Fraction of valence electrons free for conduction

In summary, the conversation discussed the process of calculating the intrinsic carrier concentration for an intrinsic semiconductor. The first part clarified that n=p=ni in an undoped semiconductor, and the second part described the steps for finding the number of atoms in one cubic centimeter of silicon and dividing it by the number of valence electrons to calculate the concentration. It was also mentioned that intrinsic carrier concentration is calculated from the sum of electrons and holes. The only objection to the method was retracted.
  • #1
jisbon
476
30
Homework Statement
The intrisinc carrier concentration of single crystal sillcon is ##(1.5*10^{10})## per cubic centimeter.
Bandgap is 1.1eV and density is 2.33g per cubic centimeter.
How many free electron and holes are there per cubic centimeter.
Determine the fraction of valence electrons that are free for conduction at room temperature. Explain where electrons get sufficient energy from in order to be free.
Relevant Equations
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For the first part, since this is a intrisinc semiconductor, n=p= intrisinc carrier concentration. Hence free electrons and hole = ##(1.5*10^{10})## per cubic centimeter.

As for part 2, here are my steps. But I'm not sure if it's correct.
I first find the number of atoms of one cubic centimeter of sillcon, (using density and atomic mass of sillcon) then multiply it by 4 (since there is 4 valence electrons per sillcon atom?) I then take :
##(1.5*10^{10})## divided by the number of valence electrons as calculated above. Is this thought process correct? Thanks
 
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  • #2
The process you describe is correct except, in both parts of the question, you need to consider what it means that there are both electrons AND holes.
 
  • #3
Cutter Ketch said:
The process you describe is correct except, in both parts of the question, you need to consider what it means that there are both electrons AND holes.
For the first part, isn't the n=p for an intrisinc semi conductor? Hence my answer theoretically should be correct for part 1?

For the second part, since there are only considering fraction of electrons, won't it only be just the concentration of electrons divided by the total number of electrons? Why are holes being mentioned?

Thanks
 
  • #4
You keep saying half of them are holes and half of them are electrons (true) and then ignoring that fact when someone asks “how many electrons are there”
 
  • #5
Cutter Ketch said:
You keep saying half of them are holes and half of them are electrons (true) and then ignoring that fact when someone asks “how many electrons are there”

Pardon, but I stated that n = p = ni in an undoped semiconductor, which means three of them have the same values?
 
  • #6
Cutter Ketch said:
You keep saying half of them are holes and half of them are electrons (true) and then ignoring that fact when someone asks “how many electrons are there”
Are you implying that intrinsic carrier concentration is calculated from the sum of electrons and holes? Not an area I know anything about, but that doesn't seem to match what I read on the web.
 
  • #7
haruspex said:
Are you implying that intrinsic carrier concentration is calculated from the sum of electrons and holes? Not an area I know anything about, but that doesn't seem to match what I read on the web.

That is EXACTLY what I am implying! (and, as it happens, I am exactly wrong, so I’ll shut up now)
 
  • #8
Was a bit confused by the replies... Was just wondering if anyone could check my concepts and tell me if the steps I did were proper. Thanks
 
  • #9
jisbon said:
Was a bit confused by the replies... Was just wondering if anyone could check my concepts and tell me if the steps I did were proper. Thanks
Cutter Ketch only had one objection to your method, and he has withdrawn that. Your method is fine as far as I can tell, but it's not a subject I know.
 

What is meant by "fraction of valence electrons free for conduction"?

The fraction of valence electrons free for conduction refers to the percentage of electrons in an atom's outermost energy level (valence shell) that are able to move freely and contribute to electrical conductivity.

How is the fraction of valence electrons free for conduction calculated?

The fraction of valence electrons free for conduction is calculated by dividing the number of free electrons in the valence shell by the total number of valence electrons in the atom.

What factors affect the fraction of valence electrons free for conduction?

The fraction of valence electrons free for conduction is affected by the type of material, temperature, and external electric fields. In general, materials with more free electrons in the valence shell, such as metals, have a higher fraction of valence electrons free for conduction.

Why is the fraction of valence electrons free for conduction important?

The fraction of valence electrons free for conduction is important because it determines the electrical conductivity of a material. Materials with a higher fraction of valence electrons free for conduction are better conductors of electricity.

How does the fraction of valence electrons free for conduction relate to other properties of materials?

The fraction of valence electrons free for conduction is related to other properties of materials such as thermal conductivity, melting point, and ductility. Materials with a high fraction of valence electrons free for conduction tend to have high thermal conductivity, low melting point, and are more malleable.

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