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Question: In a conductor, if the number of conduction electrons per unit volume is \(8.5 \times {10^{28}}{m^{ ...

In a conductor, if the number of conduction electrons per unit volume is 8.5×1028m−38.5 \times {10^{28}}{m^{ - 3}}and mean free time is 25fs25fs(femtosecond), find out its approximate resistivity. (me=9.1×10−31kg)({m_e} = 9.1 \times {10^{ - 31}}kg)
(A) 10−5Ωm{10^{ - 5}}\Omega m
(B) 10−6Ωm{10^{ - 6}}\Omega m
(C) 10−7Ωm{10^{ - 7}}\Omega m
(D) 10−8Ωm{10^{ - 8}}\Omega m

Explanation

Solution

Hint Conductivity σ=ne2τm\sigma = \dfrac{{n{e^2}\tau }}{m}(Where m is the mass of electron, n is the number of density, e is the charge of an electron and τ\tau is the relaxation time or mean free time.) Resistivity (ρ)(\rho )is the reciprocal of the conductivity (σ)(\sigma ).
Formula used: σ=ne2τm\sigma = \dfrac{{n{e^2}\tau }}{m}(Where σ\sigma is the conductivity, m is the mass of electron, n is the number of density, e is the charge of an electron and τ\tau is the relaxation time or mean free time.)
Resistivity (ρ)=1σ=mne2τ(\rho ) = \dfrac{1}{\sigma } = \dfrac{m}{{n{e^2}\tau }}

Complete step by step answer
We know that current density J=nevdJ = ne{v_d}……. (i)
Now J=σEJ = \sigma E and vd=eEτm{v_d} = \dfrac{{eE\tau }}{m}
Equation (i) can be written as,
σE=ne(eEm)τ\sigma E = ne\left( {\dfrac{{eE}}{m}} \right)\tau
⇒σ=ne2τm\Rightarrow \sigma = \dfrac{{n{e^2}\tau }}{m}
(Where σ\sigma is the conductivity, m is the mass of the electron, n is the number of density, e is the charge of an electron and τ\tau is the relaxation time or mean free time.)
Now we know that resistivity (ρ)(\rho ) is the reciprocal of the conductivity (σ)(\sigma ).
Therefore, resistivity (ρ)=1σ=mne2τ(\rho ) = \dfrac{1}{\sigma } = \dfrac{m}{{n{e^2}\tau }}……. (ii)
Given,
mass of the electron (m)=9.1×10−31kg(m) = 9.1 \times {10^{ - 31}}kg
number density of electron (n)=8.5×1028m−3(n) = 8.5 \times {10^{28}}{m^{ - 3}}
mean free time (τ)=25fs=25×10−15s(\tau ) = 25fs = 25 \times {10^{ - 15}}s
And we know that charge of an electron (e)=1.6×10−19C(e) = 1.6 \times {10^{ - 19}}C
Providing the values in equation (ii) we get,
Resistivity (ρ)=9.1×10−318.5×1028×(1.6×10−19)2×25×10−15=1.6×10−8Ωm.(\rho ) = \dfrac{{9.1 \times {{10}^{ - 31}}}}{{8.5 \times {{10}^{28}} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2} \times 25 \times {{10}^{ - 15}}}} = 1.6 \times {10^{ - 8}}\Omega m.

Hence, the resistivity would be of the order of 10−8.{10^{ - 8}}.

Additional Information Mean free time is also known as relaxation time. It is the average time between two successive collisions for an electron. The relaxation time of electrons in a conductor depends on the mass of the electron, the charge of the electron, the number density and the velocities of electrons and ions.

Note Whenever these types of questions appear, remember to consider the conductivity first. As we know resistivity is the reciprocal of conductivity hence, we can find resistivity easily. Always maintain the correct unit (SI or CGS). Convert all the units in either SI or CGS. Then determine the result.