to find: http://screencast.com/t/N7R42aS4l

The full solution is here: http://screencast.com/t/y1K3hqOus4H

Homework Equations



See solution link above.

The Attempt at a Solution



This question should relatively straightforward and should just be a simple case of inputting the values into the equation given in the link above. I think we also need to pick a radius at set intervals (ie. 1,2,3,4.. nm) for use in the equation. The problem is that I always get 1.42 for my Eg(R) value. I tested my equation and this seems to be due to the value I get being so low that when I add the first term of Eg to it I always end up with that as my final answer.

Any thoughts on what I could be doing wrong?

Thanks
David
 
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  • #2
You'll need to watch your units carefully. For example, can you show what you get for just the last term ##1.8e_o^2/(\epsilon R)## for ##R = 1## nm? This looks like a Coulomb energy term. In SI units the ##e_o^2## part would have an additional factor of Coulomb's constant ##k = 1/4\pi \epsilon_o##
 
  • #3
Hi TSny,

For that last term I get [itex]\frac{1.8 * (1.602*10^-19)^2}{(12.9 * 1*10^-9)}[/itex]

= 3.58 *10-30 c2/nm

I'm not sure what you mean about an additional factor for Coulomb's constant?

Thanks!
 
  • #4
The Coulomb energy of two point charges ##e_o## is ##U = k \;e_o^2/R## where ##k ## is Coulomb's constant. Note that you are getting units of C2/nm which is not an energy unit. However, if you multiply by ##k##, you will get energy since ##k## has units of J m /C2 where J is Joules. You'll also need to convert over to eV.
 
  • #5
Hi,

Thanks for that I think I understand the logic now, I re-did the equation and now have: http://screencast.com/t/yByLIrwdi9

The problem is that no matter what I change the radius to all the answers are always 1.41...J

Sorry for all the trouble!
Thanks
 
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  • #6
Almost there. For the last term where you included ##k = 1/(4 \pi \epsilon_o)##, you need to use ##\epsilon_o = 8.85 \times 10^{-12}## SI units. For the second term, you just need to convert from J to eV. You should then be able to express the result as ##E_G(R) = 1.42 + a/R^2-b/R## where ##a## and ##b## are numerical factors (roughly of the order of 1), E is in eV, and R is in nm.
 
  • #7
Thanks! I think we might have it :)

My answer using a radius of 4 gives 1.8ev which from the graph in the answer looks about right?

A few quick questions:

How did you know how to convert the value of Eo to 8.85*10^-12?

If I had to draw a graph like this in an exam would I keep picking radii in increments of 1 until the graph levels out? How would I know what value to start with in this case?

My current working: http://screencast.com/t/oOKJqB6rmEdX
 
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  • #8
I think you have it. See if you can multiply out all your numbers and get

##E_G(R) = 1.42 + a/R^2-b/R##

where ##a## abd ##b## are fixed numbers of roughly the order of 1. Then you can easily evaluate the expression for different values of R (in nm).

Generally, you would probably want to choose values of R on the order of 1 nm. So, maybe R = 0.5, 1, 2, 3, 4, 5 nm.

For ##\epsilon_0## see the value for permitivity of free space.
 
  • #9
Thanks for all your help Tsny, I was going crazy trying to figure this out =)
 

Related to Gap energy and changing diameters

1. What is gap energy?

Gap energy refers to the energy difference between the top of the valence band and the bottom of the conduction band in a material. It is a measure of the energy required to promote an electron from the valence band to the conduction band, and is often used to characterize the electronic properties of semiconductors and insulators.

2. How does gap energy affect a material's electrical conductivity?

The larger the gap energy, the less likely it is for electrons to be excited from the valence band to the conduction band. This means that materials with larger gap energies are less conductive, as there are fewer free electrons available to carry an electric current.

3. Can the gap energy of a material be changed?

Yes, the gap energy of a material can be changed by altering its chemical composition, structure, or external conditions such as temperature or pressure. For example, doping a semiconductor with impurities can decrease its gap energy and increase its conductivity.

4. How does changing the diameter of a material affect its gap energy?

The gap energy of a material can be affected by changing its diameter, as this can alter the electronic properties of the material. For example, in quantum dots, which are tiny particles with diameters on the nanoscale, the gap energy increases as the diameter decreases due to quantum confinement effects.

5. What is the relationship between gap energy and band gap?

Gap energy and band gap are often used interchangeably, but they are slightly different. Gap energy specifically refers to the energy difference between the top of the valence band and the bottom of the conduction band, while band gap refers to the range of energies in which no electron states can exist. In other words, the band gap includes all energies that are not within the valence or conduction bands.

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