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This is the Exam of Intermediate Physics which includes Width of Central Maximum, Intensity Distribution, Horizontal Position Axis, Orders of Diffraction, Fabry-Perot Interferometer, Beam of Light, Transmission Directions etc. Key important points are: Wave Packet Model, Newtonian Particle, Wave-Particle Paradox, Uncertainty Principle, Seeing Ability, Simple Photoconductor, Electromagnetic Radiation of Frequency, Light of Wavelength
Typology: Exams
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This paper consists of 2 sections.
Section A Quantum Physics 75 marks Section B Electromagnetic Properties of Matter 75 marks
Candidates should attempt all questions.
USE A SEPARATE ANSWER BOOK FOR EACH SECTION.
In answering the questions in this paper, it is particularly important to give rea- sons for your answer. Only partial marks will be awarded for correct answers with inadequate reasons.
No written material of any kind may be taken into the examination room. Calcu- lators are permitted.
Table of constants
Avogadro’s number N (^) A = 6. 022 × 1023 mole−^1
speed of light c = 2. 998 × 108 m.s −^1
electronic charge e = 1. 602 × 10 −^19 C
electron rest mass m (^) e = 9. 110 × 10 −^31 kg
electron rest energy energy = 511 keV
electron volt 1 eV = 1. 602 × 10 −^19 J
protron rest mass m (^) p = 1. 673 × 10 −^27 kg
neutron rest mass m (^) n = 1. 675 × 10 −^27 kg
Planck’s constant h = 6. 626 × 10 −^34 J.s
Planck’s constant (reduced) ¯h = 1. 055 × 10 −^34 J.s
Boltzmann’s constant k (^) B = 1. 380 × 10 −^23 J.K −^1
Stefan’s constant σ = 5. 670 × 10 −^8 W.m−^2 .K −^4
Coulomb constant (^4) π"^10 = 8. 988 × 109 N.m^2 .C −^2
permittivity of free space " 0 = 8. 854 × 10 −^12 C 2 .N−^1 .m−^2
permeability of free space μ 0 = 4 π × 10 −^7 kg.m.C −^2
gravitational constant G = 6. 673 × 10 −^11 N.m^2 .kg −^2
atomic mass constant u = 1. 660 × 10 −^27 kg
(ii) These two results used to be known as the “wave-particle para- dox”. Explain briefly what is paradoxical about these results.
(b) (i) Describe briefly how using a wave packet as a model by which to describe an electron, “solves” the wave-particle paradox.
(ii) Describe briefly how the wave packet model leads, as a conse- quence, to the Uncertainty Principle.
(c) The “seeing ability”, or resolution, of any optical instrument is roughly the same as the wavelength being used. If the size of an atom is of the order of 0.1 nm, how fast must an electron travel to have a wavelength small enough to “see” the atom?
(15 marks)
(i) Explain, with the aid of an appropriate energy band diagram, why its resistance changes. (Be careful to state whether its resistance increases or decreases when it is illuminated.)
(ii) Explain why such a simple photoconductor will not conduct if it is illuminated by electromagnetic radiation of frequency much less than visible light.
(iii) 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 light of wavelength 140 nm? Explain your reasoning.
(b) In the kind of photovoltaic device needed to convert solar radiation to useable energy, it is necessary to slow down recombination of the electron-hole pairs created by incident photons. This cannot be done simply with a simple piece of semiconductor, but usually requires the device to be made from a p-n junction.
(i) Draw an energy band diagram for a p-n junction under conditions of forward bias.
(ii) Explain briefly, with reference to this diagram, how this photo- voltaic device slows down recombination of electron-hole pairs.
(15 marks)
(ii) Draw an energy band diagram for a simple intrinsic semiconduc- tor. Mark clearly on that diagram what quantum mechanical fea- ture distinguishes it from a conductor.
(b) Consider now a p-type semiconductor.
(i) By drawing another careful diagram, illustrate schematically the energy level structure for a p-type semiconductor. On this diagram show which energy levels are full at absolute zero temperature, and which are empty. Also on this diagram, indicate approximately where the Fermi en- ergy should be, and explain why you drew it where you did.
(ii) Explain in words, with reference to your diagram, why this semi- conductor will conduct electricity at room temperature.
(iii) Since the material is, in general, electrically neutral, explain why there are not the same number of negative charge carriers as there are positive charge carriers.
(15 marks)
potential energy
!
distance (x)!
0
!"#!!!$%
+U 0
The height of the well (U 0 ) is extremely large, so that the box may be considered impenetrable.
(i) Calculate, to four significant figures, the energies of the ground state and the first excited state for this system.
(ii) Draw, on a copy of the potential energy graph above, an energy level diagram for the electron showing the energies of the ground state and the first excited state.
(b) Another electron is contained within a similar box, but this time there is a potential hill inside the box’s walls. The height of this hill is of order ∼ 100 eV and its width of order ∼ 0. 03 nm.
potential energy
!
distance (x)!
0
+U 0
Draw careful diagrams to show what you would expect the probability density functions for the ground state and the first excited state to look like.
(c) On a copy of the previous energy level diagram, mark in where you would expect the new energies of the ground state and the first excited state to be located.
(d) At least one of these two energy levels should be relatively close to the corresponding level from part (a). Explain briefly which state it is, and why its energy is so close.
(15 marks)
Ex = −
∂x
Ey = −
∂y
Ez = −
∂z
F (^) E = qE E =
4 π" (^0)
q r 2
ˆr Φ (^) E =
E · dA =
q (^) enclosed " (^0)
V =
4 π" (^0)
p · ˆr r 2
D · dA = q (^) f W =
D · Edr
E = σ/" 0 p = qd τ = p × E U = −p · E
V =
4 π" (^0)
q r
" 0 " (^) r A d
U =
CV 2 P = np σb = P · ˆn
D = " 0 " (^) r E = "E D = σf D = " 0 E + P
P = χ (^) e " 0 E p = αE α =
n
" (^) r − 1 " (^) r + 2
P = np (coth(pE/kT ) − 1 /(pE/kT )) i =
dq dt
i = nqv (^) drif t A VH =
Bi net
nq
J = nqv (^) drif t R = V /I ρ = E/J
σ = 1/ρ R = ρL/A ρ =
m e^2 nτ
Φ (^) B =
B · dA F (^) B = qv × B μ = N iAnˆ
ωc = qB/m r (^) L =
mv (^) ⊥ qB
B = μ 0 (H + M)
τ (^) B = μ × B dF (^) B = idL × B B = μH W = −μ · B
M = χH W = V
HdB B = μ (^) r μ 0 H
μ (^0)
E × B dB =
μ (^0) 4 π
ids × r r 3
p (^) rad = I/c or p (^) rad = 2I/c
(a) We found in class that the electric field of a dipole decreases as 1 /r 3 at large distances. Explain why the field decays more rapidly for a dipole than for a bare charge.
(b) Why are dipoles important when considering the properties of dielec- tric media?
(c) In class we derived the result σb = P · nˆ. Explain the symbols and briefly discuss the physical interpretation of this result.
(d) What is meant by “drift velocity” in the context of conduction.
(e) Explain the difference between paramagnetic, ferromagnetic, and dia- magnetic media.
(f) Briefly explain why a permanent magnet can stick motionlessly to the steel front of a refrigerator?
(30 marks)
(a) Briefly discuss this motion. You may assume that there is no electric field.
(b) Briefly discuss what is meant by a magnetic mirror.
(c) Shown is a schematic of a tokamak fusion reactor. Explain how this design manages to confine the charged particles in order to prevent them from touching the reactor walls.
(15 marks)
(a) Sketch the path followed by a negative charged particle in these fields. Be sure to indicate the direction in which the particle moves by an arrow. Briefly justify your answer (in words).