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Unformatted text preview: Physics 207 Final Exam, Spring 2010
Problem 1 (25 points) Photons with wavelength 411 can illuminate a piece of lithium metal, causing electrons to be ejected ﬁom
the lithium with a maximum kinetic energy of 0.725 eV. An electric potential V is applied to stop the
ejected electrons Recall that the potential energy of a charge q travelling across an electric potential:1 gs l/i
V. "
q (a) What is the energy of each photon? (4 points) 3/
(b) What is the wavelength of the maximum energy ejected electron? Hint: in this case electrons
behave like photons. (4 points)
(c) What is the stopping electric potential Vsmp required to stop the most energetic ejected electrons
when lithium is illuminated by 411 nm wavelength light? (4 points)
(d) What is the work function (the minimum energy required to eject an electron) for lithium? (6
points)
(e) What is the longest wavelength photon that can still eject electrons from lithium? (3 points)
(1) The energy of incident photons is now varied from 0 to 5 eV. Make a rough sketch/plot of the
magnitude of the stopping potential Vmp as a ﬁmction of incident photon energy. (4 points) Ox) E), , lA‘p’: iq_ C 1; (Q pL37‘la 3135) (37408”18) L‘a'fSwao‘Mlj
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(a) photon energy = _ (4)
(b) electron wavelength = (4)
(c) Vstop = W (4)
((1) work function = (6)
(e) longest wavelength photon = (3)
(1) Sketch [vsmgml vs photon energy I (4) Dan: ’7 Id? In Physics 207 Final Exam, Spring 2010
Problem 2 (25 points) Electrons are trapped in a two—dimensional inﬁnitely deep well with dimensions Lx = 43 51:10"12 In and Ly
= 3Lx. Assume that the electrons do not interact but do not neglect spin nor Pauli exclusion.
(a) Determine the four lowest energy levels for this trap. (8 points)
([3) The state nx=l and ny"—'6 is degenerate with (has the same energy as) another state. Determine the
quantum numbers for this other degenerate state. (4 points)
(c) Five electrons are added to this trap. Sketch how the electrons occupy the energy levels in the
ground state (IOWest energy state). (5 points)
((1) Determine the energies of the ground and ﬁrst excited states for the system of ﬁve electrons in the trap. (Spoints) a) E at?
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(b) :1," = ______________________ ny’ = _________________________________________ (4)
(c) Sketch electron distribution for ground state (5)
(d) Egraundfor 5 e—= E13! exclmifor 5 e t (8) Doom 1 AF 10 Physics 207 Final Exam, Spring 2010
Problem 3 (25 points) (Ge) is a semiconductor with an energy gap of 0.660 eV. The density of Ge is 5.32 g/cm3 and
its Geatimigiiiass is 72.6 g/mol. Consider a piece of Ge at room temperature (300K). You may assume that
the Fermi energy is exactly in the middle of the energy gap. (a) What is the probability that a state at the bottom of the conduction band (CB) is occupied? (4
point) (b) What is the probability that a state at the top of the valence band (VB) is vacant (UNoccupied)?
(4 Point) (c) The Ge is doped with arsenic so that one of every 106 Ge atoms is replaced by an arsenic (As)
atom. Each arsenic atom contributes one free electron to the conduction band. What is the
concentration of free electrons in the conduction band of this doped Ge material? (6 points) (d) Ifthe Fermi energy of the arsenic—doped Ge material described in part (c) is 0.0127 eV below the
bottom of the conduction band, what is the probability that a state at the bottom of the conduction
band is occupied at room temperature? (6 points) (e) Using parts (c) and (d), estimate the density of states (number of available states per unit volume)
at the bottom of the conduction band. Note that not all these states need to be occupied by electrons. Hint: to estimate how many people are in a hotel you need the probability that each
room is occupied as well as the number of rooms. (5 points) ' a) CB ﬂ 6) , __ l __
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(a) probability that state at bottom of CB is occupied in Ge = (4)
(b) probability that state at top of VB is UNoccupied in Ge = (4)
(c) concentration of free electrons in CB in doped Ge = (6)
(d) probability that state at bottom of CB is occupied in doped Ge = (6)
(e) density of states at bottom of conduction band = (5) TL...“ A ‘1‘1ﬂ Physics 207 Final Exam, Spring 2010
Short Problem 81 (15 points) A guitar E string with length of 0.650 m giggles at a fundamental frequency of 330 Hz. One end of the string is ﬁxed at x=0 m and the other at x=1e§§ m. The amplitude of the string oscillation is 1.45 mm. (a) What is the wavelength of this note in air? (4 points) (b) Sketch the displacement patterns and calculate the frequencies of the three lowest frequency modes
for the E suing. (5 points) (c) Determine the displacement D(x,t) of the string in the fundamental mode as a function of x and t. (6 points) , : ’U , esaw
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(b) sketch displacement patterns above, f1 = ﬂ; = f3 = __________ (5)
(c) D{x,t) * (5) Physics 207 Final Exam, Spring 2010
Short Problem 82 (15 points) A supersonic airplane travelling horizontally at Mach 1.75 passes directly over an observer at an altitude of 1200 m. The airplane produces a shock wave as it travels. (a) What is the angle that the shock wave’s leading edge makes with respect to the horizontal? (4 points) (1)) What is the horizontal distance that the airplane will have traveled away from the observer when the
shock wave hits the observer? (6 points) (c) How much time has elapsed between the moment the airplane is directly above the observer and the arrival of the shock wave at the observer’s position? (5 points) Jr
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(b) horizontal distance = (6)
(c) time =W(5) nanA K ”4' 1n Physics 207 Final Exam, Spring 2010 Short Problem S3 (15 points) Linearly polarized light ﬁom a laser produces a collimated laser beam (non—diverging with circular cross section) with an intensity of 777 W/m2 is incident on a series of two polarizers. The laser beam travels horizontally and is initially polarized with its electric ﬁeld E0 oscillating in the vertical direction. The polarization axis of the ﬁrst polarizer P1 is tilted 12° to the left of vertical while polarization axis of the second polarizer P2 is tilted 34° to the right of vertical. (:1) Calculate the magnitude (NOT rms value) of the electric ﬁeld E0 produced by laser beam before it
goes through P1. (4 points) (b) Calculate the intensity 11 and electric ﬁeld E produced by laser beam after it passes through P1. (5
points) (c) Calculate the intensity 12 and electric ﬁeld E2 produced by laser beam after it passes through P2. (6
points) ‘ Place your answers here
(a) 50 = (4)
(1)) E1 = ______—______ 11 = (5)
(e) Ez=wlz=m (6) Dan; ’7 “cm Physics 207 Final Exam, Spring 2010
Short Problem S4 (15 points) A source S is placed 5 .00 cm to the left of a spherical MIRROR with a 15.0 cm radius of curvature. (a) Determine the focal length (sign and magnitude) of the mirror and indicate the focal point (or focus)
on the ﬁgure below. (5 points)
(b) Draw two principal rays in the ﬁgure below to determine the location of the image. (5 points) (c) Calculate the image position algebraically. (5 points) Cs’vvifr 0M 5!}le cogent: ‘LO LVl/\l:(_,l/\ lls'hL jgfs “piPd eﬁfwvil‘WFV‘Mf
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(0) Calculated image position = (5) Th...“ 0 n31“ Physics 207 Final Exam, Spring 2010 Short Problem SS (15 points) Light with wavelength 633 nm passes through two slits which are 0.525 mm apart. The light passing
through the slits is projected on a screen that is 4.25 m away ﬁom the slits. You may assume that the slit
widths are negligibly small. The center of the screen is at y=0 m. (:1) Draw the intensity as a ﬁmction of position y on the screen. (5 points) (b) What is the position of the ﬁrst non—central maximum? (5 points) (c) What is the position of the ﬁrst noncentral minimum? (5 points) 7t=633 nm wan: War A 0,513 ﬁlo—3M wax
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(b) y(1St noncentral max) = (5) (c) ya“ noncentral min) = W (5) Dana ﬂ {.3111 Physics 207 Final Exam, Spring 2010
Short Problem 86 (15 points) Jupiter’s Great Red Spot (GRS) is due to a huge storm that is approximately 28,000 km wide. We would like to build a refracting telescope in order to resolve this spot, i.e., be able to distinguish the left and right edges of the spot. Jupiter can be as close as 629 million km away ﬁom the earth. Assume that the light coming from the spot is red with a wavelength of 600 nm. (a) Calculate the angular size of the GRS as seen from earth. This is just the angle from left edge to the
right edge of the GRS. (6 points) (b) What is the minimum diameter required for a telescope objective lens to be able to just resolve the GRS? Assume that the resolution is determined by diffraction eﬁ‘eets. (7 points) (c) Besides resolution, name one other advantage of using larger diameter telescope objectives. (2 points) {Ear—S «9 <9: 5, ”US, 000 km
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Place your answers here:
(a) angular size of GRS as seen from earth = (6)
(b) minimum telescope objective lens diameter = (7)
(c) Another advantage of large diameter telescope objective lenses? (2) DunD mnmn ...
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