QM - Etop of electron distribution of a semiconductor

AI Thread Summary
The discussion focuses on determining the energy level above the conduction band edge (Ec) where electron distribution peaks in a nondegenerate semiconductor, specifically GaAs with a bandgap (Eg) of 1.42 eV at 300K. The user proposes setting the Fermi energy level at 3kT above Ec and plans to calculate the electron distribution (n) using the equations provided. They intend to equate n to the integral of the product of the density of states (g_c(E)) and the Fermi function (f(E)), seeking to find the upper limit of this integral, termed Etop. The user struggles with evaluating the integral involving the square root and exponential functions but discovers that taking the derivative of g_c(E) * f(E) reveals the minimum points of electron distribution, including Etop. This approach ultimately leads to the solution for Etop as required by the homework question.
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Homework Statement



I'm trying to find energy level above Ec where electron distribution makes a peak for a nondegenerate semiconductor. For this case we may take GaAs having Eg = 1.42eV at T = 300K.

Homework Equations


m_e=single isotrophic effective mass or m_0
energy states, g_{c}(E) = \frac{m_{e}\ast\sqrt{2m_{e}(E-E_{c})}}{pi^2 * hbar^3}
fermi function for a nondegenerate semiconductor, f(E) = exp((E_f-E)/kT)
electron distribution, n=N_{c}*exp((Ef-Ec)/kT) and N_{c}=4.21\ast10^{17} cm^-3

The Attempt at a Solution


I think I'll give a fermi energy level equal to 3kT above Ec where semi.con. is still nondegenerate. Then I'll calculate n. Afterwards I'll equate n to \int g_{c}(E)*f(E)*dE taking a limit to 99 % of n. By that I intend to find top limit of the integral which must be the Etop.
But i do not how to evaluate a integral such as \sqrt{E}*exp(c*E)
ps: partial integral is not working.
Is there another {easy :( }approach?
 
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I found the answer:
derivative of
g_{c}(E) * f(E)
gives the minimum points of electron distribution
one of them is E_{c} and the other is E_{top} which is asked by the question ;)
 
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