Calculate the Eo given that Nd=1.5*10^10cm^-3, Na=1.5*10^10cm^-3 at temperature 300K?(a) 1.5*10^10eV(b) 0.256eV(c) 0eV(d) 4.14*10^-21eVThis question was posed to me in my homework.Origin of the question is Band Structure of an Open-Circuited p-n junction topic in division Semiconductor-Diode Characteristics of Electronic Devices & Circuits

Calculate the Eo given that Nd=1.5*10^10cm^-3, Na=1.5*10^10cm^-3 at te

1.	Calculate the Eo given that Nd=1.510^10cm^-3, Na=1.510^10cm^-3 at temperature 300K?(a) 1.510^10eV(b) 0.256eV(c) 0eV(d) 4.1410^-21eVThis question was posed to me in my homework.Origin of the question is Band Structure of an Open-Circuited p-n junction topic in division Semiconductor-Diode Characteristics of Electronic Devices & Circuits
Answer» The correct OPTION is (C) 0eV The best I can explain: Eo=kTln((NdNa)/(ni)^2) Substituting k=1.3810^-23/K, T=300k and the VALUES ofNd,NAAND ni, We get Eo=0eV.

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