AP Physics Quantum Mechanics

Transcription

1 AP Phyic Quantum Mechanic All of thi ection i tuff that you want to pay attention to. Okay? At the turn of the century, phyic wa in a terrible fix. What wa the problem? Well, it wa with the behavior of electron. Claical phyic had gotten quite good at dealing with motion, orbit, &tc. A phyicit could it down and figure out exactly what and where an electron hould be doing. Unfortunately, the rotten electron did not oblige the phyicit with doing what it wa uppoed to be doing. Ernet Rutherford pictured the atom a a ort of miniature olar ytem. The electron circled the nucleu in orbit, jut like the planet circle the un. According to claical, Newtonian phyic, a centripetal force act on the electron, accelerating it toward the nucleu. Thi force i caued by the charge difference between the poitive nucleu and the negative electron. The problem wa thi: accelerated electron emitted light. Thu a the electron were accelerated toward the nucleu, they hould radiate photon of light. Thi would caue them to loe energy, which would give them a maller orbit, they would emit light, move cloer to the nucleu a their orbit get maller, and o on. In about 10-9 econd the electron would piral into the nucleu. Thi meant that atom a we know them could not exit in a table tate. Rutherford Model Unfortunately for claical phyic, but fortunately for the univere, thi doe not happen. Atom actually do manage to hang around for time in exce of a billionth of a econd. Actually they eem quite capable of hanging around for billion of year. Clearly omething wa going on with the behavior of electron that did not fit into claical kinematic. Black Body Problem: All object in the univere contantly emit electromagnetic wave. Thi i a fact of nature. You, a you read thi, are quite happily emitting maive number of photon from all over your body into the univere. You, blighter that you are, have done thi your entire life! People motly emit long wavelength infrared. The army ha thee night viion cope that detect the IR ignature given off by humanoid, thu enabling the oldier to ee in the dark. The frequency of the emitted electromagnetic wave i a function of temperature and only temperature. The higher the temperature of the body, the higher the frequency (and the horter 573

2 the wavelength) of the emitted electromagnetic wave. It doen t matter what the object i made from, for a given temperature it will emit a given frequency of light. Alo of coure, the higher the frequency of the emitted photon, the greater their energy. At the turn of the century phyicit were looking at black bodie. A black body i one of thoe ideal thing that phyicit love to invent. (One definition of claical phyic i that it deal with elephant with zero volume, no friction, acting independently of gravity. Do you think that i fair?) The black body i defined a an object that aborb all the light that i incident on it. Imagine that you have a hollow object any hollow object will do. Let ay you have a white platic ball. It white on the outide and white on the inide becaue it made of white platic. Okay? Got it? Now you make a mall hole in the ball. Having done that, you now look into the hole what do you ee? Well, the whole look black. The inide of the ball look black, even though we know that it i actually white. So what the deal? The interior of the ball behave a if were a black body - it aborb jut about all the electromagnetic wave that enter it. Viible light i aborbed and doe not get out. In fact the only radiation that doe come out of the hole from the inide of the thing i infrared. The infrared i emitted becaue of the temperature of the ball. Thu doe the tupid hole in the ball behave a a black body. Star are conidered to be black bodie (even though they certainly aren t black). Planet can be treated a black bodie a well. Below i a graph made up of the emiion from a blackbody at three different temperature. The area under the curve repreent the total radiation. Each curve ha a peak wavelength thi i the wavelength at which mot of the energy i emitted. For K you can ee that the amount of radiation i much greater than for the lower temperature curve. The K curve alo ha a peak wavelength, but it ha a greater value than the one for the higher temperature the wavelength of the emitted radiation i longer, which mean that it ha a lower frequency. The 000 K curve peak wavelength i much maller in amplitude and longer in wavelength. It frequency i lower a well. Intenity Wavelength The general rule i that the intenity and frequency of emitted radiation increae with temperature. Thi i een in tar and planet. Planet, which are very cool, don t even emit 574

3 viible light, they can only manage infrared. Cool tar give off motly red light, warmer tar give off yellow light, and the hottet tar give off blue light. Claical mechanic cannot explain thee curve. It work for the long wavelength, but a the wavelength decreae (and the frequency increae), claical mechanic prediction become very bad, very bad indeed. In fact the old theory predict that the intenity of the emitted radiation will approach infinity a the wavelength near zero. You can ee that thi doe not happen. The curve how that a the wavelength get cloe to zero, the intenity alo approache zero. Thi i one of your baic contradiction. Since claical mechanic cannot explain what actually happen, phyicit were forced to abandon the law of Newton and develop a new theory that would explain the data. The theory that came out of thi i known a quantum mechanic. Max Planck ( ) wa a German phyicit who pent a great many year trying to puzzle out thi problem. Planck wa trying to find a fundamental law that would decribe the energy emitted by blackbodie. He eventually got the job done, but to get hi law to work, he had to aume that the radiation, which everyone knew wa a wave, wa actually made up of little packet of energy (which everyone knew wa not true). He called thee packet quanta Intenity Claical Theory Planck Theory (ingular) and quantum (plural). There wa no evidence for the quanta, except that they made hi law work. Uing thi cobbled up thing, he wa able to explain the blackbody radiation curve and calculate accurate energy for emitted radiation. In other word, the cobbled up thing actually worked! Planck believed that the quanta were merely an artificial, mathematical device without reality that jut happened to make hi equation yield accurate reult. Planck equation (which we have een before remember that the Phyic Teacher promied to reviit it, well, that time i here) i: E hf Wavelength Here, E i energy, f i the frequency, and h i known a Planck' contant. It value i: 34 h6.63 x10 or 15 h ev You will have both of thee value when you take the AP Phyic tet. 575

4 What energy i carried by a photon of electromagnetic radiation that ha a frequency of 1.55 x Hz? E hf 6.63 x x x x10 The energy of a photon can alo be calculated a a function of wavelength. Wavelength i related to frequency by: v f Thi i the equation for the peed of the wave. c v f o cf f E hf E hc E hc You won t be provided with thi equation, o you need to be able to get there on your own. A photon ha a wavelength of 550 nm. How much energy doe thi repreent in oule? c hc v f o cf f E hf E 34 3 m 6.63 x x10 4 E x x x10 m Since the value of Planck contant multiplied by the peed of light i itelf a contant, we can treat hc a a contant. (Save u ome work!) Two uch value, uing different unit, will be provided to you on the AP Tet: 5 hc 1.99 x10 m hc 1.4 x10 3 ev nm Thi make olving the above problem a lot eaier. To wit: c hc 1.99 x10 m 4 v f f E hf 3.6 x x10 m 5 576

5 Planck' theory came to be called the quantum theory and proved o important, that it i conidered to be a waterhed in cience. All phyic before Planck equation i called claical phyic and all phyic afterward i known a modern phyic. But what did all thi mean? Momentum and Light: You need to be able to calculate the momentum of a photon a a function of it frequency or wavelength. Okay, let do a typical problem. The momentum for a photon i given by thi equation: E hf pc What i the momentum of a photon that ha a wavelength of 455 nm? c hc hc v f f E hf E pc pc h 34 kgm 1 p 6.63 x x10 m kg m p x x kg m Photon and Power of a Source: Imagine that you have a ource of light that i rated at a certain power level. It produce photon of only one frequency. So, how many photon per econd would it produce? Thi i pretty imple. Power i imply the rate that energy i produced. The energy i in the form of photon. All you have to do i calculate the amount of energy produced in one econd. Then determine the amount to energy one photon repreent. Then divide the total energy by the energy per photon. Thi give you the number of photon in a econd. That lat part i really jut a dimenional analyi problem, ain t it? Okay, here a problem. Let go for it. A 505 nm light ource produce 0.50 W. How many photon per econd doe it kick out? E P E Pt t c hc v f f E hf 577

6 x10 m 19 E x x x10 m x10 photon 6.35 x10 photon 3.94 x10 photon Photoelectric Effect: Toward the end of the 19 th Century (in 1887 to be exact) Heinrich Hertz dicovered that certain metal would emit electron when light wa incident on them. Thi wa the firt intance of light interacting with matter and wa very myteriou. In 1905 Albert Eintein, a 3 rd Cla Technical Expert in the Swi Patent Office, the obcure phyicit (although he wa not a phyicit at the time, he wa a bureaucrat) mentioned before, publihed a paper which provided the explanation for the effect. The light wa actually made up of mall particle - Planck little bundle of energy he called the quanta. Thee particle are now called photon. The urface electron were bound to the metal with a mall amount of energy. Some of the incident photon would enter the urface, mack into atom of the metal and be totally aborbed. They would give their energy to an electron, which, if the aborbed energy wa great enough, could then break free from the atom. You can think of the photoelectric effect a being the reult of colliion between photon and electron, which knock the electron out of the metal. The amount of energy binding the electron to the metal i called the work function. The ymbol for thi i the Greek letter. Work Function Recall that: E hf Thi i the energy of the photon. The electron that ha been knocked out of the metal ha ome amount of kinetic energy. Thi kinetic energy ha to be le than the photon energy becaue ome of the energy added to the ytem wa ued to break the electron free of the metal (thi amount of energy i given by the work function). So the photon ha to provide more energy than the work function if the electron i to be et free. The maximum kinetic energy that an electron can have i jut the difference between the energy of the work function (the energy that bind the electron to the metal) and the energy of the photon. hf Thi equation will be provided to you on the AP Phyic Tet. 578

7 Each metal ha it own value for the work function. A handome table of uch value for elected metal ha been helpfully provided to you. Dear Doctor Science, What i a laer beam made of? -- Lauren Grace from Toledo, OH Dr. Science repond: Normal light i compried of zillion of photon. Laer light i made of futon, which are fat, tuffed photon with a zipper down the ide. Some have a foam core and thee are often mitakenly referred to a mu meon, which i jut a fancy oriental term for futon. A in retail advertiing, Science often give the proaic a new name to make it eem like thing are really happening when, in fact, everyone i jut playing Tetri on their office computer and waiting for lunch. Work Function for ome Different Metal Metal Sodium (Na).8 Aluminum (Al) 4.08 Copper (Cu) 4.70 Cobalt (Co) 3.90 Zinc (Zn) 4.31 Silver (Ag) 4.73 Platinum (Pt) 6.35 Lead (Pb) 4.14 Iron (Fe) 4.50 Work Function (ev) Wavelength and the Photoelectric Effect: We have an equation that relate the electron energy to frequency, but what about the wavelength of the photon? For ome reaon phyicit are not very fond of wavelength and prefer them frequencie. The frequency and wavelength are related by the peed of light. So when we want to find the value for the frequency we get: v f o c f we can ubtitute thi into the Planck equation and get: c hc E hf E h f c You can then plug thi into the photoelectric equation for the energy term: hc hf Of coure, thi equation you will not have for the AP Phyic tet. What wa trange about all thi i that the effect i baed on the energy of the photon, a function of it frequency or wavelength. The intenity of the light how trong the beam i, doe matter, but only if the frequency of the photon i high enough. Photon which have too low a frequency (or too long a wavelength) will not knock any electron looe no matter how intene the light i. The intenity i really a meaure of the number of photon that will be incident on the urface in a given amount of time. So if the frequency i large enough to caue the effect and 579

8 you increae the intenity, you will increae the photocurrent becaue there will be more photon hitting the metal to knock looe more electron. The kinetic energy of the electron i alo independent of the intenity of the light. More intene light will dilodge more electron, o the current will increae, but the kinetic energy of the electron will all be limited to the ame value (the maximum kinetic energy). What i the maximum kinetic energy of a photoelectron that ha been liberated from a ilver metal urface by a photon that ha a frequency of 3.13 x Hz? hf 4.14 x10 ev 3.13 x ev 8. ev What i the velocity of a photoelectron that ha been liberated from a zinc metal urface by a photon that ha a wavelength of 75 nm? Conulting the table, we find that the work function for zinc i 4.31 ev. We can ue thi and the wavelength of the incident photon to find the kinetic energy of the ejected electron. We can then olve for their velocity. c c hc vc f f E hf h hc 1.4 x10 3 ev nm K Max ev x ev 4.31eV 75 nm K 4.51eV 4.31eV 0.00 ev Max x ev 0.3 x10 3. x10 1eV 0 kg m 3.x10 1 K K mv v v 31 m 9.11 x10 kg m m m v x x10.65 x10 Typical Photoelectric Effect Experiment: A typical laboratory etup for a photoelectric experiment would conit of a metal plate that the light will be incident upon. Thi plate i called the emitter (E). Acro from the emitter i a plate called the collector (C). The 580

9 emitter i connected to the negative terminal of a variable dc voltage ource and the collector i connected to the poitive terminal of the ource. An ammeter i placed in erie with the collector/battery and a voltmeter i placed in parallel with the photoelectric element. A typical circuit i hown below. E e C A V Variable power upply Incident light trike the emitter, which caue photoelectric electron to be emitted. The electron are attracted to the poitively charged collector and a current i etablihed. We can then meaure the current and voltage. If the wavelength of the incident light i varied, but the intenity of the light i kept contant, then we get a graph of current VS. wavelength that look like thi: Current Wavelength Notice that the current i emitted only for wavelength le than l 0. For longer wavelength, no current i emitted. Thee repreent photon that don t have enough energy to knock the electron out of the metal. Thi maximum wavelength, l 0, i called the photoelectric threhold wavelength. l 0 581

10 The amazing thing i that the current doe not depend on the intenity of the light. Thi eem to make no ene. You would think that if you made the light brighter, you ought to get a bigger photoelectric current. It another intance where claical phyic fell apart. According to claical phyic, the incident wave would provide the energy to knock the electron out of the metal. The greater the intenity of the light, the more electron ought to be knocked looe. But thi didn t happen. For a great many wavelength, no photoelectric current would flow no matter how large the intenity. Current doe depend on intenity, but only for wavelength that caue the photoelectric effect. Effect of Collector Voltage: If the poitive voltage on the collector i increaed we oon get a maximum amount of current. If the intenity i increaed, we alo ee an increae in the current. But even for zero voltage on the collector, ome current will flow. But what happen if you make the collector voltage negative intead of poitive? The electron will be repelled from the collector. If the voltage i mall, ome will till make it, but a the voltage get more negative it become harder and harder for the electron to bridge the gap. More and more are turned away and the current fall off. At ome voltage value, none of the electron make it to the collector and current i zero. Thi i hown in a graph of Voltage VS. Current below. Current High Intenity Low Intenity -DV S Applied Voltage At V the current top completely. If V i le than or equal to V no electron reach the collector and all electron are repelled. V i called the topping potential. The topping potential i independent of the intenity of the light! The electron i accelerated through the electric field between the collector and emitter. The energy it gain i equal to the potential energy the electron tart with. The equation for thi energy i in the Electricity and Magnetim ection of the equation heet. It i given a 58

11 U qv E. So at the topping potential, the potential energy of the electron i equal the maximum kinetic energy. We can write: qv Note here that K Max i alo independent of the intenity of the light! If we look at a graph of frequency VS. kinetic energy, we ee it ha a traight line. There i a minimum frequency before the electron have any kinetic energy. The minimum frequency i called the cutoff frequency. The photon with a frequency le than f C don t have enough energy to dilodge the electron from the metal. The lope of the graph i h, Planck contant. The value for the cutoff frequency i imply the intercept on the x axi. The equation for kinetic energy a a function of frequency i: hf Thi i a linear equation and the value for it can be found from the graph. KE fc f Finding the Cutoff Frequency: The cutoff frequency i the minimum frequency that will generate photoelectron. So we ue the max kinetic energy equation. The minimum frequency occur when the kinetic energy i zero. hf hf f fc h So the cutoff frequency i: h 583

12 f C h A odium photoelectric urface with work function.3 ev i illuminated by electromagnetic radiation and emit electron. The electron travel toward a negatively charged cathode and complete the circuit hown above. The potential difference upplied by the power upply i increaed, and when it reache 4.5 V, no electron reach the cathode. (a) For the electron emitted from the odium urface, calculate the following. i. The maximum kinetic energy. qv K 1.6 x10 C 4.5V 7. x10 or Max 19 1eV 7. x ev x10 ii. The peed at thi maximum kinetic energy. 1 K mv 19 kg m 7. x10 K v 31 m 9.11 x10 kg 1.6 x10 6 m (b) Calculate the wavelength of the radiation that i incident on the odium urface. c c hc E hf c f f E h c c c h h h K Max 3 hc 1.4 x10 ev nm x10 nm 18 nm K 4.5 ev.3 ev Max 584

13 (c) Calculate the minimum frequency of light that will caue photoemiion from thi odium urface. hf But 0 for the minimum frequency, o 0 hf hf f h.3 ev f x10 Hz 5.55 x10 Hz x10 ev What i the cutoff wavelength for a copper metal urface? C hc 1.4 x10 3 ev nm 0.64 x10 3 nm 64 nm 4.70 ev Note that thi wavelength i much maller than viible light, o no photoelectric effect for copper with viible light thi would have to be like ultraviolet light. Finding the Work Function: To find the work function, et the kinetic energy to zero a above and olve for. The frequency i the cutoff wavelength, which i the minimum frequency. Recall that the work function i the minimum energy needed to break an electron out of the metal urface. K hf 0 hf hf Max nm light i incident on a metal urface. The topping potential i found to be V. (a) Find the work function for thi material and (b) the longet wavelength that will eject electron from the metal. (a) work function: c c hc E hf c f f E h K hc hc hc Max qv qv qv Now, when we plug in our value, we will tick in the ymbol e for the charge of an electron. Thi will get u ev a a unit, eventually. 585

14 1.4 x 10 3 ev nm e V x 10 ev ev 500 nm.48 ev ev.04 ev (b) longet wavelength Set the maximum kinetic energy equal to zero to get the longet wavelength. hc hc hc hc x10 3 ev nm x10 3 nm 608 nm 04 ev Stuff Claical Mechanic or Wave Theory Cannot Explain: No electron are emitted if the light frequency fall below ome cutoff frequency, f C Maximum kinetic energy i independent of the light intenity The electron are emitted almot intantaneouly KE Max increae with increaing frequency a it i function of hf Happen o fat becaue it i a one to one photon/electron deal Stopping Potential veru frequency: Stopping potential i a function of frequency. The higher the frequency, the higher the topping potential. On the following graph topping potential i plotted along the y axi while frequency i along the x axi. A you can ee, the intercept on the x axi repreent the topping potential for the cutoff frequency. At the cutoff frequency, the total energy i equal to the work function, o hf So you can find the work function if the cutoff frequency i known. At the topping potential, the energy of the electron i equal to the electron charge time the voltage. Thi i the potential energy gained by an electron in the field at the topping potential. E qv 586

15 But thi energy ha to equal the energy of the photon, o: hf qv q i the charge of an electron, o we can plug in e for the electron charge and write the equation a: olving for h/e hf ev h e V f Thu, h/e i imply the lope of the graph. V f c threhold frequency or cutoff frequency f AP Tet Item: In a photoelectric experiment, light i incident on a metal urface. Electron are ejected from the urface, producing a current in a circuit. A revere potential i applied in the circuit and adjuted until the current drop to zero. That potential at which the current drop to zero i called the topping potential. The data obtained for a range of frequencie are graphed below. 587

16 a. For a frequency of light that ha a topping potential of 3.0 volt, what i the maximum kinetic energy of the ejected photoelectron? Set the maximum kinetic energy equal to the potential energy gained by the electron in the electric field at the topping potential. K qv e 3.0V 3.0 ev Max b. From the graph and the value of the electron charge, determine an experimental value for Planck' contant. h Slope of graph i: e 19 h V 0V 1.6 x10 CV 0V h e Hz Hz x h 0.64 x x10 or x10 h V V1 V 0V e V h 14 e f f Hz x10 ev h 0.40 x10 ev 4.0 x10 ev x10 b. From the graph, determine the work function for the metal. 588

17 The graph i a traight line o: y mx b Plug in the value on the graph for the term in the traight line equation: h ymxb V f b ev hf b e But ev i K Max hf b o y intercept i work function Thi i.0 ev d. On the axe above, draw the expected graph for a different metal urface with a threhold frequency of 6.0 x hertz. More Important Stuff: If you thought that thi ection of the handout wa over, you are adly mitaken. There lot more tuff for you to tudy, mater, imprint on your brain, worry about, weat over, and be all around concerned with. So here we go. x-ray Production: x-ray, you will recall, are a group of electromagnetic wave, part of the old electromagnetic pectrum. But where do they come from? Target Filament High Voltage x-ray 589

18 One ource, and thi i how they were firt dicovered, i from bombarding a metal urface with high energy electron. What happen i that the electron collide with one of the metal atom. The electron ha enough energy to remove one of the inner-hell electron. When thi happen, an outer-hell electron mut drop down and fill the vacant energy hell. To do thi, it mut loe a large amount of energy, typically in exce of 10 5 ev. Thi energy i emitted in the form of a high-energy photon. Typical wavelength would be between and 0.1 nm. Thi i the x- ray region of the good old electromagnetic pectrum. Wilhelm Roentgen ( ) dicovered x-ray. He baically contructed a cathode ray tube. Thi i a large, long gla tube that i evacuated o that there i a vacuum on the inide of the thing. TV picture tube and computer monitor are baically cathode ray tube. Anyway, at one end of the tube wa a mall filament called a cathode. When electric current wa paed through the cathode, it gave off electron. Thi i called thermonic emiion. Anyway, thee electron are then accelerated to a high velocity with an electric field. Typical potential difference in uch a tube would be around to V. The high velocity electron collide with a metal target atom and x-ray are given off. Roentgen of coure didn t know that thi would happen. What he did notice wa that a phophorecent creen everal feet from hi tube began to glow brightly when the tube wa lit off. The glow continued even when he tuck a piece of wood between the tube and the creen. He concluded that ome very penetrating type of radiation wa being given off. Since he didn t know what it wa, he called them x-ray. x meaning unknown. Many phyic teacher in the pat had cathode ray tube that they would ue for demontration, not realizing that they were producing maive amount of x-ray. x-ray are produced by TV tube and computer monitor a well. The gla on the face of the tube contain lead, which aborb x-ray. Before they began doing thi, it wa dangerou to it too cloe to a TV et. Thee day it okay it only hurt your eye. 590

19 AP Phyic - Nuclear E Atomic ma unit -- unified ma unit, u 1 1u ma of C 1atom 1 7 1u x10 kg Find Binding E of deuterium Ma of tritium nucleu: u mp u mn u Add mae together: m u u u p Binding E: u u u MeV u MeV 1 u Ma of tritium i le than ma of part. Ma difference repreent energy releaed 591

20 Calculate the energy releaed when 1.05 kg of U-35 undergoe fiion. Each fiion produce 08 MeV. 3 1mol 6.0 x10 nuclei 3 N 1750 g 44.8 x10 nuclei 35 g 1 mol MeV 10 ev 1.6 x x10 nuclei 1nuclei 1MeV 1eV x x10 59

21 An untable nucleu that i initially at ret decay into a nucleu of fermium-5 containing 100 proton and 15 neutron and an alpha particle that ha a kinetic energy of 8.4 MeV. The atomic mae of helium-4 and fermium-5 are u and u, repectively. a. What i the atomic number of the original untable nucleu? Z 10 b. What i the velocity of the alpha particle? (Neglect relativitic effect for thi calculation.) 1 K mv K ev ev 7 kg 7 m4 u kg u 1 kg m K v.0110 m kg 7 m c. Where doe the kinetic energy of the alpha particle come from? Explain briefly. Ma Equivalence: The original nucleu decay into the product particle and energy. Energy Conervation: Potential or binding energy wa converted into kinetic energy. 593

22 d. Suppoe that the fermium-5 nucleu could undergo a decay in which a - particle wa produced. How would thi affect the atomic number of the nucleu? Explain briefly. Atomic number increae by one. A neutron convert into a proton and an electron Fm Md e A polonium nucleu of atomic number 84 and ma number 10 decay to a nucleu of lead by the emiion of an alpha particle of ma atomic ma unit and kinetic energy 5.5 MeV. (1 u = MeV/c = 1.66 x 10-7 kg.) a. Determine each of the following. i. The atomic number of the lead nucleu Number of Proton 84 8 ii. The ma number of the lead nucleu Number of Nucleon b. Determine the ma difference between the polonium nucleu and the lead nucleu, taking into account the kinetic energy of the alpha particle but ignoring the recoil energy of the lead nucleu. E The kinetic energy of the alpha particle i the ma difference of the two nuclei. E 5.5MeV m m c c 7 MeV kg 30 m kg c MeV c mc 594

23 K c. Determine the peed of the alpha particle. A claical (nonrelativitic) approximation i adequate. 1 mv ev K 5.5 MeV MeV 1eV v K m v m The alpha particle i cattered from a gold nucleu (atomic number 79) in a "head-on" colliion. d. Write an equation that could be ued to determine the ditance of cloet approach of the alpha particle to the gold nucleu. It i not neceary to actually olve thi equation. e. At cloet approach KE goe to zero and electric potential goe to max (Throw omething up and KE i zero while PE i max. KE become U E ) 1 KE U E mv mv qv 1 q q k r q r k mv 595