Expression for thermal conductivity of a metal (Derivation)

EXPRESSION FOR THERMAL CONDUCTIVITY OF A METAL (Derivation)

Consider two cross-sections A and B of a metal rod separated by a distance λ.

Let A be at a high temperature (T) and B at a low temperature (T-dT).

Now, heat flows from A to B by the free electrons (Fig. 2.6)

Conduction electron per unit volume = n

Average velocity of the electrons = v

During the movement of the electrons in the metal rod, the elastic collision takes place. Hence, the electrons near A lose their kinetic energy while electrons near B gain kinetic energy.

At A, average kinetic energy of an electron = 3/2 kT ....(1)

K. E. = ½ mv² = 3/2 kT)

where k - Boltzmann's constant

T- Temperature at A.

At B, average kinetic energy of the electron

= 3/2 k (T-dT) ........(2)

The excess of kinetic energy carried by the electron from A to B

 = 3/2 kT – 3/2 k (T-dT)

= 3/2 kT- 3/2 kT+3/ 2 kdT= 3/2 kdT...(3)

Number of electrons crossing per unit area per unit time from A to B

= 1/6 nv..... (4)

The excess of energy carried from A to B per unit area in unit time.

= 1/6 nv × 3/2 kdT

=1/4 nvk dT ...............(5)

Similarly, the deficient of energy carried from B to A per unit area per unit time

= - 1/4 nv k dT......(6)

Let us assume that there is an equal probability for the electrons to move in all '6' directions as shown in the fig. 2.7.

Each electron travels with thermal velocity v and n is the free electron per unit volume (density). Then, on an average 1/6 nv electrons travel in any one of the directions per unit area per unit time.

Hence, the net amount of energy transferred from A to B per unit area per unit time

Q = ½ n v k dT ...(7)

But, from the definition of thermal conductivity, the amount of heat conducted per unit area per unit time

Q = K dT / λ dT    Q = K dT/ dx Here, λ = dx

Note: The students are not expected to write the part of the derivation given in the box in the examination.

1/2  n c k dT = K dT / λ

K = ½ n v k λ ...(8)

We know that for the metals

relaxation time (τ )= collision time (τ_c)

i.e., τ = τ_c = λ/v

τv = λ ..(9)

Substituting the eqn (9) in the eqn (8), we have

K = 1/2 n v κ τ υ

K = 1/2 n v² k τ...(10)

The equation (10) is the expression for the thermal conductivity of a metal.

Wiedemann - Franz Law

Statement

The ratio of thermal conductivity (K) to electrical conductivity (σ) is directly proportional to the absolute temperature (T). This ratio is constant for all metals at a given temperature.

K / σ ∝  T

K / σ = LT

where L is a proportiona y constant. It is known as Lorentz number. Its value is 1.12 × 10^-8 W Ω K^-2

Derivation

Wiedemann - Franz law is derived from the expressions of thermal and electrical conductivities of a metal.

We know that

Thermal conductivity of the metal

K = 1/2  nv²k τ ...(1)

Electrical conductivity of the metal

σ =  ne² τ / m ...(2)

Thermal conductivity / Electrical conductivity = K / σ = 1/2  nv²k τ / ne² τ

K / σ = ½  mv² k / e² ....(3)

The kinetic energy of the electron is given by

1/2  mv² =3/2 kT...(4)

Substituting eqn (4) in eqn (3), we have

K / σ = LT  ...(5)

where L =3/2 (k/e)² is a constant and it is known as Lorentz number.

K / σ ∝ T ...(6)

Thus, it is proved that the ratio of thermal conductivity to electrical conductivity of the metal is directly proportional to the absolute temperature of the metal.

This law is verified experimentally and it is found to hold good at normal temperature. But, this law is not applicable at very low temperature.

Conclusion

Wiedemann - Franz law clearly shows that if a metal has high thermal conductivity, it should also have high electrical conductivity.

Among the metals, the best electrical conductors (silver, copper, aluminium) are also the best conductors of heat.

Lorentz Number

The ratio of thermal conductivity (K) to the product of electrical conductivity (∝) and absolute temperature (T) of the metal is a constant. It is known as Lorentz number and it is given by

L = K / σ T

Consider the expression L = 3/2 (k/e)²

Substituting for Boltzmann's constant k = 1.38 × 10- ²³JK^-1 and the charge of the electron e = 1.602 × 10-¹⁹ coulomb, we get Lorentz number as

L = 1.12 × 10^-8 WΩK^-2

It is found that the value of Lorentz number determined using classical free electron theory is only half of the experimental value i.e., 2.44 × 10^-8 WΩK^-2 experimental and theoretical values of Lorentz number is one of the failures of the classical theory. It is rectified in quantum theory.