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

In summary, the conversation discusses finding the densities of states and volume density of states for germanium with specific energy levels. The first part involves using a formula to calculate the density of states, while the second part requires knowledge of the effective mass of electrons and holes in germanium. The purpose of multiplying by *10^-22 is to account for the volume of the sample being used. The final result may be different from the example given due to potential errors in the calculations or different values being used.
  • #1
orangeincup
123
0

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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  • #2
This should be posted in the advanced physics forum.
 
  • #3
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.
 
  • #4
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.)
 

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

1. What is Fermi energy?

Fermi energy, also known as Fermi level, is the highest occupied energy level in a material at absolute zero temperature.

2. How is Fermi energy related to the electronic band structure?

Fermi energy is a key parameter in the electronic band structure of a material. It determines the boundary between the filled and unfilled energy states in a material, and is crucial in understanding the electrical and thermal properties of materials.

3. What factors affect the Fermi energy of a material?

The Fermi energy of a material is affected by its composition, temperature, and external forces such as pressure or magnetic fields. It also depends on the type of material, as different materials have different electronic band structures and therefore different Fermi energies.

4. How is Fermi energy measured experimentally?

Fermi energy can be measured using various experimental techniques such as X-ray photoelectron spectroscopy, angle-resolved photoemission spectroscopy, and electrical conductivity measurements. These methods involve analyzing the energy distribution and behavior of electrons in a material.

5. What is the significance of Fermi energy in understanding material properties?

Fermi energy plays a crucial role in determining the electrical and thermal properties of materials. It also affects the behavior of electrons, such as their mobility and ability to conduct electricity. Understanding Fermi energy is essential for studying the electronic structure and properties of materials.

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