Raffles Institution (J C)
JC2 H2 Physics 2009
Physics Tutorial : 20
LASERS AND SEMICONDUCTORS
Self-Check Questions :
Part A : LASERS.
S1. What does LASER stand for ?
Light Amplification by Stimulated Emission of Radiation.
S2. List the characteristics of laser light.
Laser light is highly coherent, unidirectional (or collimated) and monochromatic.
S3. Explain the difference between spontaneous emission and stimulated
emission of radiation from atoms ?
In spontaneous emission, a photon is emitted by an atom randomly and in any direction,
without any external stimulation, whereas in the case of stimulated emission, an
incoming photon, whose energy is exactly equal to the excitation energy of the atom,
induces the excited atom to fall to a lower energy level and releases a photon in the
process. This photon released is similar to the one which induces its emission. The two
photons are emitted at the same time and in the same direction.
S3. What is meant by a ‘population Inversion ‘ of atoms ?
Population inversion is a situation in which for a given sample of atoms of the same
element, there are more atoms in the excited state than in the ground state.
S4. Why is a ‘population Inversion ‘ necessary in the construction of a laser ?
‘Population inversion’ of atoms in the metastable state is necessary in order that the
photons produced by the first stimulated emission events can further trigger more
excited atoms to produce identical photons. In this way, a chain reaction of stimulated
emissions can occur to build up a laser beam of a sufficiently high intensity.
S5. Explain the use of the two mirrors placed at the ends of a laser.
The mirrors at the ends of the laser face each other and are parallel. They reflect the
photons produced by stimulated emission back into the lasing medium and along the
axis of the laser cavity so that more stimulated emissions can occur to produce the laser
beam. One mirror is totally reflecting while the other is partially reflecting in order to allow
the laser beam to emerge.
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Part B : SEMICONDUCTORS
S6. Name two examples of Intrinsic Semiconductors.
Silicon and germanium
S7. What is meant by ‘doping’ of a semiconductor ? Hence how may
( I ) a p – type semiconductor be obtained ?
( 2 ) an n – type semiconductor be obtained ?
Give one example for each of the doped semiconductors.
Doping is a process in which a small number of atoms of an element of valency 3 or 5
are added into the lattice of silicon (or germanium) with the purpose of increasing the
number density of free charge carriers which will ultimately increase the electrical
conductivity of the semiconductor. Example : An impurity concentration of 1 part in 108
increases the conductivity of germanium by a factor of 12 at room temperature.
A p-type semiconductor would be formed if silicon is doped with aluminium (valency 3)
and an n-type semiconductor would be obtained if silicon is doped with phosphorous
(valency 5). In the p-type material, the number density of holes is significantly increased,
thus making holes the majority carriers. In the n-type material, the number density of
electrons is increased and these are the majority carriers.
One way of understanding how hole number density outnumbers electron number
density (as is the case for p-type material) is that since aluminium atoms have only 3
valence electrons to form covalent bonds with 4 neighbouring silicon atoms, there will be
a missing electron in one of the 4 covalent bonds. This missing electron is a hole which
can accept an electron from another covalent bond. Thus compared with pure silicon,
there are now more holes in the lattice. Another way of understanding this effect is by
using the band theory --- see S9.
S8. Compare the electrical conductivities of metals and semiconductors using
the ‘Band-Theory’ of energy levels.
In a metal, the highest occupied energy band is not full. Hence, electrons in this band
can easily gain just a little energy to occupy other empty levels within the same band. In
other words, the electrons can easily move through the lattice when a potential
difference is applied. However, in a semiconductor such as silicon, the highest occupied
band (called the valence band) is completely filled at low temperatures. Thus electrons
within the same band cannot move. The next available band, called the conduction band
has plenty of empty levels and is separated from the VB by an energy gap of only1.2 eV.
Electrons at the top of the VB can easily gain 1.2 eV of energy to cross the gap over to
the CB. In doing so, an equal number of holes will be created in the VB. Thus both
electrons in the CB and holes in the VB contribute to current flow if a potential difference
is applied across the semiconductor. The conductivity of a semiconductor therefore
depends very much on the number density of charge carriers (includes both holes and
electrons) and this number naturally increases with temperature. As the number density
of charge carriers (electrons) in a metal is very high (~1029 m—3 for copper) compared
with that in an intrinsic semiconductor (~1016 m—3 for silicon ), the electrical conductivity
of metals is much higher than that of semiconductors.
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S9. In terms of ‘Band Theory’, why is the electrical conductivity of a doped
semiconductor much higher than that of a pure semiconductor ?
In the case of n-type silicon, the presence of dopant atoms introduces donor states
within the energy gap lying very close to the bottom of the conduction band. Electrons
from the donor states easily gain the small amount of energy required to go into the CB.
(example: Ed is 0.05 eV below the CB for silicon doped with phosphorous). At room
temperature, typically all the electrons from the donor states are in the CB. Thus the
number density of charge carriers is significantly increased compared with that of the
intrinsic semiconductor.
Similarly in the case of p-type semiconductor, the presence of dopant atoms introduces
acceptor states into the energy gap and they lie very close just above the top of the
valence band. Electrons from the top of the VB readily gain the small amount of energy
required to go into the acceptor states, thus creating holes in the VB. As a result, the
number density of charge carriers in the p-type material increases and this explains why
the conductivity increases.
S10. What is meant by the ‘depletion layer (or zone) ’ in a p-n junction ? Explain
its formation.
When a p-type semiconductor is jammed against an n-type semiconductor forming a
plane junction, electrons from the n-side close to the junction will diffuse over to the pside because there are few electrons there. In so doing, positive charge accumulates on
the n-side. Similarly, holes from the p-side close to the junction will also diffuse to the nside, causing negative charge to appear on the p-side. The electrons diffusing from the
n-side can recombine with holes on the p-side and results in a loss of free electrons and
free holes (the charge carriers in both types of materials) over a certain width at the p-n
junction. The appearance of positive charge on the n-side and negative charge on the pside produces an electric field directed from the n-side to p-side and we can associate
this electric field with a potential difference across the junction. The diffusion of charge
carriers across the junction will eventually stop when the potential difference is high
enough to prevent further movement of charge carriers across the junction. The region
over the p-n junction in which there are no free charge carriers is called the depletion
layer (or zone).
S11. Why does a p-n junction conduct electricity predominantly in only one
direction ?
In forward bias, since the p-side is made more positive than the n-side, the height of the
contact potential V0 is reduced and hence more of the majority charge-carriers can now
surmount this smaller barrier. Thus the diffusion current increases markedly and this
explains why current flows from the p-side to the n-side during forward bias. In reverse
bias, the p-side is made negative with respect to the n-side. This causes the contact
potential to increase and no electrons can move from the n-side to the p-side. In other
words the majority charge carriers cannot flow through the junction. The minority charge
carriers can flow across the junction, but because their number is very small, the
resulting current is negligible. Thus the p-n junction only conducts current predominantly
in one direction, that is; from the p-side to the n-side during forward bias.
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S12. Sketch the Current – Voltage Characteristics of a typical diode.
I/mA
4
3
2
1
-1
Reverse-bias -
0
1
Forward-bias +
2
V/V
Discussion Questions :
Part A : LASERS
[Data given : Speed of light = 3.0 x 108 m s—1 , Planck constant = 6.6 x 10—34 Js]
1. A pulsed laser emits light at a wavelength of 694.4 nm. The pulse duration is 12.0 ps,
and the energy per pulse is 0.150 J.
(a) What is the length of the pulse?
(b) How many photons are emitted in each pulse?
Solution :
(a)
(b)
Length of pulse = (12.0 x 10—12 ) x 3.0 x 108 = 3.60 mm
number of photons = 0.150 x 694.4 x 10—9 / (6.6 x 10—34 x 3.0 x 108 )
= 5.26 x 1017
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2.
In a ruby laser, light of wavelength 550 nm from a xenon flash lamp is used to excite
the chromium, Cr3+ ions in the ruby from ground state E1 to state E3. In subsequent
de-excitations, laser light is emitted. Which of the following statements regarding this
laser is incorrect?
2.25 eV
E3
1.79 eV
E2
Optical pumping
(550 nm photons)
Ground state
E1
A
E3 cannot be the metastable state because, if it is, then there will be no net
production of light when equilibrium is reached, since stimulated absorption and
stimulated emission will then occur at the same rate because the number of
electrons in E3 and E1 will be the same at steady state.
B
E3 is the metastable state because, having a longer lifetime than a normal
excited state, the metastable state allows the accumulation of excited electrons,
resulting in population inversion and net light production.
C
E2 is the metastable state because it is not subjected to stimulated emission
caused by the 550 nm photons used in optical pumping, and so allows the
accumulation of excited electrons to achieve population inversion.
D
The transition from state E2 to E1 produces the laser light.
2. B -- Option A explains the answer. (You can also refer to the discussion on two-level,
three-level and four-level laser systems found at the end of the Appendix of the lecture
notes for a clearer understanding).
3.
Which of the following statements concerning a laser system is false ?
A
An external energy source is needed to create population inversion.
B
The laser beam produced is coherent and of a single wavelength.
C
By changing the reflective coefficient of the partially reflecting mirror, the
intensity of the laser beam can be varied.
D
High-end laser systems can produce perfectly collimated beams, i.e. beams that
would not spread.
3. D -- Diffraction will cause a spreading of the light beam.
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Part B : SEMICONDUCTORS
4.
The diagram shows the two highest energy bands of 3 materials.
(II)
(I)
(III)
Which of the following is a possible material for each set of energy bands shown ?
5.
(I)
(II)
(III)
A
Germanium
Plastic
Silver
B
Tungsten
Gold
Silicon
C
Wood
Silicon
Copper
D
Nickel
Germanium
Water
Which of the following statements is correct ?
A
An n-type semiconductor has a net negative charge.
B
A p-type semiconductor conducts through holes only.
C
A net current always flows across the p-n junction due to the junction electric
field.
D
The junction electric field always points from the n-type to the p-type regions
across the depletion region.
5. D – The n-side is positive and the p-side is negative.
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6. When a photon enters the depletion zone of a p-n junction, the photon can scatter from
the valence electrons there, transferring part of its energy to each electron, which then
jumps to the conduction band. Thus, the photon creates electron-hole pairs. For this
reason, the junctions are often used as radiation detectors, especially in the x-ray and
gamma - ray regions of the electromagnetic spectrum. Suppose a single gamma - ray
photon of frequency 1.6 x 1020 Hz transfers its energy to electrons in multiple
scattering events inside a semiconductor with an energy gap of 1.1 eV, until all the
energy is transferred.
Assuming that each electron jumps the gap from the top of the valence band to the
bottom of the conduction band, find the number of electron-hole pairs created by the
process.
Solution :
The number of electron-hole pairs = energy of gamma ray photon/energy gap
= (6.6 x 10—34 )(1.6 x 1020 )/(1.1 x 1.6 x 10—19 )
= 6.0 x 105
7. A potassium chloride crystal has an energy band gap of 7.6 eV above the topmost
occupied band, which is full. Is this crystal opaque or transparent to (a) light of
wavelength 140 nm ? and (b) violet light ?
Solution :
(a)
For light of wavelength 140 nm, the photon energy =
(3.0 x 108 x 6.6 x 10—34 )/(140 x 10—9 x 1.6 x10—19 ) = 8.84 eV
Since this is higher than the band gap energy of 7.6 eV, the photons can be absorbed by
the material ( that is, electrons from the VB can absorb the photons and get promoted into
the CB). Thus the material is opaque to light of wavelength 140 nm.
(b) For violet light, a typical value of its wavelength is 400 nm.
The photon energy is = (3.0 x108 x 6.6 x10—34 )/(400 x 10—9 x 1.6 x10—19 )
= 3.09 eV
Since this is less than the band gap energy, the light will not be absorbed. The material is
thus transparent to violet light.
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