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
544.4k+ 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
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

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

Explain the structure of megasporangium class 12 biology CBSE

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

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

