# Effective Electron Mass

1. Aug 18, 2013

### roam

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.

2. Aug 18, 2013

### fzero

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. Aug 19, 2013

### erst

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

4. Aug 20, 2013

### roam

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
5. Aug 22, 2013

### ZapperZ

Staff Emeritus
Don't you mean m*/m_e for the LHS of the equation?

Zz.