
Define relaxation time of the free electrons drifting in a conductor: How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material.
Answer
456.5k+ views
Hint: Drift velocity is directly proportional to relaxation time. Write the formula for current flowing through the conductor. Substitute the value of drift velocity in terms of relaxation time. Use the formula for electric field applied and Ohm’s law. Get the value of R in terms of l and A.
Formula used:
${ v }_{ d } = -\dfrac { eE }{ m } \tau $
$I = -neA{ v }_{ d }$
$V = IR$
Complete step by step answer:
Relaxation time is defined as the time interval between two successive collisions of electrons in a conductor when current flows through it. It is denoted by $ \tau$.
Relation between drift velocity of the electrons (${ v }_{ d }$) and relaxation time (\tau) is given by,
${ v }_{ d } = -\dfrac { eE }{ m } \tau$ …(1)
Where, E: Electric field
M: mass of electron
E: charge of electron
Consider a conductor having length l, area of cross section A and current density n.
Current flowing through the conductor is given by,
$I = -neA{ v }_{ d }$…(2)
Now, by substituting equation.(1) in equation.(2) we get,
$I = neA\dfrac { eE }{ m } \tau$
$\therefore I = \dfrac { n{ e }^{ 2 }AE }{ m } \tau$ …(3)
Electric field applied is given by,
$E = \dfrac { V }{ l }$ …(4)
Substituting equation. (3) in equation.(4) gives,
$I =\dfrac { n{ e }^{ 2 }AV\tau }{ ml }$
Now, by rearranging the equation we get,
$\dfrac { V }{ I } = \dfrac { ml }{ n{ e }^{ 2 }A\tau }$ …(5)
According to Ohm’s Law,
$V = IR$
Rearranging the above equation we get,
$R = \dfrac { V }{ I }$
Now, substituting the equation. (5) in above equation we get,
$R = \dfrac { ml }{ n{ e }^{ 2 }A\tau }$
Let $\rho = \dfrac { m }{ n{ e }^{ 2 }\tau }$
$\therefore R = \rho \dfrac { l }{ A }$
Where, $\rho$ : Resistivity of the material
Therefore, the expression for electrical resistivity is given by $\rho = \dfrac { m }{ n{ e }^{ 2 }\tau }$.
Note:
Electrical resistivity is inversely proportional to relaxation time. Reciprocal resistivity gives conductivity. It is denoted by $\sigma$. Therefore, $\sigma = \dfrac { n{ e }^{ 2 }\tau }{ m }$.
As the temperature of the conductor is decreased, the number of electron collisions also decreases. As a result, relaxation time gets increased. Thus we can say, relaxation time is dependent on temperature.
Formula used:
${ v }_{ d } = -\dfrac { eE }{ m } \tau $
$I = -neA{ v }_{ d }$
$V = IR$
Complete step by step answer:
Relaxation time is defined as the time interval between two successive collisions of electrons in a conductor when current flows through it. It is denoted by $ \tau$.
Relation between drift velocity of the electrons (${ v }_{ d }$) and relaxation time (\tau) is given by,
${ v }_{ d } = -\dfrac { eE }{ m } \tau$ …(1)
Where, E: Electric field
M: mass of electron
E: charge of electron

Consider a conductor having length l, area of cross section A and current density n.
Current flowing through the conductor is given by,
$I = -neA{ v }_{ d }$…(2)
Now, by substituting equation.(1) in equation.(2) we get,
$I = neA\dfrac { eE }{ m } \tau$
$\therefore I = \dfrac { n{ e }^{ 2 }AE }{ m } \tau$ …(3)
Electric field applied is given by,
$E = \dfrac { V }{ l }$ …(4)
Substituting equation. (3) in equation.(4) gives,
$I =\dfrac { n{ e }^{ 2 }AV\tau }{ ml }$
Now, by rearranging the equation we get,
$\dfrac { V }{ I } = \dfrac { ml }{ n{ e }^{ 2 }A\tau }$ …(5)
According to Ohm’s Law,
$V = IR$
Rearranging the above equation we get,
$R = \dfrac { V }{ I }$
Now, substituting the equation. (5) in above equation we get,
$R = \dfrac { ml }{ n{ e }^{ 2 }A\tau }$
Let $\rho = \dfrac { m }{ n{ e }^{ 2 }\tau }$
$\therefore R = \rho \dfrac { l }{ A }$
Where, $\rho$ : Resistivity of the material
Therefore, the expression for electrical resistivity is given by $\rho = \dfrac { m }{ n{ e }^{ 2 }\tau }$.
Note:
Electrical resistivity is inversely proportional to relaxation time. Reciprocal resistivity gives conductivity. It is denoted by $\sigma$. Therefore, $\sigma = \dfrac { n{ e }^{ 2 }\tau }{ m }$.
As the temperature of the conductor is decreased, the number of electron collisions also decreases. As a result, relaxation time gets increased. Thus we can say, relaxation time is dependent on temperature.
Recently Updated Pages
Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Why is the cell called the structural and functional class 12 biology CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

What are the major means of transport Explain each class 12 social science CBSE
