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Effective Electron Mass 
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#1
Aug1813, 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 2^{nd} 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. 


#2
Aug1813, 04:04 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

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.



#3
Aug1913, 09:39 PM

P: 21

Yup, you're done if that's all the information provided. Put m_0 in the denominator.



#4
Aug2013, 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}## 


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