* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download RTD Part 4 - County Central High School
Anti-gravity wikipedia , lookup
Conservation of energy wikipedia , lookup
Photon polarization wikipedia , lookup
Renormalization wikipedia , lookup
Elementary particle wikipedia , lookup
Nuclear fusion wikipedia , lookup
History of subatomic physics wikipedia , lookup
Nuclear drip line wikipedia , lookup
Atomic nucleus wikipedia , lookup
Chien-Shiung Wu wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Use the following information to answer the next question. Decay Curve for Radioisotope A 15 ..- ~ ~ !:Ill ·a= ·; e 50 f! .!! t = !Jl 25 0.0 2.0 1.0 3.0 Time (h) The graph illustrates the decay of a radioactive isotope. !Numerical Respons~ lt97.j The time required for a 40.0 g sample to decay to 1.25 g is h. (Record your two-digit answer in the numerical-response section on the answer sheet.) !Numerical Respons~ lt9S.j For a 768 g sample of an unknown radioactive element, 48.0 g remain after I 0.2 h. The h. half-life ofthe element is (Record your three-digit answer in the numerical-response section on the answer sheet.) - 71 - Use the following information to answer the next questions. In medical diagnosis, a patient may be injected with a radioactive isotope. As the isotope decays, the gamma ray emissions are detected and a computer builds images of the patient's blood flow and organs. A radioactive isotope commonly used in medical diagnosis is technetium-99. This isotope has a half-life of 6.00 h and decays to a stable isotope by gamma ray emission. 199. If the biological processes that might eliminate some of the technetium-99 from the body are ignored, the maximum percentage of radioactive technetium-99 that could still be present in a patient's system 24.0 h after injection is A. 12.5% B. 6.25% 2.00% D. 0.841% c. Use the following information to answer the next two questions. Cobalt-60 is a common radiation source used in cancer treatment. The half-life of cobalt-60 is 5.2 years. A cobalt-60 nucleus decays by emitting a beta negative particle and a gamma photon. 200. Which of the following equations describes the decay of cobalt-60? + -~P + 60Ni + -~P + 28 A. 60co --t 60Ni 27 28 y +v B. 60co --t 27 y + c. 60co --t 60Fe 27 26 y +v D. 60co --t 27 y +v + +~p + 60Fe + +~p + 26 v !Numerical ReSJ!ODS~ j201.j The percentage of cobalt-60 remaining after 15.6 years is %. (Record your three-digit answer in the numerical-response section on the answer sheet) - 72- Use the following information to answer the next three questions. A particular nuclear fission reaction ofuranium-235 is represented by 2 ~~u + 5n ~ 1 t§cs + ~~X+ 35n where element X is unknown. 202. The value of cd in the above reaction can be identified using the physics principle of conservation of A. B. C. D. charge nucleons momentum mass-energy !Numerical Respons~ J2o3.l The fission product in this reaction is represented by ~~X. The values of a, b, and c, are, respectively _ _ , __ , __ , and __ . (Record all four digits ofyour answer in the numerical-response section on the answer sheet) 204. In the above fission reaction, the mass of the reactants is 236.05 atomic mass units, and the mass of the products is 235.86 atomic mass units. Which of the following explanations best describes the change in mass that occurs in this nuclear fission reaction? A. Mass and energy are equivalent, and energy has been converted into mass in this reaction. B. Mass and energy are equivalent, and mass has been converted into energy in this reaction. C. Mass and energy are equivalent, and the missing mass is due to inaccurate laboratory measuring equipment. D. Neutrinos that are given off in the fission reaction are undetectable, which accounts for the differences in mass of the detectable components of the reaction. - 73- Use the following additional information to answer the next three questions. A cyclotron is a particle accelerator used to investigate subatomic structure. Magnetic fields are used to control the path of charged particles within a cyclotron. 205. The radius of the path followed by charged particles moving perpendicularly through the magnetic field of a cyclotron could be reduced by A. B. C. D. using particles with a greater mass increasing the speed of the particles using particles with a smaller charge increasing the strength of the magnetic field 206. The period T for a particle of charge q in a magnetic field of strength B is A. 2rrm qB B. rrm qB C. qB 27r D . .!!..!!._ rrm 207. An alpha particle travels in a direction perpendicular to a magnetic field of strength 1.6 T. If the alpha particle experiences a force of magnitude 1.1 x 10- 13 N, then its measured speed will be A. B. C. D. 2.1 4.3 2.1 4.3 x 10-7 m/s x 10-7 m/s x I 0 5 m/s x 105 m/s - 74- Use the following information to answer the next four questions. Fusion Research Interest in nuclear fusion is growing because of the amount of energy available from nuclear reactions. A major difficulty in producing a nuclear fusion reaction is that in order for nuclei to fuse, the nuclei must possess a large amount of kinetic energy. Under most circumstances, 0.25 MeV per nucleus is sufficient. At such high energies, the nuclear fuel is called a plasma. The average kinetic energy of a nucleus within a plasma can be found using 3 = -zbT Ek where Tis the temperature of the plasma, in Kelvin, and b is a physical constant equal to 1.4 x w-23 JIK. One method of obtaining the temperatures necessary for fusion is to use a high-intensity laser to heat a small cluster of nuclei. One such laser emits a 1.0 x 10 15 W pulse of ultraviolet radiation that lasts for 1.0 x 10- 12 s. The wavelength of this laser is 280 nm. A Fusion Reaction Equation fH + fH ~ _x_ + neutron 208. The missing product, X, in the fusion reaction given above is A. ~He B. ~He C. {H D. ~He 209. The main reason that the nuclei need to have such large kinetic energies is that A. fusion releases large amounts of energy B. fission must occur before fusion can occur C. this kinetic energy is converted into nuclear energy D. the nuclei must overcome a strong electrostatic repulsion - 75- 210. When the average kinetic energy of the nuclei in a plasma is 0.25 MeV, then the temperature is A. 1.9 B. 2.9 4.3 D. 1.2 c. X X X X 109 K 109 K 109 K 1028 K 211. The energy of a single photon of the ultraviolet laser is 10-'~ 0 J A. 1.9 X B. 7.1 x w-zs 1 c. x w-27 1 X 10- 19 J 1.0 D. 7.1 - 76- Use the following information to answer the next seven questions. There are many different types of propulsion engines for satellites. One type of ion propulsion thrust chamber and the satellite to which it is attached are described below. The cylindrical thrust chamber of the engine has a central spike. Electromagnets are used to produce a non-uniform magnetic field directed radially toward the spike. A virtual cathode consisting of trapped electrons is located at the rear of the thrust chamber. An electric field exists between the anode and the virtual cathode. Positive xenon ions enter the thrust chamber at the anode and accelerate toward the virtual cathode, which results in thrust on the satellite. As the xenon ions pass through the virtual cathode, they pick up electrons and neutral xenon atoms fly out of the chamber. Diagram 1: Thrust Chamber in Engine Thrust chamber Engine Location where xenon ions enter .. Satellite Exhaust of xenon atoms motion Diagram II: Cross Section of Thrust Chamber Location where Vn1Ual cathode Anode Central spike field direction RegionY Thrust Chamber Specifications Magnetic field intensity at the location where the xenon ions enter Electric field intensity at the location where the xenon ions enter Mass of one xenon ion, Xe+ Exit speed of neutral xenon atom with respect to the thrust chamber -77- 0.0200 T 1. 00 x 10 4 VIm 2. }9 X 10-2S kg 1. 5 x 104 rnls 212. In diagram II on the previous page, the direction of the electric field in region Y is A. B. C. D. to the left to the right into the page out of the page 213. As the xenon ions, Xe +, move through region Y, as labelled in diagram II on the previous page, they experience both electric and magnetic forces. The direction of the magnetic force that they experience is A. B. C. D. into the page out of the page toward the top of the page toward the bottom of the page 214. The xenon ions, Xe +, enter the thrust chamber at a negligible speed. The minimum distance between the anode and the virtual cathode that is required to produce the exit speed is A. B. C. D. 1.2 x 10- 16 m 1.0 x 10-6 m 1.5 x 10~ 2 m 1.4 x 10 12 m !Numerical Respons~ 1215.1 While in the thrust chamber, notation, of a. b x a xenon ion experiences an impulse, expressed in scientific 10-cd kg·m/s. The values of a, b, c, and dare-~~'-~·~~' and (Record all four digits of your answer in the numerical-response section on the answer sheet) - 78- Use the following additional information to answer the next question. Xenon ions, mion, reach the virtual cathode with a speed of v 1• When a xenon ion collides with a stationary electron, me, in the virtual cathode, the xenon atom, matom, formed has a speed of vz. 216. The relationship between v2 and v 1 can be expressed as A. v 2 me) Vt = (mion+ matom c. Vz mion) - Vt = (matom B. v 2 matom ) V 1 = ( mion+me D. Vz v1 = (matom) mwn Use the following additional information to answer the next two questions. Ion Propulsion Engine and Satellite Specifications Average thrust applied by the engine to the satellite Mass of satellite and propulsion system Speed of xenon atom exiting the thrust chamber Mass of xenon atom 0.200 N 2.5 X }03 kg 1.5 x 104 m/s 2. 19 x w-25 kg 217. The length of time, in hours, that this type of ion propulsion engine must be in operation in order to increase the speed of the satellite and propulsion system by 12. 0 ml s is A. 0.0240 h B. 41.7 h c. 250 h D. 1.50 X 10 5 h 218. The number of xenon atoms that would have to be discharged as exhaust in order to increase the speed of the satellite and propulsion system described above by I. 00 ml s is A. B. C. D. 5.1 7.8 1.6 7.6 x x x x I 0 19 atoms 1021 atoms 1022 atoms 1023 atoms - 79- Use the following information to answer the next four questions. A negatively charged, graphite-coated sphere is suspended from the ceiling on an insulating string in the region between oppositely charged parallel plates, as illustrated below. The plates are 20.0 em apart and are maintained at an electrical potential difference of 3.1 x 102 V. The charged sphere experiences an electrical force of 8.4 x 10-7 N. Negatively Positively charged plate charged plate Graphite coated sphere 219. One way to give the graphite-coated .~phere a negative charge is to touch it with a --=--- charged rod. This process is called charging by zz The statement above is completed by the infonnation in row i Row ii A. positively induction B. positively conduction c. negatively induction D. negatively conduction !Numerical Respons~ j220.j The charge on the graphite-coated sphere, expressed in scientific notation, is a. b x 1o- cd C. The va1ues of a, b, c, and dare _ _ , _ _ , _ _ , and (Record all four digits of your ansv:er in the numerical-response section on the ans;ver sheet) - 80- 221. Which of the following scale diagrams is the free-body diagram for the negatively charged sphere? A. B. F, Fg c. D. Fr FT \: \L Fe Fe ~ Use the following additional information to anslver the next question. The charged plates are now removed. The positively charged plate is replaced by the north pole of a strong magnet and the negatively charged plate is replaced by the south pole of a strong magnet. The system is allowed to reach equilibrium. 222. As a result of the magnetic field, the negatively charged, graphite-coated sphere will A. B. C. D. swing back and forth between the magnetic poles be deflected toward the magnetic north pole be deflected toward the magnetic south pole hang midway between the magnetic poles - 81 - Use the following additional information to answer the next question. Two small metal spheres are fixed to insulated stands and given static charges of -4.00 x 10-6 C and + 2. 00 x 10-6 C, respectively. The spheres are then placed 0. 500 m apart. Point P is halfway between the charged spheres. 223. At point P, the magnitude of the electric field caused by the two charged spheres is A. 8.63 B. 2.88 c. 2.16 D. 7.19 X X X X 10 5 N/C 105 N/C 105 N/C 104 N/C Use the following information to answer the next six questions. One Solar Nuclear Fusion Reaction Equation iH + iH ~ ~He + &n + Energy Representation of Nuclei Involved in This Fusion Reaction Deuterium CFI <;I Tritium '\../ ."+". Legend • Intermediate product ~ Alpha partide / .a ~ '\.. e Neutron G) Proton Neutron One way to harness this energy on Earth is to use a nuclear fusion reactor. One of the problems in terrestrial fusion reactors is the very high energy required to overcome the electrostatic repulsive force between the deuterium ions and the tritium ions. A particular reactor design uses magnetic fields in a process called magnetic confinement to keep the ions inside the reactor. However, neutrons escape magnetic confinement. These neutrons are captured by a shield called a lithium blanket. - 82- 224. Energy is released in this nuclear fusion reaction because the A. B. C. D. free neutron has a high energy number of protons remains the same number of nucleons remains the same mass of the alpha particle and neutron is less than the mass of the intermediate product !Numerical Respons~ ~25.1 At a particular instant, the electrostatic force that the deuterium ion exerts on the tritium ion is 23. 3 N. The distance between the centres of the two ions, expressed in scientific X } o-w m. notation, is (Record your three-digit answer in the numerical-response section on the answer sheet) Use the following additional information to answer the next question. A positively charged deuterium ion enters a magnetic field directed out of the page, as shown below. • • • Represents a magnetic field directed out of the page 226. The direction of the magnetic deflecting force that acts on the positively charged deuterium ion as it just enters the magnetic field is A. into the page B. out of the page C. toward the top of the page D. toward the bottom of the page 227. The neutron produced in the fusion reaction escapes the magnetic confinement because A. B. C. D. neutral particles are not deflected by magnetic fields the neutron is moving so fast that it escapes the magnetic field the energy produced in the nuclear reaction is enough to cause the neutron to escape conservation of momentum requires that the neutron has to be pushed in the opposite direction to that of the helium produced - 83- 228. As a particular neutron travelling at 5. 21 x 106 m/s hits the lithium blanket and stops, it experiences an impulse of ii and the neutron-lithium collision is classified as The statement above is completed by the information in row Row i A. -8.70 x 10-21 N·s elastic B. -8.70 x 10-21 N-s inelastic c. -2.27 X 10- 14 J elastic D. -2.27 X 10- 14 J inelastic ll 229. Which of the following equations most likely describes a neutron-lithium collision? Bt• A. ~n + 7t• 3 I ~ 3 I B. 0tn + 4Li 3 ~ 3 I c. + 3 I ~ I Be D. ~n + 4t• 3 I ~ ~Be ~n 7t• St· - 84- Use the following information to answer the next three questions. One type ofbreathalyzer involves illuminating a photocell (photoelectric surface) with infrared (IR) radiation of wavelength 9.50 x 10-6m. Alcohol molecules absorb infrared radiation. A breathalyzer circuit is illustrated below. Ammeter Photocell · r-: Breath sample '' IRSource The ammeter in the breathalyzer is calibrated to register a maximum reading with no alcohol sample between the detector and the IR source. 230. A breath sample containing alcohol is introduced into the analyzer. If it absorbs 50% of the radiation emitted by the infrared source, the current in the ammeter will be A. B. C. D. halved doubled the same quartered !Numerical Respons~ ~ A current of 4.71 10~ 3 A passes through the ammeter for 30.2 s. The number of electrons that pass through the ammeter in that time, expressed in scientific notation, is a.b x 1Ocd. The values of a, b, c, and dare _ _ , _ _ , _ _ , and x (Record all four digits of your answer in the numerical-response section on the answer sheet) !Numerical Respons~ !232.1 The energy of a photon of infrared radiation from this source, expressed in scientific notation, is b X 1o~w J. The value of b is - - - - (Record your three-digit answer in the numerical-response section on the answer sheet.) - 85- Use the following information to answer the next four questions. A student follows three procedures to study the properties of laser light. She uses a laser that emits monochromatic light that has a wavelength of 634 nm. Procedure Observation The student first shines the laser light I through a crystal that has an index of The path of the refracted ray is recorded. refraction of 1.53. The student shines the laser light II through a diffraction grating that has 5.00 x An interference pattern is projected onto a screen. 105 lines/m. The student shines the laser light upon III a photovoltaic cell that is connected to No electrical current is measured. an ammeter. 233. The energy of one photon emitted by the laser is A. 4.20 R 4.20 c. 3.14 D. 3.14 X x x x 10-4° J w-31 J w-28 J w- 19 J 234. In procedure I, the wavelength of the laser hght in the crystal is --=---· The speed of the laser light in the crystal is ii than its speed in air. The statement above is completed by the information in row ii i Row A. 4.14 B. 4.14 x 10-7 m greater c. 9. 70 less D. 9.70 x 10-7 m X X 10-7 m 10-? m less greater !Numerical Respons~ ~35.1 In procedure II, the angle between the central maximum and the first bright spot ofthe interference pattern is o (Record your three-digit ans\ver in the numerical-response section on the answer sheet.) - 86- 236. In order to produce an electrical current in procedure III, the student must use electromagnetic radiation that has a i wavelength or a photovoltaic plate that has a ii work function than those she actually used in procedure III. The statement above is completed by the information in row Row i ii A. shorter larger B. longer larger c. shorter smaller D. longer smaller Use the following information to answer the next five questions. The element ununquadium (i~lUuq) has been created by fusing calcium ions (~gca) with plutonium nuclei CZ~!Pu). The calcium ions are doubly charged(+ 2e) and have a mass of7.96 x 10-26 kg. To accelerate these ions to a high enough energy to fuse with plutonium, they are repeatedly accelerated by an electric potential difference. They are contained in a magnetic field between these accelerations. In one stage of the acceleration process, calcium ions enter the accelerating chamber at a speed of 1.00 x 106 m/sand exit it at a speed of2.75 x 106 m/s. They immediately enter a magnetic field and follow a path that has a radius of 1.24 m. 237. Which of the following equations could be the nuclear reaction equation for the fusion of calcium and plutonium in the production ofununquadium? A. 48Ca 20 + 244pu 94 B. 48Ca 20 + 244pu 94 c. ~ i~lUuq +~ex: ~ 2s9u uq 114 + 3 -1o~ + v + 2 ~!Pu i~lUuq + 3 5n D. 48Ca + 3 5n ~ i~lUuq 20 + 244pu 94 ~gca ~ !Numerical Res~ons~ 1238.1 The strength of the magnetic field used to contain the calcium ions, expressed in scientific notation, is a. be x 10--d T. The values of a, b, c, and dare _ _ , _ _ , and (Record all four digits of your answer in the numerical-response section on the answer sheet) - 87- 239. The electric potential difference in the accelerating chamber is A. B. C. D. 1.24 X 1.52 X 8.16 X 9.38 X 105 v 105 v 105 V 105 v Use the following additional information to answer the next two questions. The decay chain ofununquadium-289 is shown below. Ununquadium-289 has a half-life of 30.4 s. 240. The decay particles X and Yare A. B. C. D. both alpha particles both beta positive particles a beta positive particle and an alpha particle, respectively an alpha particle and a beta positive particle, respectively 241. If 1.00 )lg of ununquadium-289 is initially produced, the mass ofununquadium-289 remaining after 1.00 min will be A. 0.255 )lg B. 0.507 )lg c. 0. 703 )lg D. 0.977 11g - 88- _.:;, Use the following information to answer the next four questions. Photon-Matter Interactions When a photon passes through matter, it interacts with the atoms and their electrons. There are four important interactions with matter that a photon can undergo. I. The photon may be scattered by an electron and in the process lose some energy, transferring momentum and energy to the electron. II. The photon may move an electron out of an atom, and in the process, the photon disappears (the photoelectric effect). III. The photon may move an electron to a higher energy state in the atom, and in the process, the photon disappears. IV. A photon may actually create matter. The most common process, called pair production, is the production of an electron and a positron. A positron has the same mass as an electron, but it has the opposite charge. In addition, a massive particle, such as an atomic nucleus, must gain some of the photon's initial momentum. (See the diagram below.) e\Positron) Photon Atomic nucleus e- (Electron) 242. The name given to interaction I is A. B. C. D. Lenz's Law X-ray production the Compton effect the de Broglie hypothesis 243. The curved paths of the particles in the pair production diagram result from the electron and positron moving through an external magnetic field. In this diagram, the direction of the magnetic field causing these paths to curve is A. to the left B. to the right C. into the page D. out of the page - 89- 244. The reason that pair production occurs, rather than the production of a single electron, is that the production of a single electron would violate the Law of Conservation of A. B. C. D. Mass Charge Energy Momentum 245. During pair production, the speed of the electron or of the positron can be calculated by measuring the radius of the circular path it travels within the magnetic field. The speed of a charged particle moving in a circular path in a uniform magnetic field is given by qBr A. v = m B. v = Bqrm m Bqr C. v = rB D. v = qm - 90- Use the following information to answer the next five questions. A sample of iodine-131 has an initial mass of 76.0 mg. The activity of the sample is measured and the amount of iodine-131 remaining in the sample is determined. The following graph was obtained. Mass oflodine-131 Versus Elapsed Time 80_0 -Q.ll 70_0 e - 4"') ~ 60_0 50.0 o_o . . . . . . . ~-.,.......---'-1 25_0 15_0 20.0 -+-~~-r-~~-'-+~~~ o_o 5.0 10_0 Time (days) A particular nucleus of iodine-131 decays by emitting a beta particle that travels at 2.34 x 10 5 m/sand gamma radiation that has a wavelength of 5.36 x 1o- 12 m. Extra momentum and kinetic energy are carried off by a neutrino. !Numerical Respons~ 1246.1 After 48.0 days the amount of iodine-131 that remains in the sample is mg. (Record your three-digit answer in the numerical-response section on the answer sheet.) 247. The energy emitted as gamma radiation during the transmutation of an iodine-131 nucleus IS A. 3.55 X 10-4 5 J B. 2.68 X 10-27 J c. 1.24 x 1o-22 1 D.3.71x10 14 J - 91 - Use the following additional information to answer the next question. The momentum of the gamma ray photon and the beta particle can be calculated. The momentum of a gamma ray photon (y) is determined by the equation h A p=- 248. For the decay of iodine-131, the relationship between the magnitude of the momentum of the gamma ray photon (py) and the magnitude of the momentum of the beta particle (pp) can be represented by the equation A. B. C. D. py=-pp py=pp Pr =(I. 72 x 10-3 ) x pp Pr = (5.80 x 102) x pp 249. The equation for this radioactive decay is A. 1311 --+ 1 ~iSb 53 + beta +gamma + neutrino B. 1311 --+ 1 ~~Xe 53 + beta + gamma + neutrino c. 1311 --+ 1 ~~1 53 D. 1311 --+ 1 ~iXe 53 + beta + gamma + neutrino + beta + gamma + neutrino 250. To protect lab technicians from harmful radiation, the equipment used in this experiment should be shielded with A. B. C. D. lead to stop the y radiation paper to stop the ~particles an electric field to stop the y radiation a magnetic field to stop the ~ particles -92- Use the following information to answer the next three questions. In 1997, the Mars Pathfinder Mission included a robotic rover called Sojourner. Sojourner carried a variety of instruments to analyze the Martian soil, rocks, and air. One of Sojourner's instruments bombarded rocks with alpha particles to produce an "alpha-proton" reaction. One example of an alpha-proton reaction occurs when an alpha particle enters a sodium atom and knocks a proton out of its nucleus. The nuclear reaction equation can be written as follows. ~He+ nNa ~ iP + ~~X Scientists on Earth sent instructions to Sojourner via radio waves. Because it took 10 min for the signals to reach the robot, the scientists instructed it to move only a few centimeters at a time. INumerical Respons~ 1251.1 In the nuclear reaction equation above, the values of a, b, c, and dare ___ , ___ , , and (Record all four digits of your answer in the numerical-response section on the answer sheet) 252. In the nuclear reaction equation above, element X is A. B. C. D. Al Mg Ne Fe !Numerical Respons~ 1253.1 When the signals were sent to Sojourner, the distance between Earth and Mars, expressed in scientific notation, was a.b , and x I Ocd m. The values of a, b, c, and dare ___ , ___ , (Record all four digits of your answer in the numerical-response section on the answer sheet) - 93- Use the following information to answer the next two questions. Energy States of Chromium in a Ruby Crystal Laser Photon ----il--l.. ~ 0.0 kV -f5··-~- Photocathode P'ho1oeklcUOll-+---.., +0.10 kV +0.40 kV +0.60 leV + O.BOkV + 1.0 kV +U.kV When the scintillation crystal is struck by gamma radiation, it emits photons with a wavelength of 4.93 x 10~ 7 m. Some of these photons strike the photocathode. As a result, electrons are ejected from the bottom surface of the photocathode. The work function of the photocathode is 2.10 e V. 254. The maximum kinetic energy of the electrons ejected from the photocathode is A. B. C. D. 6.7 X 3.4 X 4.0 X 7.4 X 10-20 J 10- 19 J J0- 19 J 10- 19 J -94- 255. The electrons leaving the photocathode are attracted by the 0.20 kV electrode. The maximum speed they attain is A. B. C. D. 8.6 x 8.4 x 7.0 x 2.1 X 10 5 m/s 10 6 m/s 10 13 m/s 10 15 m/s Numerical Respons~ 1256.1 Northern lights are often observed in Alberta skies. The most common colour, green, has a wavelength of 558 nm. When a collision occurs between energetic electrons and oxygen atoms in the upper atmosphere, the oxygen atoms are excited. To cause the most common colour of northern lights, the electrons must be travelling with a minimum speed, expressed in scientific notation, is x 1Ow m/s. (Record your three-digit answer in the numerical-response section on the answer sheet.) -95- Use the following information to answer the next four questions. Physicists have produced "optical cooling" by shining a laser onto glass that contains ytterbium ions (Yb 3+). The glass with ytterbium ions absorbs the laser photons and radiates photons with a shorter wavelength, as shown below. This process decreases the temperature of the glass with ytterbium ions. Laser photons Emitted photons IV\J\r- Glass with A.= 1 008 nm Yb3+ 1\N\r..t=995 nm "t Temperature decrease ""0.03° c One theory suggests that the cooling occurs because of electron movement between energy levels in the ytterbium ions, as shown below. If a ground state electron in an ytterbium ion absorbs a small amount of thermal energy, it moves to the second energy level (n = 2). The ion then absorbs the laser photon, which moves the electron to the excited state (n = 3). The cooling occurs when the ytterbium ion emits a photon. n =3 Exdtt>d stat£> Laser photons .., A~l~ f n~2 Emitted photons 1\N\r..t= 995 nm abso;::,~l Ground state :.L_ n = I 257. When the glass cools, the ions lose both the thermal energy and the energy that was absorbed from the laser photons. The electron energy level transition that occurs is from energy level A. B. C. D. n = 3 ton= n = 3 ton= n = 2 ton= n = 2 ton= 2 I 1 3 - 96- !Numerical Respons~ ~58.1 The frequency of the laser photons, expressed in scientific notation, is a.b x 1ocd Hz. The values of a, b, c, and dare _ _ , _ _ , _ _ , and (Record all four digits of your answer in the numerical-response section on the answer sheet) 259. The energy difference between a laser photon and an emitted photon is A. 2.oo x B. 1.97 X C. 2.58 X D. 8.62 x 1o- 19 J 10- 19 J 10-21 J 1o-33 J 260. Visible light has frequencies that range between 4.3 x 10 14 Hz (red) and 7.5 x 10 14 Hz (violet). Which of the following statements best describes the absorbed laser photon and the emitted photon in the optical cooling experiment? A. B. C. D. Both photons are in the infrared range. Both photons are in the ultraviolet range. Both photons are in the visible light range. One photon is in the visible light range, and one is not in the visible light range. - 97- Use the following information to answer the next two questions. An electron beam enters the region between two oppositely charged parallel plates near the negatively charged plate, as shown below. The plates are 0.150 m long and are 0. 040 m apart. There is an electrical potential difference of 60.0 V across the two plates which creates an electric field in the region. The path of the electrons just touches the edge of one of the plates as the beam exits the region between the parallel plates. Initial path of---liO' beam of electrons -1 1 0.040m 60.0V I ~-------0.150 m -------~ Physics Principles: I. Uniform motion (balanced forces) 2. Uniformly accelerated motion (unbalanced forces) 3. Circular motion (unbalanced forces) 4. Work-energy theorem 5. Conservation ofmomentum 6. Conservation of energy 7. Conservation of mass-energy 8. Conservation of charge 9. Conservation ofnucleons 10. Wave-particle duality !Numerical Respons~ ~ The two physics principles required to detennine the minimum initial speed of an electron in the beam, in the order of their use are _ _ and _ _ . (Record your two-digit answer in the numerical-response section on the answer sheet) 1262.1 The minimum initial speed of an electron in the beam, expressed in scientific notation, is - - - - x 10"' m/s. (Record your three-digit answer in the numerical-response section on the answer sheet) - 98- Use the following information to answer the next two questions. Electrons are accelerated from rest by an electric potential difference of 3. 00 x 102 V. These electrons enter a region that contains mutually perpendicular electric and magnetic fields. The electric field has a magnitude of8.50 x 10 5 N/C and the magnetic field has a magnitude of 2.00 x 10- 1 T. 3.00 X 102V ~~~~ Positively charged plate Path of electrons I x x x x~x X X Negatively charged plate x Represents a magnetic field directed into the page. Physics 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Principles: Uniform motion (balanced forces) Uniformly accelerated motion (unbalanced forces) Circular motion (unbalanced forces) Work-energy theorem Conservation ofmomentum Conservation of energy Conservation of mass-energy Conservation of charge Conservation ofnucleons Wave-particle duality !Numerical Respons~ 1263.1 The two physics principles required to detennine the instantaneous acceleration experienced by an electron as it first enters the region containing the perpendicular fields, in the order of their use are~~ and~-· (Record your two-digit answer in the numerical-response section on the answer sheet) 1264.1 The instantaneous acceleration experienced by an electron as it first enters the region containing the perpendicular fields, expressed in scientific notation, IS X 10" m/s 2 . (Record your three-digit answer in the numerical-response section on the answer sheet) - 99- Use the following information to answer the next two questions. The sun produces energy through nuclear fusion. In one particular reaction, energy is released when a hydrogen-2 nucleus fuses with a hydrogen-3 nucleus. This produces a helium-5 nucleus that is unstable and that decays to a helium-4 nucleus and a neutron. The fusion reaction chain is iH + fH ~ ~He ~ 6n + ~He. The masses of two of these particles are given in the table below. Isotope Notation Mass (x 10·27 kg) Helium-4 ~He 6.648 84 Neutron 6n 1.674 93 Particle The decay of helium-5 to helium-4 and a neutron forms an isolated system. In this system, the energy equivalence of the mass defect is observed as an increase in the system's kinetic energy. A helium-5 nucleus, at rest, decays. Both the neutron and the helium-4 nucleus move away from the location of the decay. The helium-4 nucleus has a momentum of I. 903 06 x 10-20 N·s and a kinetic energy of 2. 723 50 x 10- 14 J. Physics 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Principles: Uniform motion (balanced forces) Uniformly accelerated motion (unbalanced forces) Circular motion (unbalanced forces) Work-energy theorem Conservation ofmomentum Conservation of energy Conservation of mass-energy Conservation of charge Conservation of nucleons Wave-particle duality !Numerical Respons~ !265.1 The two physics principles required to determine the mass of a helium-5 nucleus, in the order of their use are ~~ and ~~· (Record your two-digit answer in the numerical-response section on the answer sheet) !266.1 The mass of a helium-5 nucleus, expressed in scientific notation, is x I o-" kg. (Record your three-digit answer in the numerical-response section on the answer sheet) - 100-