Question

A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free electron per atom. The density of aluminum is 2.7~grams/cm^3, and the aluminum molar mass is 27 g. What is the electron number density (the number of electrons per unit volume) in the wire

Answer 1

Answer: The electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$.

Explanation:

Given: Current = 5.0 A

Area = $4.0 \times 10^{-6} m^{2}$

Density = 2.7 $g/cm^{3}$, Molar mass = 27 g

The electron density is calculated as follows.

$n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}$

where,

$\rho$ = density

M = molar mass

$N_{A}$ = Avogadro's number

Substitute the values into above formula as follows.

$n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}$

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$.