题目

EESM5200 (L2) Quiz 01

多项填空题

In this question, assume Ex is 0.2eV above EV, the bandgap is 1.1eV, and kT=0.025eV. F(E) denotes the Fermi-Dirac function. Give numerical answers up to 3 decimal places in the following questions. a) If EF is positioned at an energy Ex in a semiconductor, what is the probability of finding an electron in a state at an energy of Ex+kT? [Fill in the blank], b) The probability of a state being filled at Ex+kT is equal to the probability of a state being empty at Ex+3kT. How far is the Fermi level located above EV in eV? [Fill in the blank], eV c) It is known that F(EF + 3kT) = 1 - F(EF + x). What is the value of x in eV? (Hint: be careful with the sign) [Fill in the blank], eV d) Assume the equivalent density of state at the conduction band NC=3x1019cm-3. What is the number electron density at the conduction band of the material if EF is located at 9kT below EC? Use the scientific "e" notation in your answer if you can. [Fill in the blank], cm-3

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标准答案

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思路分析

We begin by restating the problem components and the given values to anchor the reasoning for each blank. - Ex is 0.2 eV above EV, so Ex − EV = 0.2 eV. - Bandgap Eg = 1.1 eV, so EC − EV = 1.1 eV. - kT = 0.025 eV. - F(E) is the Fermi-Dirac distribution: F(E) = 1 / [exp((E − EF)/kT) + 1]. a) If EF is positioned at Ex, what is F(Ex + kT)? - With EF = Ex, the energy difference (Ex + kT) − EF = kT. Therefore, F(Ex + kT) = 1 / [exp(1) + 1] = 1 / (e + 1) ≈ 1 / 3.718 ≈ 0.269. - Among the provided numerical options for this blank, 0.275 is the closest to the exact value (0.269). The other values are clearly far from 0.269: 0.438 is too large, 0.075 is too small, and 3.7e16 is not a probability. - Note: if you round to three deci......Login to view full explanation