*Notes to a video lecture on http://www.unizor.com*

__Speed of Electrons__

*Electric current*is a movement of electrons. We know from experience that, when we turn on the switch, the lights in the room are lit practically immediately. Does it mean that electrons from one terminal of a switch go to the light fixture and back to another terminal of a switch that fast?

No.

Let's calculate the real speed of electrons, first, theoretically and then in some practical case.

Assume, the

*amperage*of the electric current going through a wire, that is the number of

*coulombs*of electric charge going through a wire per second, is

*, and the wire has cross-section area*

**I***.*

**A**Assume further that we know all the physical characteristics of a material our wire is made of, which will be introduced as needed.

Based on this information, our plan is to determine the number of electrons going through the wire per unit of time and, knowing the density of electrons per linear unit of length in the wire, determine the linear speed of these electrons.

Obviously, to determine the linear density of electrons, we will need physical characteristics of a wire.

The number of electrons going through a wire per unit of time is easily determined from the

*amperage*

*. Since*

**I***represents the number of*

**I***coulombs*of electric charge going through a wire per second, we just have to divide this by the charge of a single electron in

*coulombs*

*.*

**e=−1.60217646·10**C^{−19}So, the number of electrons going through a wire per second is

*.*

**N**_{e}= I / eNow we will determine the linear density of electrons in the wire.

First of all, we have to know how many active electrons in an atom of material our wire is made of participate in the transfer of electric charge, because not all electrons of each atom are freely moving in the electric field, but only those on the outer orbit. Let's assume, this number is

*.*

**n**_{e}Using this number, we convert the number of electrons participating in the transfer of electric charge

*into the number of atoms*

**N**_{e}*in that part of a wire occupied by all electrons transferring the given charge per second*

**N**_{atoms}*.*

**I***.*

**N**_{atoms}= N_{e}/ n_{e}= I / (n_{e}·e)Next, from the number of atoms we will find their mass and, using the density of wire material, the volume.

Dividing the volume by a cross-section of a wire, we will get the length of a segment of wire occupied by those electrons transferring charge per second, which is the

**speed of electrons**or

**drift**.

Knowing the number of atoms, to get to their mass, we will use the

*Avogadro number*

**N**_{A}=6.02214076·10^{23}that represents the number of particles in one

*mole*of a substance. One

*mole*of material our wire is made of is the number of grams equal to its

*atomic mass*

*, known for any material used for a wire.*

**m**_{a}If

*atoms have mass of*

**N**_{A}*gram,*

**m**_{a}*have total mass*

**N**_{atoms}*.*

**M**

= m_{atoms}= m_{a}· N_{atoms}/ N_{A}== m

_{a}·I / (n_{e}·e·N_{A})Knowing the total mass

*of all atoms that contain all electrons traveling through a wire per second, we calculate the volume*

**M**_{atoms}*by using the*

**V**_{atoms}*density*

*of a material the wire is made of.*

**ρ**

**V**

= m_{atoms}= M_{atoms}/ ρ == m

_{a}·I / (n_{e}·e·N_{A}·ρ)Dividing the volume

*by the cross-section area of a wire, we will get the length of the wire*

**V**_{atoms}*occupied by electrons traveling through it in one second*

**L**

**L = V**

= m_{atoms}/ A == m

_{a}·I / (n_{e}·e·N_{A}·ρ·A)where

*-*

**m**_{a}*atomic mass*of wire's material (assuming it's one atom molecules, like copper)

*-*

**I***electric current - amperage*in the wire

*-*

**n**_{e}*number of active electrons in each atom*of wire's material that participate in the transfer of electric charge

*-*

**e***electric charge of one electron*

*-*

**N**_{A}*number of atoms in 1 mol of conducting material - Avogadro Number*

*-*

**ρ***density*of wire's material

*-*

**A***cross-section area*of a wire

The above formula represents the length of a wire occupied by all active electrons traveling through it during one second,

**which is the speed of movement of electrons making up an electric current**, called

**drift**.

Let get to practical examples.

Assume, the

*voltage*or the difference of

*electric potential*between two ends of a copper wire is maintained at

**E****110V**(standard voltage for apartments in the USA).

This wire connects a lamp that consumes

**120W**of electric power

*(or*

**P***wattage*).

Let the cross-section area of a wire be

**3 mm²**.

First of all, let's calculate the amount of electricity moving through the wire per unit of time -

*amperage*.

**I**As we know, the

*amperage*, multiplied by

**I***voltage*, is the electric power

**E***(*

**P***wattage*).

Therefore,

**E = 110**V

**P = 120**W

**I = P/E = 60W/110V ≅ 1.09**AThe

*atomic number*of copper is

*. It means, the atom of copper has 29 protons in the nucleus and 29 electrons orbiting a nucleus. These 29 electrons are in four orbits: 2+8+18+1.*

**Z=29**The outer orbit has only one electron that participates in the movement of electric charge, so the number of active electrons in an atom of copper is

*.*

**n**_{e}=1The nucleus of an atom of copper has 29 protons and 34 or 36 neutrons, its

*atomic mass*is

**m**g/mol_{a}= 63.546The electric charge of one electron is

*.*

**e = −1.60217646·10**C^{−19}The Avogadro number is

**N**_{A}=6.02214076·10^{23}Density of copper is

**ρ = 8.96**g/cm³**= 0.00896**g/mm³Cross-section area of a wire is

**A = 3**mm²Using the formula above with values listed, we obtain

**L = m**_{a}·I / (n_{e}·e·N_{A}·ρ·A)we get

**L ≅ 0.0267**mm/secThis is the speed of electrons traveling along a copper wire in this case.

Pretty slow!

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