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I'm trying to get some basic principles sorted in my head and I cant 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 foranyideas on this.

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# Why is the effective mass of electron different for metals

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