Is Fermi-Dirac Probability Part B Solvable?

In summary, my textbook gives the probability distribution for holes as not W(p) as it is for electrons, but as 1 - W(p). Without a believable rationale, but I'm sure it's correct, because later on they use that expression for deriving the totality of free carriers for electrons and for holes and get roughly the same number for each, which has to be correct. So, bottom line, if you use 1 - W(p) for the holes, and use Wv in lieu of Wc, and proceedig exactly as you did in part (a), you will get your answer. Do that and let us know what you come up with.If we assume the number of electrons and the holes are the same
  • #1
falcon555
12
0
Hi dear friends
Please reffer to my work , I did part ( a )
Can someone help me to solve part( b )
Please
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  • #2
(a) is correctly done.
(b):
My textbook gives the probability distribution for holes as not W(p) as it is for electrons, but as 1 - W(p). Without a believable rationale, but I'm sure it's correct, because later on they use that expression for deriving the totality of free carriers for electrons and for holes and get roughly the same number for each, which has to be correct.

So, bottom line, if you use 1 - W(p) for the holes, and use Wv in lieu of Wc, and proceedig exactly as you did in part (a), you will get your answer. Do that and let us know what you come up with.
 
  • #3
If we assume the number of electrons and the holes are the same ,then we suppose to get the same value

20160911_151058.jpg


Or you mean to say

1 - 3.99 × 10^(-10) = 2.99 × 10^(-10)
 
  • #4
falcon555 said:
If we assume the number of electrons and the holes are the same ,then we suppose to get the same value
View attachment 105749
Right. The answer for (b) is the same as the answer for (a).

But you derived p(Wh) = {1 - exp(WG/kT)}-1 when it should be {1 + exp(WG/kT)}-1 which may be just a typo.
Or you mean to say
1 - 3.99 × 10^(-10) = 2.99 × 10^(-10)
I don't think anyone would say that! (Look at that statement more carefully! :smile:)
 
  • #5
Ok
It means without any extra calculation we can asume that both the values are the same as a rule of thumb, right?
Because the holes and the electrons are generated in pairs, right?
 
  • #6
falcon555 said:
Ok
It means without any extra calculation we can asume that both the values are the same as a rule of thumb, right?
Because the holes and the electrons are generated in pairs, right?
That's what I always suspected, but I decided to make sure by going with the formulas. I wasn't sure if an electron at the very bottom of the conduction band implied a hole at the very top of the valence band until I worked the formulas.

It's certainly true that, in an intrinsic (undoped) semiconductor like Si, a hole is created when and only when an electron is kicked up to the conduction band, and vice-versa.
 
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Likes falcon555
  • #7
Thank you very much bro for helping me.
You are great.

Cheers
 
  • #8
falcon555 said:
Thank you very much bro for helping me.
You are great.

Cheers
You're very welcome. Educational for me too!
P.S. is that your cat? Lovely boy!
 

FAQ: Is Fermi-Dirac Probability Part B Solvable?

What is Fermi-Dirac probability?

Fermi-Dirac probability is a mathematical concept used in quantum mechanics to describe the probability of finding a fermion (particles with half-integer spin) in a particular energy state. It takes into account the exclusion principle, which states that no two identical fermions can occupy the same quantum state simultaneously.

How is Fermi-Dirac probability different from other types of probability?

Fermi-Dirac probability is unique in that it follows the Pauli exclusion principle, which affects the probability of finding fermions in a particular energy state. Other types of probability, such as classical or quantum probability, do not take into account this principle.

What is the formula for calculating Fermi-Dirac probability?

The formula for Fermi-Dirac probability is P(E) = 1 / (e^(E-EF) / kT + 1), where P(E) is the probability of finding a fermion in a particular energy state E, EF is the Fermi energy, k is the Boltzmann constant, and T is the temperature.

What is the significance of Fermi energy in Fermi-Dirac probability?

The Fermi energy (EF) is the highest occupied energy state in a system of fermions at absolute zero temperature. It serves as a reference point for calculating the probability of finding fermions in other energy states.

How is Fermi-Dirac probability used in real-world applications?

Fermi-Dirac probability is used in many areas of physics, including condensed matter physics, statistical mechanics, and quantum field theory. It is also used in the study of electronic devices, such as transistors and semiconductors, and in understanding the behavior of particles in the early universe.

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