Question

A high electron mobility transistor (HEMT) controls large currents by applying a small voltage to a thin sheet of electrons. The density and mobility of the electrons in the sheet are critical for the operation of the HEMT. HEMTs consisting of AlGaN/GaN/Si are being studied because they promise better performance at higher currents, temperatures, and frequencies than conventional silicon HEMTs can achieve. In one study, the Hall effect was used to measure the density of electrons in one of these new HEMTs. When a current of \(10.0 \mu\) A flows through the length of the electron sheet, which is \(1.00 \mathrm{~mm}\) long, \(0.300 \mathrm{~mm}\) wide, and \(10.0 \mathrm{nm}\) thick, a magnetic field of 1.00 T perpendicular to the sheet produces a voltage of \(0.680 \mathrm{mV}\) across the width of the sheet. What is the density of electrons in the sheet?

Short answer

Answer: The density of electrons in the sheet is \(30.64 \times 10^{21} \frac{A \cdot T}{V \cdot m \cdot C}\).

Step-by-step solution

Identify the given values

In this problem, we are given the current \(I = 10.0 \mu A\), length \(l = 1.00 \mathrm{~mm}\), width \(w = 0.300 \mathrm{~mm}\), thickness \(t = 10.0 \mathrm{nm}\), magnetic field \(B = 1.00 T\), and Hall voltage \(V_H = 0.680 \mathrm{mV}\). We need to find the density of electrons, \(n\).

Calculate the Hall coefficient

Using the Hall effect formula, we can find the Hall coefficient \(R_H\): \(V_H = R_H \times \frac{I \times B}{w}\) Now, let's solve for \(R_H\): \(R_H = \frac{V_H \times w}{I \times B}\) Substitute the given values: \(R_H = \frac{0.680 \times 10^{-3} V \times 0.300 \times 10^{-3} m}{10.0 \times 10^{-6} A \times 1.00 T}\)

Calculate the Hall coefficient value

After substituting the values, we get: \(R_H = \frac{0.204 \times 10^{-6} V \cdot m}{10^{-5} A \cdot T}\) \(R_H = 20.4 \times 10^{-3} \frac{V \cdot m}{A \cdot T}\)

Get the density of electrons

Knowing the Hall coefficient, we can find the density of electrons using the formula: \(n = \frac{-1}{R_H \times e}\) where \(e = 1.6 \times 10^{-19} C\) (the charge of an electron) Now, substitute the values: \(n = \frac{-1}{20.4 \times 10^{-3} \frac{V \cdot m}{A \cdot T} \times 1.6 \times 10^{-19} C}\)

Calculate the density of electrons

After substituting the values, we get: \(n = \frac{-1}{32.64 \times 10^{-22} \frac{V \cdot m \cdot C}{A \cdot T}}\) \(n = -30.64 \times 10^{21} \frac{A \cdot T}{V \cdot m \cdot C}\) Since the density of electrons cannot be negative, we can assume that the negative sign is due to the Hall effect formula accounting for the sign associated with the electron charge. Therefore, the magnitude of the electron density is the required answer. \(n = 30.64 \times 10^{21} \frac{A \cdot T}{V \cdot m \cdot C}\) Hence, the density of electrons in the sheet is \(30.64 \times 10^{21} \frac{A \cdot T}{V \cdot m \cdot C}\).