- #1
rwooduk
- 762
- 59
... and semiconductors?
I'm trying to get some basic principles sorted in my head and I can't seem to find a straight answer to this. The lecturer drew a dispersion graph to explain it but I'm still a little confused.
I understand since ##m^{*}\propto \frac{1}{\frac{\partial^2 E}{\partial k^2}}## then the curvature of the E vs k graph will determine the effective mass. BUT a metal will have curvature, a semi conductor will have curvature, so why should the effective mass be different?
Is there a physical scenario that I'm missing here?
Thanks for any ideas on this.
I'm trying to get some basic principles sorted in my head and I can't seem to find a straight answer to this. The lecturer drew a dispersion graph to explain it but I'm still a little confused.
I understand since ##m^{*}\propto \frac{1}{\frac{\partial^2 E}{\partial k^2}}## then the curvature of the E vs k graph will determine the effective mass. BUT a metal will have curvature, a semi conductor will have curvature, so why should the effective mass be different?
Is there a physical scenario that I'm missing here?
Thanks for any ideas on this.