Question

The area of cross-section, length, and density of a piece of metal of atomic weight $$60$$ are $${10}^{-6}{m}^2$$ , $$1m$$ and $$5\times {10}^{3}kg{m}^{-3}$$ respectively. If every atom contributes one free electron, find the drift velocity (in $$mm{s}^{-1}$$ ) of electrons in the metal when a current of $$16A$$ passes through it. Avogadro’s number is $$N_A=6\times {10}^{23}/mol$$ and charge on an electron is $$e=1.6\times {10}^{-19}C$$

Answer 1

Hint: We can make use of the relation between the current density and the drift velocity of the electron(s). Current density is the amount of current travelling per unit cross-section area. Before proceeding, we’ll need to find the number of atoms per unit volume of the substance.

Formula Used:
$$n={\displaystyle \frac{m}{M}}\times {\displaystyle \frac{N_A}{V}}$$
$$J={\displaystyle \frac{I}{A}}$$
$$v_d={\displaystyle \frac{J}{en}}$$

Complete step by step answer:
As discussed in the hint, we’ll begin with finding the number of atoms in unit volume of the substance, which will be $$n={\displaystyle \frac{m}{M}}\times {\displaystyle \frac{N_A}{V}}$$ where $$m$$ is the mass of the metal, $$M$$ is the atomic weight of the substance, $$N_A$$ is the Avogadro’s number and $$V$$ is the volume of the metal
Substituting the values in the above equation, we get
$$\begin{array}{rl}& n={\displaystyle \frac{6\times {10}^{23}/mol}{60}}\times \rho (\because \rho ={\displaystyle \frac{mass}{volume}})\\ & \Rightarrow n={10}^{22}\times 5\times {10}^3\\ & \Rightarrow n=5\times {10}^{25}\end{array}$$
We can find the number of electrons from the no. of atoms per unit volume since we’ve been told that each metal atom contributes one free electron.
Current density is given as $$J={\displaystyle \frac{I}{A}}$$
Substituting the values, we get
$$\begin{array}{rl}& J={\displaystyle \frac{16}{{10}^{-6}}}\\ & \Rightarrow J=1.6\times {10}^{7}A/m^2\end{array}$$
Drift velocity can be calculated as follows
$$\begin{array}{rl}& v_d={\displaystyle \frac{J}{en}}\\ & \Rightarrow v_d={\displaystyle \frac{1.6\times {10}^7}{1.6\times {10}^{-19}\times 5\times {10}^{28}}}\\ & \Rightarrow v_d=0.002m/s=2mm{s}^{-1}\end{array}$$

Note: An alternative method of solving this problem is using the relation for current in a conductor, which is $$I=n\times e\times A\times v_d$$ , where the symbols have their meaning as stated in the solution. The benefit of using this relation is that we won’t have to find the current density. Also if the question had mentioned that every atom furnishes two electrons, then the answer would be reduced by half.