Effective Electron Mass

  1. 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)? :confused:

    Any suggestion or correction is appreciated.
     
  2. jcsd
  3. fzero

    fzero 2,805
    Science Advisor
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    Gold Member

    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.
     
  4. Yup, you're done if that's all the information provided. Put m_0 in the denominator.
     
  5. 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}##
     
    Last edited: Aug 20, 2013
  6. ZapperZ

    ZapperZ 30,152
    Staff Emeritus
    Science Advisor
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    Don't you mean m*/m_e for the LHS of the equation?

    Zz.
     
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