Anna university solved problems

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 × 10^-8 m, and average thermal velocity of electron is 1.6 × 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 × 10⁶ ms^{- 1}

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

Mass of an electron m = 9.11 × 10-³¹ kg

Solution

We know that σ =ne²λ / mv  ( τ = λ/v)

Substituting the given values, we have

σ = 5.9 × 1⁰⁷ mho m^-1

 

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 τ = 10^-14 S

Temperature T= 300 K

Electron concentration n = 6 × 10²⁸ m^{- 3}

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

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

Boltzmann's constant k = 1.38 × 10^-23 Jk^-1

Solution

We know that σ = ne² τ / m

Substituting the given values, we have

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

Thermal conductivity K = π²/3  nk²τT/m

 (By quantum free electron theory)

Substituting the given values, we have

L =2.44 × 10^-8 W Ω K^-2

 

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 ρ = 1.54 × 10^-8 Ωm

Solution

We know that the electrical conductivity of a metal

σ = n e²τ /m   [ρ = 1/σ]

ρ = m / ne²τ  or τ =  m / ne²ρ

Substituting the given values, we have

τ = 3.98 × 10^-14 s

Problem 2.7

A uniform silver wire has a resistivity of 1.34 × 10^-8 Ωm 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 of the material. (A.U. May 2009)

Given data

Resistivity of the wire ρ = 1.34 × 10^-8 Ωm

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

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

Solution

Electrical conductivity σ = ne²τ / m

Or 1/ρ = ne² τ / m    [σ = 1/ρ]

Τ = m/ρne²

Substituting the given values, we have

τ  = 4.57 × 10^-14 s

Drift velocity is given by v_d = e E τ / m

v_d  = 0.804 ms^-1

Mobility μ = v_d / E =0.804/ 10² =  0.804 × 10^-2 m² V^-1s^-1

μ = 8.04 × 10^-3 m² V^-1 s^-1

 

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 0.02Ω carries a current of 15 A.

 Given mobility μ = 4.3 x 10^-3m²V^-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 x 10^-3m²V^-1s^-1

Solution

Voltage drop V across the wire is

V = IR = 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

v_d = μE = 4.3 × 10^-3 × 0.15 = 0.645 × 10^-3

v_{d =} 0.645 × 10^-3 ms^-1

We know that 3/2 kT = ½ mv²

where v is thermal velocity

v² = 3kT / m

Substituting the given values, we have

V = 1.17 × 10⁵ms^-1

 

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

v_d = I / neA (J= ne v_d and J=I/A)

Substituting the given values, we have

vd = 7.4 × 10^-5 ms^-1

 

Problem 2.10

A metallic wire has a resistivity of 1.42 × 10^-8 Ωm an electric field of 0.14 V/m. Find (i) average drift velocity and (ii) mean collision time, assuming that there are 6 x 10²⁸ electrons / m³. (A.U. April 2015)

Given data

Electric field E = 0.14 Vm^-1

Resistivity ρ = 1.42 × 10²⁸  m^-3

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

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

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

Solution:

The resistivity of a metal is given by,

ρ  = m / ne² τ

Substituting the given values, we have

V_d = 1.03 × 10^-13ms^-1

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= eV (A.U. June 2010)

Given data

Mass of electron m = 9.1 × 10^-31kg

 Energy interval ∆E= 0.01 eV

Planck's constant h = 6.63 x 10^-34 Js

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

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^-19J

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

Substituting the given values, we have

n = 4.14 x 10²⁵ m^-3