Question

In a circuit, we are using conducting wires made from Manganese.
(i) If we assume there are 3 free electrons per an atom of manganese, what is its
electron density?
(ii) How much current flows through a cylindrical manganese wire of volume 27 cm3,
length 3cm if the circuit is switched on for 5 seconds?

Answer 1

i. The electron density of maganese is 2.44 × 10²⁸ electrons/m³

ii. The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

To solve the question, we need to know what electron density is.

What is electron density?

This is the number of free electrons per unit volume of material.

To find electron density, we need to find the atom density which is the number of atoms per unit volume, n.

Atom density of manganese

So, the atom density of manganese is n = Nρ/A where

• N = avogadro's number = 6.022 × 10²³ atoms/mol,
• ρ = density of manganese = 7430 kg/m³ and
• A = atomic mass of manganese = 0.0594 kg/mol

So, n = Nρ/A

n = 6.022 × 10²³ atoms/mol × 7430 kg/m³/0.05494 kg/mol

n = 44743.46 × 10²³ atoms/mol × kg/m³/0.05494 kg/mol

n = 81440.59 × 10²³ atoms/m³

n = 8.144059 × 10²⁷ atoms/m³

n ≅ 8.14 × 10²⁷ atoms/m³

i. Electron density of manganese

The electron density of maganese is 2.44 × 10²⁸ electrons/m³

So, the electron density of manganese n' = n" × n where

• n" = number of free electrons per atom = 3 and
• n = atom density = 8.14 × 10²⁷ atoms/m³

So, n' = n" × n

= 3 electrons per atom × 8.14 × 10²⁷ atoms/m³

= 24.42 × 10²⁷ electrons/m³

= 2.442 × 10²⁸ electrons/m³

≅ 2.44 × 10²⁸ electrons/m³

So, the electron density of maganese is 2.44 × 10²⁸ electrons/m³

ii. How much current flows through the cylindrical maganese wire?

The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

The current flowing through the wire is given by i = n'eV/t where

• n = electron density = 2.44 × 10²⁸ electrons/m³,
• e = electron charge = 1.602 × 10⁻¹⁹ C,
• V = volume of wire = 27 cm³ = 27 × 10⁻⁶ m³ and
• t = time current flows = 5 s

So, substituting the values of the variables into the equation, we have

i = n'eV/t

i = 2.44 × 10²⁸ electrons/m³ × 1.602 × 10⁻¹⁹ C × 27 × 10⁻⁶ m³/5 s

i = 105.54 × 10³ C/5 s

i = 21.11 × 10³ C/s

i = 21.11 × 10³ A

i = 21.11 kA

So, the current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

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