Question

The resistivity of copper at room temperature is $1.7 \times {10^{ - 8}}ohm$-$meter$. If the density of mobile electrons is $8.4 \times {10^{28}}{m^{ - 3}}$, the relaxation time for free electrons in copper is: (mass of electron $9 \times {10^{ - 11}}kg$, 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: Relaxation time is defined as the time interval between two successive collisions of electrons in a conductor when current flows through it. It is directly proportional to drift velocity.

Complete step by step answer:
Current through a conductor flows because of the electric field$\left( E \right)$ applied across its length. It can be calculated by,
$E = \dfrac{V}{l}$
Where $V = $potential difference across the conductor and
$l = $length of the conductor
Relaxation time$\left( \tau \right)$ is defined as the time interval between two successive collisions of electrons in a conductor when current flows through it.
Relation between drift velocity$\left( {{v_d}} \right)$ and relaxation time$\left( \tau \right)$ is given by,
${v_d} = - \dfrac{{eE}}{m}\tau $
Where, $e = $charge of electron
$E = $Electric field
$m = $Mass of electron
Let us assume that the length of the copper conductor through which the current is flowing is $L$, area of cross-section is $A$ and its current density is $n$.
$\Rightarrow I = - neA{v_d}$
Substituting the value of ${v_d}$ from the previous equation,
$\Rightarrow I = neA\dfrac{{eE}}{m}\tau $
$\Rightarrow I = \dfrac{{n{e^2}AE}}{m}\tau $
Substituting the value of $E$ from fist equation,
$\Rightarrow I = \dfrac{{n{e^2}AV}}{{ml}}\tau $
$\Rightarrow \dfrac{V}{I} = \dfrac{{mL}}{{n{e^2}A\tau }}$...................(1)
Now according to Ohm’s law,
$\Rightarrow V = IR$
Where, $R = $resistance of the conductor.
$\Rightarrow R = \dfrac{V}{I}$...........(2)
Resistance can also be calculated by,
$\Rightarrow R = \rho \dfrac{l}{A}$...............(3)
Where, $\rho = $resistivity of the conductor.
Substituting equation two in equation one,
$\Rightarrow R = \dfrac{{mL}}{{n{e^2}A\tau }}$...........(4)
Substituting equation three in equation four,
$\Rightarrow \rho \dfrac{L}{A} = \dfrac{{mL}}{{n{e^2}A\tau }}$
$ \Rightarrow \rho = \dfrac{m}{{n{e^2}\tau }}$
$ \Rightarrow \tau = \dfrac{m}{{n{e^2}\rho }}$
Substituting the values given in the question in the above equation,
$\Rightarrow \tau = \dfrac{{9 \times {{10}^{ - 11}}}}{{8.4 \times {{10}^{28}} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2} \times 1.7 \times {{10}^{ - 8}}}}$
$ \Rightarrow \tau = \dfrac{{9 \times {{10}^{ - 11}}}}{{8.4 \times {{10}^{28}} \times 2.56 \times {{10}^{ - 38}} \times 1.7 \times {{10}^{ - 8}}}}$
$\therefore \tau = 2.5 \times {10^{ - 14}}s$

Hence option A is the correct answer.

Note: Resistivity is a temperature dependent quantity. It decreases as temperature increases and vice-versa. Since relaxation time is inversely proportional to resistivity. Thus, relaxation time increases as temperature increases and vice-versa.