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 * 10^(28) m^(-3)`.

Step-by-step explanation:

Given: Current = 5.0 A

Area =
`4.0 * 10^(-6) m^(2)`

Density = 2.7
`g/cm^(3)`, Molar mass = 27 g

The electron density is calculated as follows.

`n = (density)/(mass per atom)\\= (\rho)/((M)/(N_(A)))\\`

where,

`\rho` = density

M = molar mass

`N_(A)` = Avogadro's number

Substitute the values into above formula as follows.

`n = (\rho * N_(A))/(M)\\= (2.7 g/cm^(3) * 6.02 * 10^(23)/mol)/(27 g/mol)\\= (16.254 * 10^(23))/(27) cm^(3)\\= 0.602 * 10^(23) * (10^(6) cm^(3))/(1 m^(3))\\= 6.0 * 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 * 10^(28) m^(-3)`.