What is the Effective Mass Equation for Excited Electrons in Units of Angstroms?

[rwooduk]

Homework Statement

Homework Equations

None.

The Attempt at a Solution

I understand the question, for a phonon it can excite an electron from the valance band to the conduction band between any values of k. If it were a photon it could only excite the electron at specific k values.

Anyway so looking at the graph minimum/maximum looks (i tried differentiating Ec and Ev and setting to zero to find the exact minimum and maximum but it ended up getting complicated) like it is at k=0 for Ev and k=1 for Ec.

So now I need to put k=1 into the equation for Ec. And this is what I'm struggling with, so I put k=1 in and get $$E_{c} = 11 - 2\AA^{-2} + \AA^{-4} = 11 - \frac{2}{(10^{-10})^{2}}+ \frac{1}{(10^{-10})^{4}}$$ which must be wrong as the angstrom terms are really small and Ec goes massive.

Any advice on this would be appreciated.

edit the red AA's are angstroms, don't know why its not working

 

[BvU]

No, in the expression for Ec is Ec in eV and k is in $\unicode{x212B}$.
So when you differentiate, you do so in $\unicode{x212B}$ units. (in fact in the dimensionless units $u = k/\unicode{x212B}$ ).
You get $-4u+4u^3 = 0 \Leftrightarrow u(u^2-1) = 0$

And the $E_c$ expression is scaled too, in the dimensionless quantity y = E/eV.

Often in graphs you see axis annotations like $E_c/{\rm eV}$ and $k * \unicode{x212B}^2$ and those are the numbers (dimensionless !) at the tick marks. I like that and find it more correct than e.g. ##E (eV)

 

[rwooduk]

No, in the expression for Ec is Ec in eV and k is in $\unicode{x212B}$.
So when you differentiate, you do so in $\unicode{x212B}$ units. (in fact in the dimensionless units $u = k/\unicode{x212B}$ ).
You get $-4u+4u^3 = 0 \Leftrightarrow u(u^2-1) = 0$

And the $E_c$ expression is scaled too, in the dimensionless quantity y = E/eV.

Often in graphs you see axis annotations like $E_c/{\rm eV}$ and $k * \unicode{x212B}^2$ and those are the numbers (dimensionless !) at the tick marks. I like that and find it more correct than e.g. ##E (eV)

Ahh so the Angstroms cancel I see, and thanks for the tip on the derivative!

Thanks! I get it now but why then on the graph does it say k ($\unicode{x212B}^{-1}$) not k($\unicode{x212B}$)

 

[BvU]

Yes, (unfortunately) it's customary to write e.g Distance (km) instead of Distance/km as axis title. And the unit of $k$ is $\unicode{x212B}^{-1}$.

 

[rwooduk]

Yes, (unfortunately) it's customary to write e.g Distance (km) instead of Distance/km as axis title. And the unit of $k$ is $\unicode{x212B}^{-1}$.

Just one more question that's related if I may, if I had something like: $$q=\frac{\omega }{\upsilon }= \frac{36GHz}{458m/s}=786\cdot 10^{5}$$ could you then write it as $$786\cdot 10^{5}\cdot \frac{\unicode{x212B}}{\unicode{x212B}}=0.008 \unicode{x212B}^{-1}$$

 

[BvU]

This looks really weird. Normally $\omega$ is not expressed in GHz (dimension 1/s) but in radians/sec. Makes a factor $2\pi$ difference, as in $\omega = 2\pi f\;$ or $\omega = 2\pi \nu\;$. Don't know what $q$ stands for, or what its dimension is.

Aha ! enlightened myself in the modern fashion by googling wavenumber and found $\displaystyle k = {2\pi\over \lambda}= {2\pi\over v_p}={\omega\over v_p}$. This simply has the dimension of length^-1.

With your $\displaystyle q=\frac{\omega }{\upsilon }= \frac{36GHz}{458m/s}=786\cdot 10^{5}\; {\rm \bf m^{-1}}$ you can get inverse $\unicode{x212B}$ by simply multiplying by 1 !

This 1 here taking the form $\displaystyle 1\, {\rm m} \over 10^{10} \, {\displaystyle \unicode{x212B}}$ or $\left ( 10^{10} \, {\displaystyle \unicode{x212B}} \right ) ^{-1} \over {\rm m}^{-1}$ , in other words, you get $$
q=786\cdot 10^{5}\; {\rm \bf m^{-1}} = 786\cdot 10^{5} \, {\rm m^{-1}}\; 10^{-10} \, {{\displaystyle \unicode{x212B}} ^{-1} \over \displaystyle{\rm m}^{-1}} = 786\cdot 10^{-5}\,
{\displaystyle \unicode{x212B}} ^{-1}$$

--

 

[rwooduk]

That's awesome thanks! Will just have to practice this a little more.

 

[rwooduk]

No, in the expression for Ec is Ec in eV and k is in $\unicode{x212B}$.
So when you differentiate, you do so in $\unicode{x212B}$ units. (in fact in the dimensionless units $u = k/\unicode{x212B}$ ).
You get $-4u+4u^3 = 0 \Leftrightarrow u(u^2-1) = 0$

And the $E_c$ expression is scaled too, in the dimensionless quantity y = E/eV.

Coming back to the derivative, if you set it equal to zero it makes sense however on the first part of the question below:

You would have for the effective mass $$m^{*}= \frac{\hbar^{2}}{\left ( \frac{\partial^2 E}{\partial k^2} \right )}$$

If we apply what you said earlier we get $$\frac{\partial^2 E_{c}}{\partial u^2} = -4 + 12u$$ but we need $$ \frac{\partial^2 E}{\partial k^2}$$

 

[rwooduk]

also in a separate question

if you set u=k/A then differentiate twice for the E_c term the u vanishes!

exam on Wednesday, can anyone help further?

 

[rwooduk]

I'll update this for anyone interested:

 

[BvU]

Coming back to the derivative, if you set it equal to zero it makes sense however on the first part of the question below:

You would have for the effective mass $$m^{*}= \frac{\hbar^{2}}{\left ( \frac{\partial^2 E}{\partial k^2} \right )}$$

If we apply what you said earlier we get $$\frac{\partial^2 E_{c}}{\partial u^2} = -4 + 12u$$ but we need $$ \frac{\partial^2 E}{\partial k^2}$$

With ${\partial \over \partial k} = {\partial \over \partial u} {du \over dk}\$ and $\ u = { k\over \unicode{x212B}}$ you get $$
m^* = {\displaystyle \hbar^2 \over {\partial^2 \left ( E/eV \right ) \over \partial u^2} \left(du \over dk \right ) ^2 eV } = {\hbar^2 \over {(4-12u)} \; {\rm eV \,\unicode{x212B}^2 }} $$
which is also dimensionally correct: ${\rm (Js)^2 \over J m^2 } = {\rm kg}$

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also in a separate question

if you set u=k/A then differentiate twice for the E_c term the u vanishes!

exam on Wednesday, can anyone help further?

It's becoming a rather intermixed mess of exercises now...

The u may vanish, but there still is a coefficient -5 10^-19 ( a constant) !
(In this exercise A3 I wonder why there appear coefficients of the order of magnitude 10^-19 and $(\unicode{x212B})^2$ instead of the ^-2 we were used to...)

But in your post # 10 yet other values again. A bit confusing for me, but I hope you have it clear by now. Looks like you do.

Good luck tomorrow !

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