How Does Band Gap Energy Change with Nanoparticle Radius?

[dscot]

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

 

[TSny]

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 $k = 1/4\pi \epsilon_o$

 

[dscot]

Hi TSny,

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

= 3.58 *10^-30 c²/nm

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

Thanks!

 

[TSny]

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.

 

[dscot]

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

 

[TSny]

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.

 

[dscot]

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

 

[TSny]

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.

 

[dscot]

Thanks for all your help Tsny, I was going crazy trying to figure this out =)