Question

Every atom makes one free electron in copper. If $1.1$ ampere current is flowing in the wire of copper having $1\,mm$ diameter, then the drift velocity (approx) will be (density of copper $ = 9 \times {10^3}kg/{m^3}$ and atomic weight of copper $ = 63$ )
A. $0.1\,mm/s$
B. $0.2\,mm/s$
C. $0.3\,mm/s$
D. $0.2\,cm/s$

Answer 1

Hint:In order to answer this question, we must determine the drift velocity, which we must understand. We need to be aware of the number density, which shows how many carriers are included in each volume unit of the substance (it is usually expressed in carriers per cubic meter). The drift velocity of an electron in a unit electric field is referred to as the "mobility of the electron".

Formula Used:
Number of atoms per unit volume,
$n = \dfrac{{\rho {N_A}}}{M}$
Here, $\rho {N_A}$ is the density of copper and $M$ is the atomic weight of the copper.
Cross section area of wire,
\[A = \dfrac{{\pi {d^4}}}{4}\]
Drift velocity,
${v_d} = \dfrac{I}{{Ane}}$
Here, $J$ is the current density, $n$ is the charge density, $e$ is the charge on the electron and $A$ is the area of the cross section.

Complete step by step solution:
Firstly, we find the number of copper atoms per unit volume as per the formula mentioned above,
$n = \dfrac{{\rho {N_A}}}{M}$
Putting all the consecutive values in the above equation, we get,
$n = \dfrac{{9 \times {{10}^3}kg/{m^3} \times 6.023 \times {{10}^{23}}}}{{0.063kg}}$
By doing further solution of the above equation, we get,
$n = 8.60 \times {10^{28}}{m^{ - 3}}$

Since each copper atom releases one free electron, there are $8.60 \times {10^{28}}{m^{ - 3}}$ of electrons per unit volume.
Now, we just find the Cross-section are of the wire,
\[A = \dfrac{{\pi {d^4}}}{4}\]
By putting all the values from question, we get,
\[A = \dfrac{{3.14 \times {{(0.001m)}^2}}}{4}\]
By solving the above equation, we get,
$A = 7.85 \times {10^{ - 7}}{m^2}$
Now, we get all the required values for drift velocity, so let we find drift velocity,
${v_d} = \dfrac{I}{{Ane}}$

Putting all the values from the above equation and question, we get the equation as,
${v_d} = \dfrac{{1.1A}}{{(7.85 \times {{10}^{ - 7}}{m^2})(8.6 \times {{10}^{28}}{m^{ - 3}})(1.6 \times {{10}^{ - 19}}C)}} \\ $
By doing further solution of the above equation, we get the resultant as,
${v_d} = 0.01 \times {10^{ - 2}}m/s \\ $
By simplifying as according to need we get,
${v_d} = 0.1\,mm/s$
Therefore, the correct answer is ${v_d} = 0.1mm/s$ .

Hence, the correct option is A.

Note: The drift velocity is the negligible average speed of free electrons moving in the direction of positive potential. Whereas, Relaxation Time: The relaxation time is the amount of time an electron has to move freely between two collisions with lattice ions or atoms.