- #1
abhaymv
- 9
- 0
In my college,proof of law of mass action was given thus:
(I suspect it is wrong)
Let no. of electrons per unit volume at constant temperature at an energy state E be given by:
N(E)=(1/(2π2))*[2me/ħ2](3/2)*{E-Eg}(1/2)
(1)
Probability of filling an electron in an energy state:
E=F(E)=1/(1+e((E-Ef)/(kT))
(2)
Where F(E) is called "Fermi function".
Total no. of electrons with energy between E and E+ΔE is the product :
N(E)F(E)dE
(3)
E-Ef>>kT,so F(E) can be written as:
F(E)=e-((E-Ef)/kT)
Applying this and expanding (3),
and integrating to get the total no. of electrons in the conduction band,
n=∫N(E)F(E)dE
Integration from E to ∞
we get,
n=2[mekT/2πħ2]3/2e((Ef-Eg)/kT)
Let the hole density of V.B be:
Np(E)=(1/(2π2))*[2mh/ħ2](3/2)*E(1/2)
Probability of filling a hole=1-F(E)
Total probability being 1
P(E) is then approximated as:
e((E-Ef)/kT
Is this correct?I don't think it is.Because,if I substitute random values for the power of e in the expressions,I don't get total probability as 1...(not even close)
Where is the proof wrong?
(I suspect it is wrong)
Let no. of electrons per unit volume at constant temperature at an energy state E be given by:
N(E)=(1/(2π2))*[2me/ħ2](3/2)*{E-Eg}(1/2)
(1)
Probability of filling an electron in an energy state:
E=F(E)=1/(1+e((E-Ef)/(kT))
(2)
Where F(E) is called "Fermi function".
Total no. of electrons with energy between E and E+ΔE is the product :
N(E)F(E)dE
(3)
E-Ef>>kT,so F(E) can be written as:
F(E)=e-((E-Ef)/kT)
Applying this and expanding (3),
and integrating to get the total no. of electrons in the conduction band,
n=∫N(E)F(E)dE
Integration from E to ∞
we get,
n=2[mekT/2πħ2]3/2e((Ef-Eg)/kT)
Let the hole density of V.B be:
Np(E)=(1/(2π2))*[2mh/ħ2](3/2)*E(1/2)
Probability of filling a hole=1-F(E)
Total probability being 1
P(E) is then approximated as:
e((E-Ef)/kT
Is this correct?I don't think it is.Because,if I substitute random values for the power of e in the expressions,I don't get total probability as 1...(not even close)
Where is the proof wrong?