I apologize if this is a very basic question. But I have always known it to be true that in Silicon, electrons have higher mobility than holes. From my semiconductor physics classes in first year, the thing I recalled was that mobility is inversely related to effective mass for the basic scattering model. So I assumed that electrons being more mobile must have lower effective mass. This is wrong as I learned going through the material again: electrons have HIGHER EFFECTIVE MASS than holes. So is there any explanation for this? Thanks!
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\$\begingroup\$ I’m voting to close this question because it belongs on the Physics stack \$\endgroup\$– Neil_UKCommented Sep 8, 2022 at 11:13
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1\$\begingroup\$ Here's the wiki on your topic. There, you can find that the electron can have 'negative mass'. Are you referring to this description? \$\endgroup\$– jonkCommented Sep 8, 2022 at 11:14
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\$\begingroup\$ @jonk no the electron cannot have negative mass \$\endgroup\$– Miss MulanCommented Sep 8, 2022 at 14:02
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2\$\begingroup\$ @Neil_UK Sorry, but I don't agree. In the help center it's stated that the theory and simulation of electromagnetic forces are on topic, and arguably solid state physics as employed in electronic devices falls under that category. Moreover, semiconductor behavior is a topic commonly taught at university in Electronic Engineering courses (at least here in Italy). So closing is not the right action here. OTOH, I may agree that the OP could get better answers at PHY.SE just because more experts of the field may well be dwell there. \$\endgroup\$– LorenzoDonati4Ukraine-OnStrikeCommented Sep 8, 2022 at 20:13
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The mobility is proportional to a scattering time and inversely proportional to a quasiparticle mass. The explanation you are looking for is simple: in silicon, a lower hole mass is outweighed by a higher scattering time of electron quasiparticles in lattice.
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1\$\begingroup\$ thanks. I was assuming a parameter to be constant when it is not. However, Im confused. You mean the scattering time of holes is LOWER right? because mobility is proportional to the scattering time? \$\endgroup\$ Commented Sep 8, 2022 at 19:40
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1\$\begingroup\$ Oh, sorry, corrected. Thank you. \$\endgroup\$– V.V.TCommented Sep 8, 2022 at 20:55