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Файл:Steady electric current. Tutorial
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51
The energy transferred to the lattice in a single collision of one elect-
ron with an one ion is equal to
2
222
V2m
Eλe
Vm
Eλe
~
~
~
~
~
max
=
==
2
2
k
2
m
2
mU
w
. (3.13)
The energy released in a unit volume of a conductor per unit time, that
is, the specific thermal power of the current will be equal to
2
2
ku
E
V2m
λen
w
λ
V
n
dtdV
dQ
Q
~
~
~
~
~
===
, (3.14)
where
λ
V
~
~
is the frequency of collisions of one electron with the lattice, n is
the number density of the electron gas.
Comparing (3.11) with Ohm's law in differential form (2.9), we ob-
tain an expression for the specific electrical conductivity
V2m
λne
2
~
~
=σ
. (3.15)
Thus, the classical electron theory, using the laws of classical mechanics
and the “electron gas” model, was able to explain the basic laws of steady current—Ohm’s law and the Joule–Lenz law. The obtained expression for the spe-
cific electrical conductivity of metals is in qualitative agreement with the experiment: σ increases with the growth of the concentration of free electrons and
the length of their free path, and decreases with the growth of temperature, that
is, with the growth of the velocity of thermal motion of electrons. However, the
classical electron theory of metals encountered inevitable insurmountable contradictions. Only with the advent of wave or quantum mechanics the band theory of solids was created, which was able to explain the electrical properties of
metals, dielectrics and semiconductors from a unified position.
3.3. P r o b l e m s
Problem 1. Calculate the number density (concentration) and volume
density of the electron gas in sodium, copper and aluminum at 0 °C. Compare
with similar characteristics of air under normal conditions.
Given
SI
Solution
t = 0 °C
Na, Cu, Al
T = 273 K
The concentration of the electron gas n
e
is equal to the product of the metal valence Z
and the concentration of metal atoms (ions) n
m
me
nZn =)1
.
ne – ?
ρ
e
– ?
The concentration of metal atoms nm is equal to the density of the
a
m
m
m
ρ
n =)2
A
N
mμ=
a
)3
μ
ρN
Zn
mA
e
=)4
e
mA
eee
m
μ
ρN
Zmnρ
==)5
metal
Na
Cu
Al
valence 1 2
2
density, kg/m3
971
8920
2700
molar mass, kg/mol
0.023
0.0635
0.027
metal
Na
Cu
Al
electron gas concentration, 1/m3
2.5∙1028
1.7∙1029
1.2∙1029
electron gas volume density, g/m3
22,8
155
110
metal ρm divided by the mass of one atom ma
.
Let's find the mass of atom using the molar mass μ and Avogadro num-
ber NA
.
Substituting (2) and (3) into (1), we obtain
.
By multiplying the concentration of the electron gas by the mass of
the electron, we obtain its volume density:
.
Let's perform calculations using the data below:
Answer:
For comparison: оne cubic meter of air under normal conditions contains
a number of molecules equal to the Loschmidt number NL = 2.69∙1025 1/m3.
The density of air under normal conditions is 1290 g/m3.
52
53
Problem 2. A current of 50 A flows through a straight wire 20 m long.
Determine the total momentum of the electrons in the electron gas.
Given
Solution
L = 20 m
I = 50 A
Electrons of the electron gas participate in two movements: thermal chaotic movement and ordered movement
under the influence of an electric field. The total momentum
of thermal chaotic movement is zero, since the magnitudes
and directions of the momenta of individual electrons change
chaotically over time.
P – ?
The movement of electrons under the influence of an electric field oc-
curs in the direction opposite to the electric field strength vector at an average
velocity
U
~
. The total momentum of ordered movement is non-zero, and is
equal to the product of the momentum of one electron:
Ump*
~
=)1
and the total number of free electrons (n is the concentration of electron gas):
SL nVnN ==)2
.
Let's express the average speed of ordered motion of electron us-
ing (3.5)
en
j
U =
~
)3
.
Thus, the total momentum of the electrons in the electron gas is
IL
e
m
nLS
S
I
en
m
nLS
en
j
mnLSUmN*pP =====
~
)4
.
Substitute numerical data.
Answer: Р = 5.7∙10−9 kg∙m/s.
Problem 3. In the famous experiment of Tolman1 and Stewart, a ro-
tating coil of copper wire stopped abruptly. Due to the inertial motion of the
current carriers in the coil, a current pulse arose, measured by a sensitive
galvanometer. Let the length of the copper wire be 400 m, the average diameter of the turns 35 cm, the angular velocity of the coil rotation 300 rad/s, the
1
Robert Chase Tolman (1881–1948) was an American mathematical physicist and physical chemist who specialized in statistical mechanics. Together with T. Stewart, he discovered (1916) the inertia of electrons in metals (the Tolman–Stewart effect).
54
resistance of the wire and galvanometer be 30 Ohm. The charge measured by
the galvanometer is 22 nC. Find the specific charge of the current carriers in
the copper wire using these data.
Given
SI
Solution
L = 400 m
d = 35 cm
ω = 300 rad/s
R = 30 Ω
Q = 4 nC
0.35 m
4∙10−9 C
q/m – ?
When the coil is braking with acceleration a, inertial forces act on the
current carriers:
dt
Vd
mamF
in
−=−=)1
.
Inertial forces play the role of extraneous forces, the work of extrane-
ous forces in all turns of the coil is characterized by EMF
L
dt
dV
q
m
ldF
q
1
L
in
=
= )()2
ε
.
According to Ohm's law for a closed circuit (coil + galvanometer) we
find the current in the coil (m and q are the mass and charge of current carriers)
dt
dV
R
L
q
m
R
ε
I ==)3
.
By integration we find the charge flowing through the circuit during
the braking time of the coil
0
0
V
0
V
t
0
V
R
L
q
m
dV
R
L
q
m
dt
dt
dV
R
L
q
m
IdtQ
00
−=
=
=
=)4
.
Thus, the specific charge of current carriers is equal to
RQ
LV
m
q
0
−=)5
.
55
Expressing the linear velocity V of the coil turns through the angular
velocity ω and the radius r of the turns, we obtain
RQ
rLω
m
q
0
−=)6
.
Substitute numerical data.
Answer: q/m = 1.75∙1011 C/kg.
Problem 4. The current density in a copper conductor 5 m long is
2 A/mm2. Assuming that each copper atom is doubly ionized, find the time
it takes for a free electron to move from one end of the wire to the other, as
well as the path it travels in doing so. Room temperature (~300 K).
Given
SI
Solution
L= 5 m
j = 2 A/mm2
2∙106 A/m2
The time it takes for an electron to
travel from the beginning to the end of the
wire is
U
L
t
~
=)1
.
t – ? S – ?
The average velocity of ordered motion of an electron is equal to
en
j
U =
~
)2
,
we take the unknown value of the electron gas concentration from problem 1.
During time t, the electron will travel the distance S with the average
velocity of thermal motion
t
πm
8kT
S =)3
.
Substitute numerical data.
Answer: t ≈ 19 hours; S ≈ 73∙109 m.
S e l f - c h e c k q u e s t i o n s
1. Formulate and explain the basic principles of the classical electron
theory of metals.
2. What do an ideal monatomic gas and an electron gas in metals have
in common?
3. Calculate the electron gas pressure in copper at room temperature.
4. Which particles are current carriers in metals: atoms, ions or elec-
trons?
5. What is the concentration of electron gas in metals? Give the num-
bers.
6. What determines the mean free path of electrons in the electron gas
in metals?
7. What is the average velocity of thermal motion of electrons in the
electron gas in metals?
8. What is the average velocity of the ordered movement of electrons
in the electron gas in metals under the influence of an electric field?
9. Get the expression connecting the current density with the velocity
of ordered movement of the electron gas.
10. What else besides the density of the metal and the mass of its at-
oms determines the concentration of the electron gas in metals?
11. Which bond between atoms in a solid is called a metalic bond?
What is the role of electron gas in metallic bond formation?
12. What causes electrical resistance of metals and the release of Joule
heat?
13. Give a conclusion Ohm's law in the classical electron theory of
metals.
14. Write and explain the expression for specific electrical conduc-
tivity of metals.
15. Get the output Joule–Lenz law in the classical electron theory of
metals.
16. What is the idea behind Tolman and Stewart's experiment? What
were they able to prove and measure?
17. Specify the advantages and disadvantages of the classical elec-
tronic theory of metals.
56
57
4 . E L E C T R I C C U R R E N T IN O T H E R M E D I A
For an electric current to exist, current carriers, i. e., free electric
charges, are necessary. The mechanisms of electrical conductivity in different medias are determined by the type of current carriers (electrons, ions),
the mechanisms of their generation and the properties of the environment
itself. Below, the mechanisms of electrical conductivity in semiconductors
and electrolytes are described, and the phenomenon of thermal electron emission is consi-dered.
4.1. E l e c t r i c c u r r e n t i n s e m i c o n d u c t o r s
Semiconductors are a wide class of substances characterized by electrical conductivity σ intermediate between the electrical conductivity of metals σ ~ (108−106) Ohm−1m−1 and good dielectrics σ ~ (10−8−10
−10
) Ohm−1m−1
(at room temperature).
Let us consider the most widely used semiconductors germanium (Ge)
and silicon (Si). The atoms of these elements have four valence electrons on
the outer electron shell and form crystal lattices in which the atoms are linked
to each other by covalent bonds (fig. 4.1). Ge or Si atoms, completing the
electron shell to eight electrons, form common electron pairs (covalent bond)
with neighboring atoms. Each atom has eight electrons on the outer shell,
which simultaneously also belong to neighboring atoms. Each of the two
neighboring atoms has two common electrons.
Fig. 4.1. Crystal lattice of germanium (flat analogue)
58
Intrinsic conductivity. Chemically pure semiconductors at low (close
to absolute zero) temperatures do not have mobile charge carriers (current
carriers) and only by spending some energy, for example, thermal energy,
they can be formed. When the temperature in the semiconductor increases,
individual electrons are torn away from the atoms and become mobile. When
an electron is removed from an atom, a free space is formed in the shell of
the atom—a hole. Since neighboring atoms that have common electrons constantly exchange them, this hole can be occupied by an electron of a neighboring atom. Then the hole will appear in the shell of the neighboring atom.
Since the atom was neutral before the electron was removed, the lack of an
electron causes the atom to become a positively charged particle. A vacant
place (hole) constantly and randomly moves around the crystal, thereby
transferring a positive charge numerically equal to the charge of an electron.
The mobile charge carriers in semiconductors are electrons and holes, creating an electron-hole conductivity mechanism.
Along with the generation of electron-hole pairs, the reverse process
also occurs – recombination. Free electrons and holes move randomly
around the crystal until one of the mobile electrons meets a hole (falls into
an empty space) in the electron shell of the atom. In this case, the pair of
mobile charge carriers (free electron and hole) disappears, i. e. recombination occurs.
Electrons and holes that arise in a semiconductor as a result of thermal
generation provide their own conductivity, which increases with temperature
according to an exponential law:
)exp( /kTε
A
−=
0
σσ
, (4.1)
where εА is the activation energy of conductivity, equal to the order of the
binding energy of electrons with atoms. As the temperature increases, thermal motion begins to break the bonds of electrons with atoms, and part of
them, proportional to
)exp( /kTε
A
−
, become current carriers, along with holes
that arise simultaneously.
Extrinsic conductivity. The electrical conductivity of semiconductors can be changed by introducing impurities (“doping”) into the crystal
structure. Impurities are a small amounts (of the order of 1 in 108) of pentavalent (antimony, phosphorus, or arsenic) or trivalent (boron, gallium, indium) atoms. This process is known as doping, and the resulting semiconductors are known as doped or extrinsic semiconductors. Impurities are divided into donor and acceptor. When two differently doped regions exist in
59
a single crystal, a semiconductor junction is created. Such junctions are the
basis of diodes, transistors, and most modern electronic devices.
When introducing atoms of the fifth group of the periodic table, for
example arsenic (As), into the germanium lattice, four electrons from those
in the outer shell of the impurity atom go to form common electron pairs with
four neighboring Ge atoms.
For each Ge atom, including the As atom, thanks to the shared electrons, the outer electron shell is completed to a stable (eight electrons) shell.
The fifth outer electron of the As atom turns out to be “extra”.
It is much more weakly bound to the nucleus than other electrons, and it only
takes a small amount of energy to tear it away from the As atom. The As
atom will then become positively charged (ionized). Thus, when introducing
atoms of the fifth group into the germanium crystal lattice, positively charged
“fixed” impurity ions are formed at the lattice nodes and mobile electrons
(fig. 4.2). The conductivity of such doped semiconductor will be mainly electronic. In this case, the crystal is called an n-type semiconductor (from the
word “negative”), and the impurity is called an n-type or donor impurity.
Fig. 4.2. Crystal lattice of n-type semiconductor with donor impurity
If we introduce an atom from the third group of the periodic table,
for example aluminum (Al), into the germanium lattice, it will lack one
electron to form shared electron pairs with four neighboring Ge atoms.
A vacancy arises in the system of covalent bonds – a hole carrying a positive charge (fig. 4.3). Such impurity is called a p-type impurity (from the
word “positive”) or an acceptor (receiving) impurity, and the crystal is
called a p-type semiconductor.
60
Fig. 4.3. Crystal lattice of p-type semiconductor with acceptor impurity
One of the electrons of the neighboring germanium atoms can fill
nearest hole, then the hole will move to its place. The process of electron
movement through holes can be formally considered as the movement of the
holes themselves. In a p-type semiconductor, the main current carriers are
holes.
4.2. E l e c t r i c c u r r e n t i n e l e c t r o l y t e s
Electrolytes are mainly aqueous solutions of salts, acids and alkalis.
When electrolytes dissolve under the influence of the electric field of water
molecules, the electrolyte molecules disintegrate into positively and negatively charged ions. This process is called electrolytic dissociation.
Crystal lattices of solids melt at high temperatures and disintegrate
into ions—this is how dissociation occurs during melting.
Finally, there are ionic liquids, which consist only of ions, the melting
point of which is lower than the boiling point of water, i. e. lower than 100 °C.
In particular, there are salts that are liquid at room temperature, they are
called “RTIL“ or “Room-Temperature Ionic Liquids“.
Electrolytes, unlike metals and semiconductors, have ionic conductivity. If an electric field is applied to an electrolyte, the free electric charges
(ions of both signs) present in it will create a current.
Unlike electrons in metals, ions do not have such characteristics as
“free path length” or “free path time”. The movement of ions is a continuous
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