Anna University Solved Problems in Electrical Properties of Materials

ANNA UNIVERSITY SOLVED PROBLEMS

Problem 2.4

Calculate electrical conductivity in copper if the mean free path of electrons is 4 × 10^-8 m, electron density is 8.4 × 10²⁸ m^-3 and average thermal velocity of electron is 1.6 x 10⁶ ms^-1. (A.U Dec 2012)

Given data

Mean free path of electron λ = 4 × 10^-8 m

Electron density n = 8.4 × 10²⁸ m^-3

Average thermal velocity of the electrons v = 1.6 x 10⁶ ms^-1

Charge of an electron e = 1.6 × 10^-19 coulomb

Mass of an electron m = 9.11 x 10^-31 kg

Solution

Problem 2.5

Calculate electrical and thermal conductivities for a metal with a relaxation time 10^-14 second at 300 K. Also, calculate Lorentz number using the above result (density of electrons = 6 × 10²⁸ m^-3). (A.U. June 2013)

Given data

Relaxation time t = 10^-14 s

Temperature T= 300 K

Electron concentration n = 6 x 10²⁸ m^-3

Mass of an electron m = 9.1 x 10^-31 kg

Charge of an electron e = 1.6 x 10^-19 C

Boltzmann’s constant k = 1.38 × 10^-23 JK^-1

Solution

We know that σ = n e² τ / m

Substituting the given values, we have

Electrical conductivity σ = 1.69 × 10⁷ Ω^-1 m^-1

Thermal conductivity 

(By quantum free electron theory)

Substituting the given values, we have

Lorentz number L = K / σ T

 

Problem 2.6

Find the relaxation time of conduction electrons in a metal of resistivity 1.54 × 10^-8 ohm-m if the metal has 5.8 × 10²⁸ conduction electrons / m³. (A.U. May 2008)

Given data

Number of electrons / unit volume n = 5.8 × 10²⁸ m^-3

Resistivity of the metal p = 1.54 × 10^-8 Ω m

Solution

We know that the electrical conductivity of a metal

Substituting the given values, we have

 

Problem 2.7

A uniform silver wire has a resistivity of 1.34 × 10^-8 Ω m at room temperature for an electric field of 1 volt/cm. Calculate (i) the drift velocity (ii) the mobility and (iii) the relaxation time of electrons assuming that there are 5.8 × 10²⁸ conduction electrons m^-3 of the material. (A.U. May 2009)

Given data

Resistivity of the wire p = 1.34 × 10^-8 Ω m

Electrical field E = 1 V/cm = 1 V/ 10^-2 m = 1 × 10² Vm^-1

Number of electron per unit volume n = 5.8 × 10²⁸ m^-3

Solution

Electrical conductivity 

Substituting the given values, we have

Problem 2.8

Calculate the drift velocity and thermal velocity of conduction electrons in copper at a temperature of 300 K. When a copper wire of length of 2 m and resistance 0.02 Ω carries a current of 15 A.

Given mobility μ= 4.3 × 10^-3 m² V^-1s^-1 (A.U. Jan 2014)

Given data

Temperature T = 300 K

Length of the wire L = 2 m

Resistance R = 0.02 Ω

Current I= 15 A

Mobility μ= 4.3 × 10^-3 m² V^-1s^-1

Solution

Voltage drop V across the wire is

V = I R = 15 × 0.02 = 0.3 V

Electric field E across the wire is given by

E = V/L = 0.3/2 = 0.15 Vm^-1

Drift velocity

where v is thermal velocity

Substituting the given values, we have

Problem 2.9

Find the drift velocity of the free electrons in a copper wire whose cross sectional area is 1.0 mm² when the wire carries a current of 1A. Assume that each copper atom contributes one electron to the electron gas. Given n = 8.5 × 10²⁸ m^-3 (A.U. May 2016)

Given Data

Conduction electron / m³, n = 8.5 × 10²⁸ m^-3

Charge of electron e = 1.6 × 10^-19 C

Area of cross section A = 1.0 × 10^-6 m²

Current I = 1.0 A

Solution

The drift velocity of the free electrons is given by

Substituting the given values, we have

Problem 2.10

A metallic wire has a resistivity of 1.42 × 10^-8 Ω m. For an electric field of 0.14 V/m. Find (i) average drift velocity and (ii) mean collision time, assuming that there are 6 × 10²⁸ electrons /m³.

(A.U. April 2015)

Given data

Electric field E = 0.14 Vm^-1

Resistivity p = 1.42 × 10^-8 ΩΜ

Number of electrons per unit volume n = 6 × 10²⁸ m^-3

Charge of the electron e = 1.6 × 10^-19 C

Mass of an electron m = 9.1 × 10^-31 kg

Solution:

The resistivity of a metal is given by,

Substituting the given values, we have

Density of energy states

Problem 2.11

Calculate the number of states lying in an energy interval of 0.01 eV above the Fermi level for a crystal of unit volume with Fermi energy E_F = 3 eV (A.U. June 2010)

Given data

Mass of electron m = 9.1 × 10^-31 kg

Energy interval ∆E = 0.01 eV

Planck’s constant h = 6.63 × 10^-34 Js

Fermi energy E_F = 3 eV [‘.’ 1 eV = 1.6 × 10^-19 J]

E_F = 3 × 1.6 × 10^-19 J

E_F = 4.8 × 10^-19 J

Solution

We know that ∆E = E - E_F = E_F + ∆E = (3 + 0.01) eV = 3.01 × 1.6 × 10^-19 J

E = 4.816 × 10^-19

Number of states per unit volume lying between E_F and E is given by

Substituting the given values, we have