Solution to Set 6

Problem 1.

Given

Aluminum(Al) is trivalent with

• atomic mass $m_{a}=27$ amu
• density $n=2.7g/cm^{3}$
• room temperature $T=293-296.5K$
• mean free time between electron collisions $t_{avg}=4{\rm {x}}10^{-14}$ s.

(a) Resistivity

Calculate the resistivity $\rho$ of aluminum(Al) at room temperature.

$\rho ={\frac {1}{\sigma }}\;$ $={\frac {1}{37.8\times 10^{6}Sm^{-1}}}\;$ $=2.82\times 10^{-8}\;$Ω·m

(b) Current

If a 2-V voltage is applied to the ends of an aluminum wire 10 m long and with a cross- sectional area of $1mm^{2}$

$R={\frac {\ell \cdot \rho }{A}}\;$ $={\frac {\left(10m\right)\cdot \left(2.646\times 10^{-8}\Omega \cdot m\right)}{\left(10^{-6}m\right)}}\;$ $=0.2646\Omega \;$

What is the current flowing through it?

$I={\frac {V}{R}}={\frac {2V}{0.2646\Omega }}=7.559A\;$

Problem 2

The resistivity of a certain material at room temperature is 0.02 Wm and the Hall coefficient is $5{\rm {x}}10^{-4}$ $m^{3}/C$. An electric field of 1 V/m is applied across it. Deduce all the information you can think of about this material.

Given

• Temperature $T=20^{\circ }C\;$
• Resistivity $\rho =0.02\Omega \cdot m\;$
• Hall coefficient $R_{H}=5{\rm {x}}10^{-4}{\tfrac {m^{3}}{C}}\;$
• Electric field $E=1{\tfrac {V}{m}}\;$

Deduction

• Conductivity

$\sigma ={\frac {1}{\rho }}={\frac {1}{0.02\Omega \cdot m}}=50{\tfrac {S}{m}}\;$

• Current Density

$\mathbf {J} ={\frac {E}{\rho }}={\frac {1{\tfrac {V}{m}}}{0.02\Omega \cdot m}}=50{\tfrac {A}{m^{2}}}\;$

• Magnetic Field

$R_{H}={\frac {E}{\mathbf {J} \cdot B}}\;$

$B={\frac {E}{\mathbf {J} \cdot R_{H}}}\;$ $={\frac {\left(1{\tfrac {V}{m}}\right)}{\left(50{\tfrac {A}{m^{2}}}\right)\left(5\times 10^{-4}m^{3}\right)}}\;$ $=40T\;$

Problem 3.

(a) Hall Effect Sketch

Sketch a setup used to measure the Hall effect. Label each part.

(b) Semiconductor Crystal

A semiconductor crystal is 5 mm long, 4 mm wide, and 2 mm thick. A 40mA current flows across the length of the sample after a 2-V battery is connected to the ends. When a 0.1T magnetic field is applied perpendicular to the large surface of the specimen, a Hall voltage of 15mV develops across the width of the sample.

Given

• Length $\ell$ = 5 mm = 0.005 m
• Width W = 4 mm = 0.004 m
• Thickness H = 2 mm = 0.002 m
• Current I = 40 mA = 0.04 A
• Voltage V = 2 V
• Mag Field B = 0.1 T
• Hall Volt $V_{H}$ = 15 mV = 0.015 V

You can deduce that:

• Area $A=L\times W=2.0\times 10^{-5}m^{2}$

Determine

1. Conductivity

$\rho =R\cdot {\frac {A}{\ell }}\;$ $={\frac {V}{I}}\cdot {\frac {A}{\ell }}\;$ $={\frac {2V}{0.04A}}\cdot {\frac {2\times 10^{-5}m^{2}}{0.005m}}\;$ $=0.2\Omega \cdot m\;$

$\sigma ={\frac {1}{\rho }}={\frac {1}{0.2\Omega \cdot m}}\;$ $=5{\tfrac {S}{m}}\;$

1. Carrier density

$R_{H}={\frac {E_{y}}{\mathbf {J_{x}} \cdot B}}\;$ $={\frac {V_{H}}{I\cdot {\tfrac {B}{\ell }}}}\;$ $=-{\frac {1}{ne}}\;$

$n=-{\frac {1}{e}}\cdot {\frac {V_{H}}{I\cdot {\tfrac {B}{\ell }}}}\;$ $=-{\frac {1}{-1.602\times 10^{-19}C}}\cdot {\frac {0.015V}{\left(0.04A\right)\cdot {\tfrac {\left(0.1T\right)}{\left(0.005m\right)}}}}\;$ $=1.1704\times 10^{17}m^{-3}\;$

1. Mobility

1. Fermi velocity

Problem 4

(a) Fermi Derivations

Derive the expressions for the Fermi energy, Fermi velocity, and electronic density of states for a two-dimensional free electron gas.

(b) Fermi Energy & Velocity of 2D Gas

A 2D electron gas formed in a GaAs/AlGaAs quantum well has a density of $2{\rm {x}}10^{11}cm^{-2}$. Assuming that the electrons there have the free electron mass, calculate the Fermi energy and Fermi velocity.