Question

If the resistivity of copper is $1.7 \times {10^{ - 6}}\,\Omega cm$, then the mobility of electrons in copper, if each atom of copper contributes one free electron for conduction, is [The atomic weight of copper is $63.54$ and its density is $8.96\,gc{c^{ - 1}}$ ].
(A) $23.36\,c{m^2}V_s^{ - 1}$
(B) $503.03\,c{m^2}V_s^{ - 1}$
(C) $43.25\,c{m^2}V_s^{ - 1}$
(D) $88.0\,c{m^2}V_s^{ - 1}$

Answer 1

The mobility of the electron in copper is determined by using some formula, here the density of the copper is given. By converting the density of the copper to the weight, and then by using the number of atoms formula, the number of electrons is determined. And, by using the relation between the conductivity and resistivity, the mobility of the electron can be determined.

Formula used:
The number of atoms in a material is given by,
${\text{Number of atoms = }}\dfrac{{{\text{Weight of the material}}}}{{{\text{Atomic weight of the material}}}} \times {N_a}$
Where, ${N_a}$ is the Avogadro number.
The relation between the conductivity and resistivity is,
${\text{Conductivity = }}\dfrac{1}{{{\text{Resistivity}}}} = n \times e \times m$
Where, $n$ is the number of electrons, $e$ is the charge of the electron and $m$ is the mobility of the electron.

Complete step by step answer:
Given that,
The resistivity of the copper is $1.7 \times {10^{ - 6}}\,\Omega cm$.
The atomic weight of copper is $63.54$.
The density of the copper is $8.96\,gc{c^{ - 1}}$.
Now,
From the density of the copper, by taking the copper wire with $1\,cc$ volume, then the weight of the copper is $8.96\,g$.
The number of atoms in a material is given by,
${\text{Number of atoms = }}\dfrac{{{\text{Weight of the material}}}}{{{\text{Atomic weight of the material}}}} \times {N_a}\,...........\left( 1 \right)$
By substituting the weight of the copper material, atomic weight of the copper and the Avogadro number in the equation (1), then
${\text{Number of atoms = }}\dfrac{{8.96}}{{63.5}} \times 6.02 \times {10^{23}}$ ( ${N_a} = 6.02 \times {10^{23}}$ )
By multiplying the terms in numerator, then
${\text{Number of atoms = }}\dfrac{{5.393 \times {{10}^{24}}}}{{63.5}}$
On dividing the above equation, then
${\text{Number of atoms = 0}}{\text{.849}} \times {\text{1}}{{\text{0}}^{23}}$
Therefore, the number of electrons is $0.849 \times {10^{23}}$.
Now,
${\text{Conductivity = }}\dfrac{1}{{{\text{Resistivity}}}} = n \times e \times m$
The above equation is also written as,
$\dfrac{1}{{{\text{Resistivity}}}} = n \times e \times m\,..............\left( 2 \right)$
By substituting the resistivity, number of electrons and the charge of the electron in the equation (2), then
$\dfrac{1}{{1.7 \times {{10}^{ - 6}}}} = 0.849 \times {10^{23}} \times 1.6 \times {10^{ - 19}} \times m$ (Charge of the electron is $1.6 \times {10^{ - 19}}\,C$ )
By keeping the term $m$ in one side and the other terms in other side, then
$\dfrac{1}{{1.7 \times {{10}^{ - 6}} \times 0.849 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}}} = m$
By multiplying the terms, then
$\dfrac{1}{{0.023}} = m$
On dividing the terms in the above equation, the
$m = 43.25\,c{m^2}V_s^{ - 1}$
Thus, the above equation shows the mobility of the electron.

Hence option (C) is correct.

Note:
The number of atoms derived in the calculation is taken as the number of electrons because every atom has one electron, so it is taken as the number of electrons. The Avogadro number and the charge of the electron are the constant values, so the value is substituted directly.