What is the effective mass of electrons and holes in germanium?

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Homework Help Overview

The discussion revolves around calculating the densities of states for germanium, specifically focusing on energies near the conduction and valence band edges. The original poster seeks clarification on the application of a formula related to the volume density of states and the effective mass of charge carriers in germanium.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply a formula for density of states but questions the inclusion of a volume factor and the accuracy of their calculations. They express confusion over the expected results and the conversion to eV.
  • Another participant notes the significance of the volume factor in the context of the formula's units and questions the intermediate results provided by the original poster.
  • Further, a participant highlights the need for effective mass values specific to germanium, indicating a potential gap in the original poster's information.

Discussion Status

Contextual Notes

There is an indication that the original poster may not have provided all necessary information for the calculations, particularly regarding the effective mass of electrons and holes in germanium, which is noted to differ from silicon.

orangeincup
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Homework Statement


Find the densities of states 0.08 eV above the conduction band edge and 0.08 eV below the valence band edge for germanium.

Find the volume density of states (i.e. number of states per unit volume) with energies between the conduction band edge and 0.4 eV above the conduction band edge for germanium.

Homework Equations


8pisqrt2/h^3*m^(3/2)sqrt(E-Ec)
Using this example as a reference http://ecee.colorado.edu/~bart/book/book/chapter2/pdf/ex2_3.pdf

The Attempt at a Solution


m=6*10^18, in the example they ignored the *10^-18 on their 1.08 value
(((8pisqrt2*(9.10*10^-31*6)^(3/2))/(6.626*10^-34)^3))sqrt(.08*1.6*10^-19)
=1.76*10^47 m^-3J^-1

Why should I times it by *10^-22, like in the example above? What is the purpose of that or is it an error?

In the example above I don't understand how they calculated that or reduced their value so low. I tried it with the same numbers they gave and it didn't work at all.

This value seems way too high. I need to convert it to eV.

Part 2)
Should I use .02 for this example, for my E-Ec value?
 
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This should be posted in the advanced physics forum.
 
I'm no expert on this topic, but I note that 10^-22 is the volume of the sample they are using (100 x 100 x 10 nm3) in cubic meters. I note also that the units of their formula work out to J-1 m-3, so to yield a result in J-1 multiplying by a volume is called for.

I further note that their intermediate result: 1.51 x 1056 m-3 J-1 is suspect. Plugging in their values I get an order of magnitude of 1046 rather than 1056, however their final result looks fine.
 
Are you sure you posted all of the information given in the question? You need the effective mass of electrons and holes in germanium to work that out. (The values are very different to the values for silicon.)
 

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