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Effective Electron Mass

by roam
Tags: effective, electron, mass
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roam
#1
Aug18-13, 07:59 AM
P: 895
I am trying to find the effective mass of the electron within the first Brillouin zone in a particular direction in a crystal where the energy of the electron varies with some wave vector ##E(k)=Ak^2+Bk^4##. But I need to express this as a fraction of the electron rest mass.

I know that the effective mass (from Newton's 2nd law) is given by:

##m^* = \frac{\hbar^2}{d^2E/dk^2}##

At the first Brillouin zone boundary we have ##k =\pi / a##. Also the second derivative of the E(k) is ##\frac{d^2E}{dk^2}=2A+12Bk^2##.

Substituting these in I think the effective mass is:

##m^* = \frac{\hbar^2}{2A+12B (\frac{\pi}{a})^2}##

Now, how does one express this as a fraction of the electron rest mass m (511 KeV)?

Any suggestion or correction is appreciated.
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fzero
#2
Aug18-13, 04:04 PM
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The obvious answer is to divide by ##m_e## so that you have an expression for ##m_*/m_e##. This doesn't make too much sense unless you've been given numerical values for ##A,B,a## though.
erst
#3
Aug19-13, 09:39 PM
P: 21
Yup, you're done if that's all the information provided. Put m_0 in the denominator.

roam
#4
Aug20-13, 01:41 AM
P: 895
Effective Electron Mass

Thank you very much for your inputs. Here is what I've got then for effective mass as a fraction of the electron rest mass:

##m^* = \frac{\hbar^2}{(2A+12B (\pi /a)^2) m_e}##
ZapperZ
#5
Aug22-13, 03:32 PM
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Quote Quote by roam View Post
Thank you very much for your inputs. Here is what I've got then for effective mass as a fraction of the electron rest mass:

##m^* = \frac{\hbar^2}{(2A+12B (\pi /a)^2) m_e}##
Don't you mean m*/m_e for the LHS of the equation?

Zz.


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