Question

The resistivity of copper at room temperature is $1.7 \times {10^{ - 8}}\Omega - m$. If the density of the mobile electrons is $8.4 \times {10^{28}}{m^{ - 3}}$, the relaxation time for the free electrons in copper is (mass of electron $9 \times {10^{ - 11}}kg$ and charge of electron $1.6 \times {10^{ - 19}}C$)
A. $2.5 \times {10^{ - 14}}s$
B. $2.5 \times {10^{ - 12}}s$
C. $2.5 \times {10^{ - 10}}s$
D. $2.5 \times {10^{ - 8}}s$

Answer 1

Hint: Whenever we hear the word density, the first thing that should strike our mind is the word distribution. For example, words like mass density (just called density in Science), population density, and nuclear density etc. mean that they are talking about the distribution.

Complete step-by-step answer:
Ohm’s law in general, gives us the following expression:
$R = \dfrac{V}{I}$
where R – resistance, V – voltage and I – current.
If we write the microscopic version of the Ohm’s law, we get –
$\rho = \dfrac{E}{J}$
Here, these are corresponding quantities associated with the quantities in the macroscopic or general Ohm’s law:
$\rho $- Resistivity corresponding to Resistance
$E$- electric field corresponding to Voltage
$J$- current density corresponding to Current.
It is important to understand the concept of drift velocity. Drift velocity is defined as the velocity at which the electron drifts towards the direction of the field. The expression for the current because of drift velocity is given by –
$I = neA{v_d}$
where n = electron density, e = fundamental charge of electron, A = cross-section area of the conductor and ${v_d}$= drift velocity.
The current density is defined as the current per unit area.
$J = \dfrac{I}{A}$
Substituting the equation for current, we get –
$J = \dfrac{{neA{v_d}}}{A} = ne{v_d}$
The drift velocity, ${v_d} = \dfrac{{eE\tau }}{m}$ where $\tau $ is called relaxation time and m = mass of electron.
Substituting, we get –
$
  J = ne{v_d} \\
  J = \dfrac{{ne\left( {eE\tau } \right)}}{m} \\
   \to J = \dfrac{{n{e^2}\tau }}{m}E \\
  Also, \\
  \dfrac{E}{J} = \dfrac{{n{e^2}\tau }}{m} \\
 $
Comparing this equation with the microscopic Ohm’s law – $\rho = \dfrac{E}{J}$, we get –
$\rho = \dfrac{m}{{n{e^2}\tau }}$
Given
Resistivity, $\rho = 1.7 \times {10^{ - 8}}\Omega m$
Density, $n = 8.4 \times {10^{28}}{m^{ - 3}}$
Electron charge, e= $1.6 \times {10^{ - 19}}C$
Mass of electron, m= $9 \times {10^{ - 11}}kg$
Solving for relaxation time,
$
  \rho = \dfrac{m}{{n{e^2}\tau }} \\
  \tau = \dfrac{m}{{n{e^2}\rho }} \\
  Substituting, \\
   \to \tau = \dfrac{{9 \times {{10}^{ - 11}}}}{{8.4 \times {{10}^{28}} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2} \times 1.7 \times {{10}^{ - 8}}}} \\
   \to \tau = \dfrac{{9 \times {{10}^{ - 11}}}}{{8.4 \times {{10}^{28}} \times 2.56 \times {{10}^{ - 38}} \times 1.7 \times {{10}^{ - 8}}}} \\
   \to \tau = \dfrac{{9 \times {{10}^{ - 11}}}}{{36.55 \times {{10}^{28 - 38 - 8}}}} \\
   \to \tau = 0.246 \times {10^{ - 11 - 28 + 38 + 8}} \\
   \to \tau = 0.25 \times {10^{ - 7}}\sec \\
   \to \tau = 2.5 \times {10^{ - 8}}\sec \\
 $

Hence, the correct option is Option D.

Note: The drift velocity is not constant for all the materials. It depends on a quantity called electron mobility. The electron mobility is the characteristic property of a material which determines how much it allows the electron to drift inside the material.
The drift velocity in terms of electron mobility is –
${v_d} = \mu E$
where $\mu $ is called the electron mobility and E is the electric field applied.