Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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

Last updated date: 22nd Jul 2024
Total views: 63.3k
Views today: 1.63k
Verified
63.3k+ views
Hint Conductivity $\sigma = \dfrac{{n{e^2}\tau }}{m}$(Where m is the mass of electron, n is the number of density, e is the charge of an electron and $\tau$is the relaxation time or mean free time.) Resistivity $(\rho )$is the reciprocal of the conductivity $(\sigma )$.
Formula used: $\sigma = \dfrac{{n{e^2}\tau }}{m}$(Where $\sigma$is the conductivity, m is the mass of electron, n is the number of density, e is the charge of an electron and $\tau$is the relaxation time or mean free time.)
Resistivity $(\rho ) = \dfrac{1}{\sigma } = \dfrac{m}{{n{e^2}\tau }}$

We know that current density $J = ne{v_d}$……. (i)
Now $J = \sigma E$ and ${v_d} = \dfrac{{eE\tau }}{m}$
Equation (i) can be written as,
$\sigma E = ne\left( {\dfrac{{eE}}{m}} \right)\tau$
$\Rightarrow \sigma = \dfrac{{n{e^2}\tau }}{m}$
(Where $\sigma$ is the conductivity, m is the mass of the electron, n is the number of density, e is the charge of an electron and $\tau$is the relaxation time or mean free time.)
Now we know that resistivity $(\rho )$ is the reciprocal of the conductivity $(\sigma )$.
Therefore, resistivity $(\rho ) = \dfrac{1}{\sigma } = \dfrac{m}{{n{e^2}\tau }}$……. (ii)
Given,
mass of the electron $(m) = 9.1 \times {10^{ - 31}}kg$
number density of electron $(n) = 8.5 \times {10^{28}}{m^{ - 3}}$
mean free time $(\tau ) = 25fs = 25 \times {10^{ - 15}}s$
And we know that charge of an electron $(e) = 1.6 \times {10^{ - 19}}C$
Providing the values in equation (ii) we get,
Resistivity $(\rho ) = \dfrac{{9.1 \times {{10}^{ - 31}}}}{{8.5 \times {{10}^{28}} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2} \times 25 \times {{10}^{ - 15}}}} = 1.6 \times {10^{ - 8}}\Omega m.$

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

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.