# Working in units of Angstroms

1. May 22, 2015

### rwooduk

1. The problem statement, all variables and given/known data

2. Relevant equations
None.

3. 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, dont know why its not working

2. May 22, 2015

### 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.

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} $, in other words, you get  q=786\cdot 10^{5}\; {\rm \bf m^{-1}} = 786\cdot 10^{5} \, {\rm m^{-1}}\; 10^{-10} \, {{\displaystyle \unicode{x212B}} ^{-1} \over \displaystyle{\rm m}^{-1}} = 786\cdot 10^{-5}\, {\displaystyle \unicode{x212B}} ^{-1} -- 7. May 24, 2015 ### rwooduk That's awesome thanks! Will just have to practice this a little more. 8. May 24, 2015 ### rwooduk Coming back to the derivative, if you set it equal to zero it makes sense however on the first part of the question below: You would have for the effective mass m^{*}= \frac{\hbar^{2}}{\left ( \frac{\partial^2 E}{\partial k^2} \right )} If we apply what you said earlier we get \frac{\partial^2 E_{c}}{\partial u^2} = -4 + 12u but we need  \frac{\partial^2 E}{\partial k^2} 9. May 25, 2015 ### rwooduk also in a seperate question if you set u=k/A then differentiate twice for the Ec term the u vanishes! exam on Wednesday, can anyone help further? 10. May 26, 2015 ### rwooduk I'll update this for anyone interested: 11. May 26, 2015 ### BvU With$ {\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}$-- It's becoming a rather intermixed mess of exercises now... The u may vanish, but there still is a coefficient -5 10-19 ( a constant) ! (In this exercise A3 I wonder why there appear coefficients of the order of magnitude 10-19 and$(\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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