Derive the relation between resistivity and relaxation time for a conducting wire.
Answers
formula used. = vd = − e Emτ
I=−ne Avd
V=IR
hope help
Answer:
resistivity row= m_e /( n e^2 tau )
Explanation:
please see the enclosed attachment for detailed explanation and step by step working.
This is for a metal conductor and not for a semiconductor.
Tau = T = Relaxation time period = the mean time between successive collisions of an electron when an electric field is applied across a conductor.
- e = charge on an electron.
m_e = mass of an electron.
v_d = drift velocity of an electron along a conductor in the direction of electric field E applied across its end.
V = voltage drop across the ends of the conductor.
Sigma = conductivity of the conductor.
row = resistivity of the conductor.
L = length of the conductor.
A = Area of cross section of the conductor.
j = current density vector, current per unit cross section area.
i = current flowing in the conductor.
n = the number of free electrons per unit volume in the conductor.
acceleration of an electron= a = eE/m_e.
Average velocity of an electron= drift velocity = v_d
v_d = a × Tau = eE Tau/m_e.
The number of electrons crossing a particular cross section in ∆t time= v_d ×∆t × A × n
So electric Charge crossing any cross section in unit time
= i = e × (v_d ×∆t × A × n) / ∆t = n e A v_d
Current Density j = i/A = n e v_d = n e [eE Tau/m_e ]
j = n e^2 E Tau /m_e.
We know that Current density j = - Sigma × E
So Sigma = 1/row = n e^2 Tau /m_e.
resistivity row = m_e /( n e^2 Tau )
