Answer: The electron number density (the number of electrons per unit volume) in the wire is [tex]6.0 \times 10^{28} m^{-3}[/tex].
Explanation:
Given: Current = 5.0 A
Area = [tex]4.0 \times 10^{-6} m^{2}[/tex]
Density = 2.7 [tex]g/cm^{3}[/tex], Molar mass = 27 g
The electron density is calculated as follows.
[tex]n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\[/tex]
where,
[tex]\rho[/tex] = density
M = molar mass
[tex]N_{A}[/tex] = Avogadro's number
Substitute the values into above formula as follows.
[tex]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}[/tex]
Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is [tex]6.0 \times 10^{28} m^{-3}[/tex].
