current-electricity Question 28
Question: Q. 2. What is relaxation time ? Derive an expression for resistivity of a wire in terms of number density of free electrons and relaxation time.
A[SQP I 2017-18]
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Solution:
Ans. Definition and Derivation.
[CBSE Marking Scheme, 2017]
Detailed answer :
(i) Relaxation time shows the effect of collisions among the electrons and ions or impurities on electrical conduction in a metal. It is the time taken for the drift velocity to decay $1 / e$ of its initial value.
As drift velocity increases, relaxation time decreases since the electrons move the distance in which they frequently collide faster.
(ii) When a potential difference $V$ is applied across conductor of length $l$, then drift speed of electron will result as :
$$ \begin{aligned} v_{d} & =\frac{e E \tau}{m} \ & =\frac{e V \tau}{l m} \end{aligned} $$
$$ \because\left[E=\frac{V}{l}\right] $$
The electric current through the conductor and drift speed are linked as $I=n e A v_{d}$ where,
$n=$ number density of electrons
$e=$ electronic charge
$A=$ area of cross section
$v_{d}=$ electron drift speed
$\therefore \quad I=n e A\left(\frac{e V \tau}{l m}\right)$
So, $\quad \frac{V}{I}=\frac{m l}{n e^{2} \tau A}$
At constant temperature :
$\frac{V}{I}=R$
Hence,
Comparing above expression with
$$ R=\rho \frac{l}{A} $$
where, $\rho=$ specific resistance
$$ \rho=\frac{m}{n e^{2} \tau} $$
