Physics 208 Final Exam May 12, 2008

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1 Page 1 Name: Solutions Student ID: Section #: Physics 208 Final Exam May 12, 2008 Print your name and section clearly above. If you do not know your section number, write your TA s name. Your final answer must be placed in the box provided. You must show all your work to receive full credit. If you only provide your final answer (in the box), and do not show your work, you will receive very few points. Problems will be graded on reasoning and intermediate steps as well as on the final answer. Be sure to include units, and also the direction of vectors. You are allowed two double-sided 8½ x 11 sheets of handwritten notes and no other references. The exam lasts exactly 120 minutes. Coulomb constant: k e = 9.0 "10 9 N # m 2 /C 2 Speed of light in vacuum: c = 3 "10 8 m /s Permittivity of free space:" o =1/4#k e = 8.85 $10 %12 C 2 /N & m 2 Permeability of free space: µ o = 4" #10 $7 T % m / A Planck s constant: h = "10 #34 J $ s = "10 #15 ev $ s, h = h /2% Bohr radius: a o = 0.053nm Atomic mass unit: 1 u = "10 #27 kg = MeV /c 2 Electron mass: m e = 9.11"10 #31 kg = u = 0.51 MeV /c 2 Proton mass: m p = "10 #27 kg = u = MeV /c 2 Neutron mass: m n = "10 #27 kg = u = MeV /c 2 Fundamental charge: e =1.60 "10 #19 C hc =1240eV " nm 1eV =1.602 "10 #19 J Problem 1: / 20 Problem 2: / 25 Problem 3: / 25 Problem 4: / 20 Problem 5: / 25 Problem 6: / 15 TOTAL: / 130

2 Page 2 1) [20 pts, 4 pts each] Multiple choice. Show work/explanation for full credit. a) A parallel-plate capacitor is connected to a battery and charged. The battery is then disconnected, and the plates are pulled apart, doubling the separation. How does the stored energy U after compare to stored energy before pulling the plates apart? a. U after =2U before b. U after =U before /2 c. U after =4U before d. U after =U before /4 e. U after =U before Explanation/Work: The charge Q is constant during this process, meaning that the electric field in the capacitor stays the same. So the energy density is constant. But the volume increases by a factor of 2. The stored energy increases by a factor of 2. Or you can use U=Q 2 /2C. C decreases by factor of 2, so stored energy increases by factor of two. b) In the resistor-capacitor circuit below, the switch was opened after charging the 10µF capacitor with a charge of 2x10-5 C. The switch is then closed again. What is the current through the 1 kω resistor immediately after the switch is closed? (1kΩ=1000Ω and 1µF=10-6 F) a. 0 A b A c A d A e A f. 0.1A g. 0.8A 10V R=1kΩ Explanation/Work The voltage across the capacitor before the switch is closed is V=Q/C=2V. The drop across the resistor is then 10V-2V = 8V. The current through it is then 8V/1kΩ=8mA = 0.008A C=10µF c) A hydrogen atom is in the n=3 state. What is the longest wavelength photon that can be emitted by this atom? a. 103 nm b. 122 nm c. 413 nm d. 656 nm e nm Explanation/Work The atom must make a transition to a lower-energy quantum state to emit a photon. The longest wavelength will correspond to the smallest energy difference, which would be a neighboring state. This is the n=3 to n=2 transition. Its energy is "13.6eV 3 2 " "13.6eV 2 2 =1.889eV The wavelength is " = hc E = 1240eV # nm 1.889eV = 656 nm

3 Page 3 d) Heavy water is the molecule D 2 O. It has two atoms of 2 1 H and one atom of 16 8 O. Approximately how much heavier is this than normal water H 2 O? a. 1% b. 5% c. 10% d. 20% e. 40% Explanation/Work: D 2 O has = 20 nucleons H 2 O has = 18 nucleons Since all nucleons have about the same mass, this is (20-18)/18 = 11% heavier e) You use a thin lens to project a 35 x 49 image of a 5 x 7 photo on a wall 2 meters away. What focal length lens do you need? a. 0.2 m b m c. 0.5 m d. 1.0 m e. 2.0 m Explanation/Work: The magnification is 35 /5 = 7.0, or equivalently 49 /7 =7.0 The object distance must then be 2 m / 7=0.286 m The focal length must satisfy 1 i + 1 o = 1 f " 1 f = 7 2m + 1 2m + = 8 2m So f=0.25m

4 Page 4 2) [25 points, 5 pts each] Short-answer questions a) About 3 micro-grams (containing 7.5x10 15 nuclei) of 241 Am (half-life is 432 years) is used in smoke detectors, with the alarm coming on when the alpha particles emitted from its nucleus are blocked by smoke and do not reach the detector. How many alpha particles per second are emitted by the 3 micro-grams of 241 Am.? A half-life of 432 years is (432yrs)(365days/yr)(24hrs/day)(60min/hr)(60sec/min)=1.362x10 10 sec. The corresponding decay rate is " = ln2/# 1/ 2 = $10 %11 /s. The total decays per second is then "10 #11 /s ( )( 7.5 "10 15 ) = "10 5 decays/s # particles/s = b) A solid insulating sphere of radius R=0.01m has a uniform volume charge density of ρ=0.1 C/m 3. Calculate the force F on a q=+0.1µc point particle d=0.02m from the surface of the sphere. Neglect any inductance phenomenon on the sphere due to the charged particle. (1µC=10-6 C). By Gauss law, the insulating sphere acts as a point charge located at the center of the sphere. It has charge q=0.1µc y Q = "V = " 4 3 #R3 = 0.1C /m 3 F = k Qq r = k = 0.419N ( ) 4 3 # ( 0.01m ) 3 = $10 %6 C ( ( ) "10$6 C) ( 10 $7 C) Qq ( R + d) = 9 2 "109 N # m 2 /C 2 ( 0.01m m) 2 d=0.02 m R=.01m x ρ=0. 1C/m 3 F= Direction

5 Page 5 c) Except for helium, the element with two electrons, inert gas atoms are ones that have just enough electrons to finish filling a p-shell. Find the number of electrons of the next two inert gas atoms after helium ( neon (Ne) and argon (Ar) ). In this range of atomic number the subshells fill in order of increasing angular momentum. He is 1s 2. Since we are told the shells fill in order of increasing angular momentum, need to fill 2s and 2p. This is 8 more electrons. So 10 electrons is next inert gas. Then need to fill 3s, 3p. This is again 8 more electrons, so 18 is the next inert has atom. 1st inert 2nd inert # electrons = d) Two conducting spheres are very far from each other. Sphere1: 2 cm radius and a charge of +1 mc. Sphere2: 1 cm radius and a charge of -1 mc. How much work must you do to move -1 µc of charge from Sphere 1 to Sphere 2? (1 mc = 10-3 C and 1 µc = 10-6 C) [Hint: Think about the relation between work and electric potential.] R 1 =2cm Q 1 =+1mC q=-1µc R 2 =1cm Q 2 =-1mC By Gauss law, the potential of each sphere relative to infinity is the same as if the charge were all concentrated at the center: kq/r. Work is qδv. ( ) V 1 = k Q 1 R 1 = 9 "10 9 N # m 2 /C 2 10 $3 C 2 "10 $2 m = 4.5 "108 V V 2 = k Q 2 = ( 9 "10 9 N # m 2 /C 2 ) $10$3 C R 2 1"10 $2 m = $9 "108 V W = %U = q(v 2 $V 1 ) = $10 $6 C ( )( $13.5 "10 8 J /C) =1350J Work =! e) A conducting ring centered at (x=0,y=0,z=0) carries a positive current in the direction shown. Graph the z-component of the magnetic field everywhere along the z-axis. [Hint: the B field of a current loop resembles that of a magnetic dipole.] y B z I x 0 z z

6 Page 6 3) [25 pts, 5 pts each] We did a demonstration in which a capacitor was discharged through a coil of wire by closing a switch. This caused a change in an empty beverage can inside the coil. I a) The capacitor had a capacitance of 62 µf and was charged to a potential of 6000 V before discharging. How much energy was stored in the capacitor? ( 1 µf = 10-6 F) E = 1 2 CV 2 = 1 ( 2 62 "10#6 F) ( 6000V ) 2 =1116J Energy = b) The capacitor was then discharged by closing the switch, generating a pulse of current through the coil. What was the average current, assuming that the current pulse lasted seconds and completely drained the charge from the capacitor? [Hint: think of the definition of current] The average current is the total charge that flowed, divided by the time. The total charge is the charge stored on the capacitor: Q = CV = 62 "10 #6 F is then 0.372C /0.002s =186A ( ) 6000V ( ) = 0.372C. The average current Current =

7 Page 7 c) The time-dependence of the current in the coil is shown below. Time is in milliseconds (1ms=10-3 s). Plot the current induced in the soda can as a function of time. CURRENT ( A ) TIME (ms) INDUCED CURRENT (arbitrary units) 0 TIME (ms) d) Suppose the magnetic flux through the soda can is given by " can = ( 10 #5 Wb / A)I, where I is the current in the coil. The graph of part c) shows the current in the coil I as a function of time. The resistance of the soda can is 0.1Ω. Calculate the current induced in the can 0.1 ms after the pulse starts. "# = 2 $10 %3 Wb in "t = 0.1#10 $3 s, so d"/dt = 20Wb /s. The EMF is then 20V. The induced current is 20V/0.1Ω=200A. Current = e) At right is an end view of the coil and soda can. Indicate on the drawing: i) the direction of the current induced in the can, and ii) the direction(s) of the force on the can Coil 0.1 ms after the pulse starts (refer to the current plot in part c). Explain. Positive current in the coil is in the direction of the arrow. Coil current is in direction of arrow at 0.1 ms. This creates a magnetic flux into page that is increasing. Lenz law says induced current should reduce this increase, so induced field out of page is required. This is a CCW current. Since oppositely directed current repel, the force is inward. Can I can I F

8 Page 8 4) [20 pts, 5 pts each] a) An unknown quantum system has energy levels E 1 =1.1 ev, E 2 =3.0 ev, E 3 =4.2 ev. Calculate the wavelengths of the three photons emitted by this quantum system. " 1 = " 2 = " 3 = hc 1240eV $ nm = = 652.6nm E 2 # E 1 1.9eV hc 1240eV $ nm = =1033.3nm E 3 # E 2 1.2eV hc 1240eV $ nm = = 400nm E 3 # E 1 3.1eV Wavelengths= λ 1 λ 2 λ 3 units b) You use a diffraction grating with 600 lines/mm in a spectroscope to observe the spectrum of this quantum system and observe the first six interference maxima (marked with X below). Label each interference maximum with the wavelength(s) of light, λ short, λ medium, λ long observed at that point. The wavelengths are related as λ short < λ medium < λ long. λ long λ medium grating λ short λ short, λ med, λ long λ short λ medium λ long

9 Page 9 c) Suppose λ short =400nm and λ medium =500nm. By what angle do you need to rotate the spectroscope to move from the λ short to λ medium first-order interference maxima? The line spacing is 1 mm/ 600 = mm=1667nm. Interference maxima occur at angles given by sin" = m# /d sin" = 400nm /1667nm # " =13.89 sin" = 500nm /1667nm # " =17.46 The angle difference is 3.57 Movement angle= d) Suppose you want to use a different grating to observe λ medium interference peak at a smaller angle than that obtained with the 600 lines/mm grating. Should you use a grating with more lines/mm or less lines/mm? Circle the correct answer and explain. Need less lines / mm. This makes the spacing d between the lines be greater. This makes the path length difference greater for smaller angles. Mathematically, dsin" = m#, so a bigger d gives a smaller θ More lines / mm Less lines / mm

10 Page 10 5) [25 points, 5 points each] This question is about electromagnetic (EM) waves. a) Indicate whether the following statements are true or false by circling T or F. T F An EM wave always has both E and B fields T F The E-field amplitude of a linearly polarized EM wave is constant in time. T F The E-field amplitude of a circularly polarized EM wave is constant in time. T F The EM wave speed does not depend on the medium. T F The E- and B-fields of an EM wave are always perpendicular T F The Poynting vector points in the same direction as the wave E-field. T F The vectors E and B are always in phase with each other. T F EM waves can exert pressure on a surface. T F EM waves transport energy T F The intensity of an EM wave is proportional to the square of the E-field. b) Your Madison oldies station (WOLX) has a frequency of 94.9 MHz (94.9x10 6 Hz). Calculate its wavelength in air (n=1.0). " = c / f = 3#108 m /s 94.9 #10 6 /s = 3.16m λ= c) The EM wave passes through window glass (index of refraction n=1.5) on the way into your house. Calculate the speed v of the EM wave while it is in the window glass. Speed in glass = c/n glass = 3 "10 8 m /s = 2 "10 8 m /s 1.5 speed=

11 Page 11 d) Calculate the frequency and the wavelength in the window glass. The frequency is the same as in vacuum, but the wavelength changes according to " = v / f = 1 c n f = " vac 1.5 = 2.11m f = λ = e) A typical radio needs 10 µw of radio wave power coming in the antennae in order to play the radio station. Assuming the antenna has an area of 0.1 m 2, and 10% of the incident power is absorbed, what is the electric field amplitude of the weakest EM wave the radio can pick up? (1µW=10-6 W) The magnitude of the Poynting vector r E " r B /µ o is the power / area of an EM wave. This can ( ). also be written 1 2 c" o E 2. The power hitting the antenna is then 1 2 c" o E 2 0.1m 2 ( ) = 0.1m2 ( 3"10 8 m /s)( 8.85 "10 #12 C 2 /N $ m 2 )E 2 = = ( "10 #4 N $ m /s C 2 /N 2 )E 2 = ( 10 #5 J /s) /0.1 E = N /C = V /m ( ) ( ) 10 #5 W E weakest =

12 Page 12 6) [15 pts, 5 pts each] A simple model of the nucleus of an atom is neutrons and protons confined to a one-dimensional box of size 10 fm (1 fm=10-15 m). The energies of the quantum states are given by E neutron n = n neutron 8m neutron L, E proton 2 n = n proton 8m proton L. 2 2 a) Calculate the energy of a proton in the lowest energy state of the one-dimensional box. h 2 2 h 2 E n = n 2 h 2 8mL 2 = n 2 ( hc) 2 8mc 2 L = n 2 2 ( 1240eV " nm) 2 ( )( 10 $5 nm) = #106 ev #10 6 ev b) What is the wavelength of the proton in this lowest energy state? In the ground state, ½ wavelength fits inside the box, so the wavelength is 20fm. c) The nucleus can make transitions between quantum states, emitting a photon. What type of radioactive decay is this? Explain a) alpha b) beta c) gamma d) none of the above Gamma decay is when a nucleus is left in an excited state by some other decay process, then decays to a lower-energy state by emitting a photon.

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