mse235f_2004_exam - MSE 235 320 Final Exam December 7 2004...

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Unformatted text preview: MSE 235/ 320 Final Exam December 7, 2004 15 Pages total MSE 235/320: Materials Physics Prof. Jun Nogami [:1 1 am in MSE 235 1:] 1 am in MSE 320 Print your name, Last name first: Student #: This exam is out of a total of 100 points. Read through the exam in its entirety before starting, and feel free to do the easier parts first. For all numerical problems, assume three significant figures for all numbers given, and give all answers to this accuracy. Remember to write down the units for each answer. Question Score Part1 (11) Part 11 (14) #1 (15) #2 (15) #3 (15) #4 (15) #5 (15) Total (100) MSE 235 / 320 Final Exam Part I: short answers December 7, 2004 1 point each 11 points total 15 Pages total Fill in the blanks with a two or three word description of the four features of this MH curve for a ferromagnetic material: Short answer: Orientational polarization is analogous to what form of magnetism? Electromagnetic radiation of wavelength 300 nm is called In a degenerate n-type semiconductor, the Fermi level is Above the Curie Temperature, a ferromagnetic material is In an X Ray spectrum, the LB line corresponds to an energy transition where the electron goes from a state with n: The diamond structure has to n: atoms in the unit cell. MSE 235 / 320 Final Exam December 7, 2004 15 Pages total True / False 1 point each 14 points total The dielectric constant of water is dominated by orientational polarization. The electron drift mobility in an n—type semiconductor decreases with increasing dopant density. A material with a band gap of greater than 1.77 eV will be optically transparent. The electron drift mobility in Au will decrease if it is alloyed with a small amount of Cu. For an n-type semiconductor, the number of electrons in the conduction band depends on the temperature in a range near room temperature. Minimizing thickness will decrease the voltage at which a capacitor will breakdown. It is desirable for a soft magnetic material to have large coercivity. A fenimagnetic material can be used to make a permanent magnet. In a transformer core, you should use a soft magnetic material A diamagnetic material has a small positive permeability. The relative permeability of a magnetic material is always greater than or equal to one. The relative permittivity of a dielectric material is always greater than or equal to one. The index of refraction is greater than or equal to the square root of the low frequency dielectric constant. DDDDDDDDDDDDDDH DDDDDDDUDDDDDD. A GaAs LED will emit orange light. MSE 235 / 320 F inal Exam 1a) lb) December 7, 2004 15 Pages total A sample of p—type GaAs has the Fermi level at 0.2 eV above the valence band maximum at 300K. What is the doping level? [6 pts] What are the electron and hole concentrations for this sample? [4 pts] MSE 235 / 320 Final Exam 1c) December 7, 2004 15 Pages total Assuming that you want to make a 1000 Ohm cylindrical resistor out of this GaAs, with a diameter of 1 mm, what is the length of this resistor? [5 pts] MSE 235/ 320 Final Exam December 7, 2004 15 Pages total 2) We want to design a parallel plate capacitor with C=0.1 nF, using a polymer dielectric PET Properties of PET s, (60 Hz) Dielectric strength (60 Hz) 2a) If we choose a thickness of 5 pm for the dielectric, what is the area of the capacitor? [2 pts] 2b) What is the maximum operating voltage for this capacitor? [2 pts] MSE 235 / 320 Final Exam December 7, 2004 15 Pages total 2c) If we want to use this capacitor at 500 Volts, what is the area of dielectric necessary to get C=0.1 nF, using the minimum safe thickness of the dielectric at this higher voltage? [5 PtS] 2d) Using the values from 2c), what is the power dissipated by the capacitor if it is run at 110V at 60 Hz? [6 pts] MSE 235/ 320 Final Exam December 7, 2004 15 Pages total 3 We want to design pick the thickness of a diamond coating for a Silicon surface that will make it as reflective as possible at a wavelength of 600 nm. Assume that the refractive index of Si is 3.5, and that for diamond is 2.39 3a) What percentage of incident light intensity is reflected from the uncoated Si surface? [5 pts} 3b) What is the minimum thickness diamond film that will maximize reflectivity? [10 pts] MSE 235 / 320 Final Exam December 7, 2004 15 Pages total 4 Gd is a ferromagnetic element, with properties given in the table below. T 4a) Estimate the energy loss per unit volume in one circuit of the B-H loop, assuming that it is roughly rectangular in shape [5 pts] 4b) Calculate the saturation magnetization. [5 pts] MSE 235 / 320 Final Exam December 7, 2004 15 Pages total 40) How many Bohr magnetons are contributed to the saturation magnetization by each Gd atom? [5 pt] -10. MSE 235/ 320 Final Exam December 7, 2004 15 Pages total 5) Consider the electron states in an infinite square well potential. Suppose we want to use these energy states in a device that will absorb light of a certain frequency. 5a) What is the energy of photons corresponding to a wavelength of 7L = 450 nm? [5 pts] Please state the answer in units of eV. 5b) Suppose that this photon energy corresponds to the energy spacing between the lowest two energy states of a one dimensional infinite square well. What is the width of this well? [10 pts] -11- MSE 235/ 320 Final Exam December 7, 2004 15 Pages total Useful Constants and Equations: Avogadro’s number N A = 6.023 ><1023 atomflol Planck’s constant h = 6.63 ><10'34 J - s = 4.13 ><10~15 eV r 5 Electron Volt 1 eV =1.602 ><10'19 J Boltzmann’s constant k = 1.38 ><10'23 14mm _ K = 8.62 ><10‘5 “761mm, K Electron Mass m = 9.11x10'3‘ kg Electron Charge one electron = 1.602 ><10‘19 Coulomb J Ideal gas constant R —- kNA — 8.3145 mol K Permittivity of vacuum 80 = 8.8542x10'12 % Permeability of vacuum yo = 471: X 10'7 1%” speed of light c = 3><103 '% Bohr magneton [3 = 113 = 9.2732 ><10’24 Am2 Equipartition of energy: 1 -2-kT per degree of freedom for a monatomic ideal gas: average kinetic energy per atom (KE) = 92-kT internal energy per mole U = a- RT d 3 J molar heat capacity Cm = i = -— R units [ dT 2 mol K Electrical properties: -1 R_ L c,_ P —* a - P A - enfle Electromagnetic Radiation: 2 c = iv a) = 275v c = 3 x 108 m/s wavenumber k = 7:! Photon energy E = hv 2 ha) Matter Waves [1 'h de Broglie: A = — = ———-— Where E is kinetic energy p 2mE .12. MSE 235/320 Final Exam December 7, 2004 15 Pages total Schroedinger’s Wave Equation 2 i V? + W = E‘I’ 2m +51 for non-time dependent potentials, we can write ‘I’(F,t) = 1110') X w(t) Where w(t) = e h and 1/107 ) satisfies the Time independent Schroedinger’s Equation: 2 —h ——V%+Vw=Ew 2m . 412 dzy/ In 1D: 5-”: dxz +Vlll= Elli -> d Momentum Operator: [3 = —th in 1D, [’5 = ——jh2—£ 7’1sz for a free electron E versus k = 3—— m 1-D Infinite Square Well of width a, O < x < a yr" = A" sin“; ) n = 1, 2, 3, 4, Energies 1227? 2 122 2 E” — Zma2 n . 8ma2 n Probability Density: P = M2 = ‘P*‘I’ 2 Normalization: dV = 1 Heisenberg’s Uncertainty Principle: Ap Ax 2 4L7: ~Z2me4 1 1 One electron atom: En = 880th = (—13.6 eV) Z2(717) . . . . . 1 Fermi-Dirac Distribution F(E) = W 3 312 Density of States for a 3D Metal Z( E) dE = flag-fl— JE dE . h2 3n % Fermi Energy for a metal. E F = -13. MSE 235/320 Final Exam December 7, 2004 15 Pages total Semiconductors: Selected semiconductor pro erties at T=300K -- 25 - NV . m. i 25 w 1.04x1025 6.6x102" 2.4x1019 -19 O- : enuu’e + epuh n-type a: 3I2 E —E n=Ncexp|:-‘-———( CkT 10] N =2£2m£kTJ P'type * 3'2 p: NV eXE{_.(EF—EQ] Nv = kT n =n2=NN ex ~531- p i c v Capacitance C =% Parallel Plate Capacitor A C = 808, E dielectric . . . __ _12 F Perm1tt1v1ty of vacuum 80 — 8.8542 x 10 A" Polarizability P=N%$ 1 s, =1+—Nae 8 0 Power dissipated per unit volume Wvol : (27V)£2808r’ tan6 way 3 m .14. MSE 235/ 320 Final Exam December 7, 2004 15 Pages total Magnetism Permeability of vacuum #0 = 47: X 10'7 11%” Bohr magneton fi = ,uB = 9.2732 X 10—7“4 % In vacuum I; = ital-l In a medium E =u0urfi= #0031 + M)=uo(l+ zmfil Optical Properties index of refraction n = J; 2 speed of light in a medium v= . . A. wavelength 1n a medium A = J- n reflection and transmission at normal incidence: in going from 111 to n2 2 r=n1—”2 R H2=(ni"n2) T: 4nan R+T=1 "1+n2 .15- ...
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This note was uploaded on 04/28/2008 for the course MSE 235 taught by Professor Nogami during the Fall '04 term at University of Toronto.

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mse235f_2004_exam - MSE 235 320 Final Exam December 7 2004...

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