Standard Model - Optics and Modern Physics - Solved Past Paper, Exams of Physics

Download Standard Model - Optics and Modern Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! Physics 228- Final MAY 12, 2009 Profs. Rabe and Coleman ‘Your name sticker => with exam code SIGNATURE: TURN OFF CELLPHONES NOW!!! The exam will last from 4:00 p.m. to 7:00 p.m. Use a #2 pen- cil to make entries on the answer sheet. Enter the following ID information now, before the exam starts. In the section labelled NAME (Last, First, M.1.) enter your last. “ name, then fill in the empty circle. for a blank, then enter your first name, another blank, and finally your middle initial. Under STUDENT # enter your 9-digit Identification Number. Enter 228 under COURSE, and your section number (see label above) under SEC. Under CODE enter the exam code given above. During the exam, you may use pencils, a calculator, and three handwritten 8.5 x 11 inch sheets with formulas and notes, with- out attachments. There are 30 multiple-choice questions on the exam. For ‘each question, mark only one answer on the answer sheet. There is no deduction of points for an incorrect answer, so even if you cannot work out the answer to a question, you should make an educated 10. guess. At the end of the exam, hand in the answer sheet and the cover page. Retain this question paper for future reference and study. When you are asked to open the exam, make sure that your copy contains all 30 questions. Raise your hand if this is not the case, and a proctor will help you. Also raise your hand during the exam. if you have a question. Please SIGN the cover sheet under your name sticker. A proctor will check your name sticker and your student ID some time during the exam. Please have them ready. Good luck! Useful Information c= speed of light = 3.00 x 108 m/s Ge = —e = charge on an electron = —1.602 x 10-!® Coulomibs Gp = +e = charge on a proton = +1.602 x 107! Coulombs ke = 8.99 x 10°N- m?2/C Na = Avogadro's number = 6.022 x 1075 particles/mol kg = Boltzmann’s constant = 1.38 x 10-*3 J/K Wien’s Constant = 2.898 x 10-3 m - K o = Stefan’s Constant = 5.670 x 10-® W/(m? K*) h = 6.626 x 10-*4 J-s h = 4.136 x 10-™ eV-s (in units of electron volts-second) A= £ = 1.054 x 1074s he = 1240 eV - nm = 1240 MeV - fm leV=1602x 107% J - 1 keV = 1000 eV . up = Bohr Magneton = 5.79 x 10“eV/T = 9.27 x 10-4 J/T g-factor of electron = 2.00232 1u=1 atomic mass unit = 931.5 MeV/c? m, = electron mass = 9.11 x 10-* ke mc? = electron rest energy = 0.511 MeV Mp, = proton mass = 1.67 x 107 ke Mpc” = proton rest energy = 938.27 MeV mac? = neutron rest energy = 939.57 MeV Mec? = charged pion rest energy = 139.6 MeV moc" = neutral pion rest energy = 135.0 MeV M(H) = 1.007825 u lrad = 0.01 J/Kg RBE = Relative Biological Effectiveness Temperature conversion: T (K) = T (°C)+273.15= 1.8 T (°F)+215.55 Zeroes of teraperature scale: 273.15K = 0°C =32 °F lom=10°%m 1A=10-%m=0.1 0m 1pm = 10-?m 1 mHz = 107° Hz 1 kHz = 107° Hz 1 MHz = 10+8 Hz 1 GHz = 10*° Hz Some quantum numbers: Qe] BIS Ti, |i, ull 2/3 | 1/31 0 ev | 1104 0 af 1/3 | 1/31 0 yy )oT 1) Oo —1/3 | 1/3) -1 typ] 0; 60] 1 16. il. tn a Young’s double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen that is 5 m from the slits is: a) 0.50 cm by 1.0 cm c) none of the other answers d} 0.10 cm (2) 0.25 em A diffraction grating of total width 4 cm is illuminated with light of wavelength 577 nm. The second-order principal maximum is formed at an angle of 41.25°. What is the total number of lines (slits) im the grating? a) About 12000 About 5000 About 23000 d) About 8000 e) About 19000 Two electrons with opposite spin are in an / = 0 orbital. A magnetic field of 1 T is present. What is the energy separation between the two electron levels? a) 2.9x 1078 eV 1.16 x 10-* eV c) 5.65 x 1073 eV d) 32x 10% eV e) 1.54x 10% eV At room temperature (T = 300K), the occupation probability of an electron state in the valence band of Zn with kinetic energy 10% greater than the Fermi energy (Er = 11.72 eV) is about: a) 0.94 b) 2.1 x 107? 2.0 x 10-7 d) 0.11 e) 9.2 10-3 (Sia 500%10 Gan (Gm) + 500%10 cn 6.001 m =Rm. Ay 7RA = ym SONS = 0.25 cm. -7 4 dsin® =ma ? d= 2nS7TP10 cm = L&xiQ om Bee Sin (41.25 °) N= 4em 2 22,354 ad AE=7~- at “Hew B Ae* ~ 2 00032.252 7.90232 eh B = 2.00232%5S. 748x107 ety zm Mb fe) E-Ee* LIAL eV , Wl (E-Gp Vkgt) +1) (exp ( ee dreV= 0.0299 12. 13. A diatomic molecule is in a vibrational and rotational state having energy 0.50 eV when it absorbs a photon of frequency 9.67 x 10% Hz. What is the energy of the final state of the molecule? a} 0.56eV c) 0.10eV d) 0.43eV e) 0.50eV For a diatomic molecule, it is found that the rotational states =0 and l=1 are separated in energy by 4 x 10~4 eV. The moment of inertia of the molecule is: Ca}) 1.7 x10~* kg m? b) not enough information. ce) 13 x10-% kg m? d) 4.2 x10-“" kg m? e) 8.4 x10-*7 kg m? If you were in a spaceship traveling at a constant speed close to the speed of light (with respect to earth), you might make the following observations: (I)That some of your physical dimensions were reduced. (I) That your mass was greatly increased. (111)That your pulse rate is a lot slower due to time dilation. Which of these observations could actually take place? None of these effects would occur. BD. More than one of these effects would occur. ec) (iy qd) e) (I) t Ep= 0.50eV 4 9.67% 10 "He Bo , o.50eV + 4 [ZG x1o eS * IGRI 'S 0.90e¥ vy -¥ 2 o eV - “pk 4 E,-B = A = Fx] a) Ald a 4 Yo" “4 t= yf xj eV Or -3¢, Js B13 10 eV's ~ 1,05 #20 Since you ore makina te observahoa S in he fame at whicr you ave ah reef, the tosullh are be sare a5 if you mach He obsevah ors while af Mot on eark,, 15. A particle in a box of length ZL is in the n = 3 state. Which 16. 17. diagram in the figure best describes its probability distribution pa)? @ ‘0 PG) / /\ 0 L ap Ww) PG) POX) / / / x I re fe = a 1 Pe) | WS * 0 L The average person receives a dose of 102 m rem/year. If the average RBE of the radiation received is 1.5, how much energy, in Joules, is absorbed per year by a 70 kg person? a) 0.1071 b) 4.76 0.0476 d) 7.14 e) 0.0714 (Qpeee as K-mesons have an average proper lifetime 7. How fast must they move with respect to the earth so that an earth-based experi- menter will measure an average lifetime of 1.5 7 ? 0.75c b) 0.40c c) 0.90 d) 0.50c e) 0.80c for n= 3 ¥ 3 L pee dvem = RBEx 0.01 LT J Kg here , b0-Eneroyie RBE 2 (02 «10 > rem Fo kg 4 ye Energy /yc> 1O2*10 x FO. TF =e 5 joe REELS YY xe eart baseol Bi periwon te yreasenas hferinn AT im =k 5 =i fee ti [xy > (y= (25) V= 0.75ce The following rest energies are known: use ennyy conserva ore 1700 +1700 + 939.24 4 938. 0F = 22% BIG + 2x10 +My mc? = electron rest energy = 0.511 MeV Mc = proton rest energy = 938.27MeV 23. m,C = neutron rest energy = 939.57MeV Ms = charged pion rest energy = 139.6MeV m, oc = neutral pion rest energy = 135.0MeV A few months from now, the Large Hadron Collider (LHC) at CERN in Geneva will be operating. Suppose that a head-on col- lision between protons and antiprotons, each having a kinetic en- ergy of 1700 MeV, produces the reaction ptproamt+a +h where H is the newly discovered “Higgs particle”. The Higgs is at rest and the x* and 77 depart in opposite directions, each with a kinetic energy of 110 MeV. What is the rest mass of the Higgs? 2388 MeV/c? (bh) 4777 MeV/c? ¢) 658 MeV/c? d) 1420 MeV/c? e) 3180 MeV/c? Consider the decay sequence shown. [ig] «@ [as Which of the following correctly iden- |_8¢ tifies nucleus Y? B a) 28°Bi Tz | & 3 en maif- ex °) B°Pb 8 \p d teu @ [208 2Po ¥ P| oP —? My A=2i2 z= $4 4LFT Mev 24, 25. 26. 27. In the Bohr model of the hydrogen atom, which of the following transitions emits a photon with the longest wavelength? a) n=4 —> n=2 B) n=6 —+n=7 absorbs plieton c) n=2 —n=1 ao n=6 —> n=5 e) n=5 — n=3 A radioactive source consists of 10? atoms. It is observed that 10%! atoms decay per second. What is the half-life of the radioac- tive material? 6) 2200 years b) 3200 years c) 1740 years d) 1520 years e) 107! years Protons are accelerated in a cyclotron with an internal field. of 0.2 T. If beam exits at a radius of 2 m from the center of the cyclotron, what is the energy of the protons? a) 15.3 MeV b) 14 GeV c) 38.3 MeV (a) 7.67 Mev €) 1.92 MeV A perfectly black body at 100°C emits light of intensity J. The temperature of this body is now raised to 200°C. The hotter black body now radiates light of intensity closest to: a) 2.01 b) 8.01 ce) 141 2.61 wot booking fr cmabfesk Ey-Em. Lge 6) LL ek Ey re so ampere Ee Ie Ayre rhe a 3e 38 0.0t 3 dbsorbs-ey j- Le alosarbs-@) | ah --ALs # 0.07 ti = 10” Ty, = 0.693 = 0.699 */0 5-20 x nar 3¢0082dr 269 =2, ig F years A 2 2er mv@ 2guB > v=3B5 k=zmv%=38 c mM LTE Lo A 16x00 “tay & Lap okg = $66 MeV Lett Dye? 1 /¥i3e 26 Te 373K Suppose the ratio of the density of free electrons in aluminum to 2 | to that in potassium is 3:1. If the Fermi energy in potassium is E <A" 3 1.3eV what is the Fermi energy in aluminum? & 19 eV Er me = (I.3eV¥ \(na =(|-3eV) Zh Ep per 21 3e/ 27 eV . 2 e) iL7eV (per) d) 6.7eV e) 3.96V The following masses are known: fe roms (01600 ~ 3 (1.0025) ~ #(n00sc6S))- Psa TLi 7.016004u. = ~34.25 Mey Given this information, the binding energy of JLi, in MeV, is ex- pected to be closest to: GD 39 b) 56 c) 52 d) 48 e) 43 In the diffraction pattern from a six slit diffraction grating, which phasor diagram represents the combination of electric fields from the six slits when the path length difference between light from neighboring slits is a third of a wavelength? / 0 a Cy mo a I a a ar ar YL Iv ¢) IL d) 1 ™ o e) V a ih