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Question: The resistivity of copper at room temperature is \(1.7 \times {10^{ - 8}}ohm\)-\(meter\). If the den...

The resistivity of copper at room temperature is 1.7×10−8ohm1.7 \times {10^{ - 8}}ohm-metermeter. If the density of mobile electrons is 8.4×1028m−38.4 \times {10^{28}}{m^{ - 3}}, the relaxation time for free electrons in copper is: (mass of electron 9×10−11kg9 \times {10^{ - 11}}kg, charge of electron 1.6×10−19C1.6 \times {10^{ - 19}}C)
(A) 2.5×10−14s2.5 \times {10^{ - 14}}s
(B) 2.5×10−12s2.5 \times {10^{ - 12}}s
(C) 2.5×10−10s2.5 \times {10^{ - 10}}s
(D) 2.5×10−8s2.5 \times {10^{ - 8}}s

Explanation

Solution

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(E)\left( E \right) applied across its length. It can be calculated by,
E=VlE = \dfrac{V}{l}
Where V=V = potential difference across the conductor and
l=l = length of the conductor
Relaxation time(τ)\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(vd)\left( {{v_d}} \right) and relaxation time(τ)\left( \tau \right) is given by,
vd=−eEmτ{v_d} = - \dfrac{{eE}}{m}\tau
Where, e=e = charge of electron
E=E = Electric field
m=m = Mass of electron
Let us assume that the length of the copper conductor through which the current is flowing is LL, area of cross-section is AA and its current density is nn.
⇒I=−neAvd\Rightarrow I = - neA{v_d}
Substituting the value of vd{v_d} from the previous equation,
⇒I=neAeEmτ\Rightarrow I = neA\dfrac{{eE}}{m}\tau
⇒I=ne2AEmτ\Rightarrow I = \dfrac{{n{e^2}AE}}{m}\tau
Substituting the value of EE from fist equation,
⇒I=ne2AVmlτ\Rightarrow I = \dfrac{{n{e^2}AV}}{{ml}}\tau
⇒VI=mLne2Aτ\Rightarrow \dfrac{V}{I} = \dfrac{{mL}}{{n{e^2}A\tau }}...................(1)
Now according to Ohm’s law,
⇒V=IR\Rightarrow V = IR
Where, R=R = resistance of the conductor.
⇒R=VI\Rightarrow R = \dfrac{V}{I}...........(2)
Resistance can also be calculated by,
⇒R=ρlA\Rightarrow R = \rho \dfrac{l}{A}...............(3)
Where, ρ=\rho = resistivity of the conductor.
Substituting equation two in equation one,
⇒R=mLne2Aτ\Rightarrow R = \dfrac{{mL}}{{n{e^2}A\tau }}...........(4)
Substituting equation three in equation four,
⇒ρLA=mLne2Aτ\Rightarrow \rho \dfrac{L}{A} = \dfrac{{mL}}{{n{e^2}A\tau }}
⇒ρ=mne2τ\Rightarrow \rho = \dfrac{m}{{n{e^2}\tau }}
⇒τ=mne2ρ\Rightarrow \tau = \dfrac{m}{{n{e^2}\rho }}
Substituting the values given in the question in the above equation,
⇒τ=9×10−118.4×1028×(1.6×10−19)2×1.7×10−8\Rightarrow \tau = \dfrac{{9 \times {{10}^{ - 11}}}}{{8.4 \times {{10}^{28}} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2} \times 1.7 \times {{10}^{ - 8}}}}
⇒τ=9×10−118.4×1028×2.56×10−38×1.7×10−8\Rightarrow \tau = \dfrac{{9 \times {{10}^{ - 11}}}}{{8.4 \times {{10}^{28}} \times 2.56 \times {{10}^{ - 38}} \times 1.7 \times {{10}^{ - 8}}}}
∴τ=2.5×10−14s\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.