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PHYSICS OF ENERGY AND THE ENVIRONMENT
HOME WORK 3
Released 06 February 2017
Deadline 4 pm Tuesday 14 February 2017
1
Draw an energy band diagram for each of an insulator, conductor and semiconductor and use
you diagrams to explain the differences them.
[10 marks]
For each of a conductor, insulator and/or semiconductor the valance band is full. However, for a
conductor, there additional electrons left after filling the valance band, and these can only go
into the conduction band, and hence are not bound to individual atoms and are free to be
conducted.
Neither an insulator or a semi-conductor have enough electrons to fill the valence band. Hence,
valence electrons in these materials need the input of energy to be excited into the conduction
band. The feature distinguishing between them is the amount of energy need to excite an
electron to the conduction band. For a semi-conductor, this is around 1 eV; whereas, for an
insulator it is considerably larger (~ 5 eV).
2
Explain why for an intrinsic semi-conductor the rate of electron-hole pair generation of
proportional to the product the square of the number of holes.
[10 marks]
An intrinsic semiconductor is a pure material. As such, the electron-hole pair generations are not
in are not independent of each other. They are each at the same time at the same place by the
same event. Therefore, every electron promoted to the conduction band leaves behind a hole in
the valence band. Hence, at any one time the number of ‘electrons’ and the number of ‘holes’ are
equal.
The rate of electron-hole pair generation is proportional to the number of ‘electrons’ present.
However, as electron and hole generation is not an independent process.
Hence, the rate of pair generation is also proportional to the number of holes present.
Rate in proportional to Nh and Ne, i.e., R  NhNe, and as Nh = Ne, R  Nh2.
2
With the aid appropriate diagrams, explain the mechanism of hole current generation in an
intrinsic semiconductor following the application of an external electric field.
[10 marks]
In an external electric field, once an electron-hole pair is generated (thermally of otherwise).
The liberated electron enters the conduction band, where it is accelerated by the external field
in a direction opposite to the direction of the field away from the generation site. This leaves
behind a co-called hole in the valence band. As the hole is associated with a specific atom it
cannot move. However, an electron in neighbouring atom in the direction of the field feels a
force in the opposite direction (i.e., in the direction of the hole). Furthermore, that electron also
feels an attractive electrostatic force towards the hole. These two factors combine to increase
the probability that the electron will move to fill the hole and that the most likely source will be
a the neighbouring atom in the direction of the field. Should this occur, the old hole is annihilate
and a new hole is created on the atom the electron came from. As this process repeats, over
and over again, it gives the illusion that the hole moves in the direction of the field, which may
be modelled as if there was an actual hole current.
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4
The vibrational motions of atoms within molecules occurs in the infra-red region of the
electromagnetic spectrum; and the rotational motion of molecules occur in the microwave
region. A certain molecule has a vibration that excited by radiation with a wavelength of 10.0
m and a rotation that is excited by radiation with a wavelength of 1.00 cm.
Determine percentage of molecules in the excited vibrational state if irradiated by 10 m
radiation at a temperature of (a) 30OC and (b) 300OC; and the percentage of molecules in the
excited rotational state at (c) 30OC.
[15 marks]
This question involves Boltzmann distributions amongst two states.
The general equation for the population differences is
N2/N1 = e-(E2-E1)/kT= e-(E)/kT
T = 30OC = 303 K
T = 300OC = 573 K
k = 1.38 x10-23
E = hc/
10 m => E = 19.9 x10-21 J
1 cm => E = 19.9 x10-24 J
For the vibrations at 30OC,
= e-[(19.9 exp(-2)]/[4.18 exp(-21)] = 0.0086 = 0.86 %
For the vibrations at 300OC,
= e-[(19.9 exp(-21)]/[7.91 exp(-21)] = 0.081 = 8.1 %
For the rotations at 30OC,
= e-[(19.9 exp(-24)]/[4.18 exp(-21)] = 0.995 = 99.5 %
6
The band gaps in intrinsic silicon and germanium are 1.14 eV and 0.67 eV, respectively.
Determine the number of thermally generated conduction in a 1.00 kg sample of (a) pure silicon
at both 10OC and 30OC and (b) pure germanium at 10OC and 30OC.
This question also involves Boltzmann distributions amongst two states.
The general equation for the population differences is
N2/N1 = e-(E2-E1)/kT= e-(E)/kT
k = 1.38 x10-23. Hence, kT10 = 3.909 x10-21 and kT30 = 4.186 x10-21
ESi = 1.14 eV = 1.826 x10-19 J
EGe = 0.67 eV = 1.07 x10-19 J
The fraction of Si atoms with electrons promoted to the conduction band at 10OC is
e-[(1.826 exp(-19)]/[3.909 exp(-21)] = 5.18 x10-21.
Hence, one mole of Si would have 5.18 x10-21 x 6.022 x1023 = 3,110 thermally generated
conduction electrons.
The molar mass of Si is 28.08 g/mol.
Hence, one kg of Si contains 1000/28.09 = 35.6 moles, there are 3110 x 35.6 = 111 thousand
conduction electrons in pure Si at 10OC.
The fraction of Si atoms with electrons promoted the conduction band at 10OC is
e-[(1.826 exp(-19)]/[4.186 exp(-21)] = 1.14 x10-19.
Hence, one mole of Si would have 1.14 x10-19 x 6.022 x1023 = 68,400 thermally generated
conduction electrons.
The molar mass of Si is 28.08 g/mol.
Hence, one kg of Si contains 1000/28.09 = 35.6 moles, there are 68,400 x 35.6 = 2.44 million
conduction electrons in pure Si at 30OC.
The fraction of Ge atoms with electrons promoted to the conduction band at 10OC is
e-[(1.07 exp(-19)]/[3.909 exp(-21)] = 1.300 x10-12.
Hence, one mole of Ge would have 1.300 x 10-12 x 6.022 x1023 = 780 x 109 thermally generated
conduction electrons.
The molar mass of Ge is 72.64 g/mol.
Hence, one kg of Ge contains 1000/72.62 = 13.77 moles, there are 780 x 109 x 13.77 = 10.7 x
1012 conduction electrons in pure Ge at 10OC.
The fraction of Ge atoms with electrons promoted to the conduction band at 30OC is
e-[(1.07 exp(-19)]/[4.186 exp(-21)] = 1.300 x10-12.
Hence, one mole of Ge would have 7.922 x 10-12 x 6.022 x1023 = 4.77 x 1012 thermally generated
conduction electrons.
The molar mass of Ge is 72.64 g/mol.
Hence, one kg of Ge contains 1000/72.62 = 13.77 moles, there are 780 x 109 x 13.77 = 66 x 1012
conduction electrons in pure Ge at 30OC.