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,
$E={\displaystyle \frac{V}{l}}$
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,
$v_d=-{\displaystyle \frac{eE}{m}}\tau$
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,
$\Rightarrow I=neA{\displaystyle \frac{eE}{m}}\tau$
$\Rightarrow I={\displaystyle \frac{n{e}^{2}AE}{m}}\tau$
Substituting the value of $E$ from fist equation,
$\Rightarrow I={\displaystyle \frac{n{e}^{2}AV}{ml}}\tau$
$\Rightarrow {\displaystyle \frac{V}{I}}={\displaystyle \frac{mL}{n{e}^{2}A\tau }}$...................(1)
Now according to Ohm’s law,
$\Rightarrow V=IR$
Where, $R=$resistance of the conductor.
$\Rightarrow R={\displaystyle \frac{V}{I}}$...........(2)
Resistance can also be calculated by,
$\Rightarrow R=\rho {\displaystyle \frac{l}{A}}$...............(3)
Where, $\rho =$resistivity of the conductor.
Substituting equation two in equation one,
$\Rightarrow R={\displaystyle \frac{mL}{n{e}^{2}A\tau }}$...........(4)
Substituting equation three in equation four,
$\Rightarrow \rho {\displaystyle \frac{L}{A}}={\displaystyle \frac{mL}{n{e}^{2}A\tau }}$
$\Rightarrow \rho ={\displaystyle \frac{m}{n{e}^2\tau }}$
$\Rightarrow \tau ={\displaystyle \frac{m}{n{e}^2\rho }}$
Substituting the values given in the question in the above equation,
$\Rightarrow \tau ={\displaystyle \frac{9\times {10}^{-11}}{8.4\times {10}^{28}\times {\left(1.6\times {10}^{-19}\right)}^2\times 1.7\times {10}^{-8}}}$
$\Rightarrow \tau ={\displaystyle \frac{9\times {10}^{-11}}{8.4\times {10}^{28}\times 2.56\times {10}^{-38}\times 1.7\times {10}^{-8}}}$
$\therefore \tau =2.5\times {10}^{-14}s$

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.