Question

The resistivity of Aluminum (Al) at room temperature is 2.62 x 10-8

ohm meter.
Calculate the i) Drift velocity of the electrons in a field of 50 Volt/ meter, ii) their mobility
iii) their relaxation time and mean free path on the basis of classical free electron theory.
Density of Al = 2700 kg/m3

. NA = 6.02 x 1026/kmol. Assume that each Al atom contributes 3

electron in the electron gas.

Answer 1

Answer:

A) Drift Velocity is 0.198m/s .

B) Mobility is $(\frac{0.198}{50} )=3.96*10^{-3} m^{2}/V.s$

C)Relaxation time is $2.18*10^{-25} s$ and mean free path is$2.18*10^{-25} *V_{rms}$

Explanation:

Part1:

Drift Velocity is the mean velocity of electrons within the conductor when it is exposed to the electric field.

$J=neV_{d}$

σE=$neV_{d}$

where $V_{d}$ is the drift velocity of conduction of electron of Aluminum.

n=Number of electrons per unit volume

$n=\frac{Density}{Molar\ mass\ of\ Aluminum}*N_{A} \\n=\frac{2700*10^{3} }{26.98}*N_{A} \\n=100074 N_{A}\ m^{-2}$

$e=1.6*10^{-19}$

σ=conductivity=1/ρ =  $\frac{1}{2.62*10^{-8} }$

On substituting the values,

$\frac{1}{2.62*10^{-8} }*50=100074 N_{A}*1.6*10^{-19} *V_{d}$

On solving we get $V_{d}=0.198\ m/s$

Hence 0.198m/s is the drift velocity of electrons.

Part2:

Mobility =μ=$\frac{V_{d} }{E}$

On substituting values,

μ=$(\frac{0.198}{50} )=3.96*10^{-3} m^{2}/V.s$

Part 3:

Relaxation Time,

  $\frac{m}{ne^{2}*density} \\=\frac{9.11*10^{-19}}{6.02*10^{23}*100074*e^{2} }\\ =2.18*10^{-25}sec$

Mean free path,

λ=$V_{rms}*Relaxation\ time$

λ=$2.18*10^{-25} *V_{rms}$

Hence the relaxation time is $2.18*10^{-25} s$ and mean free path is $2.18*10^{-25} *V_{rms}$.