Answers to Courseook questions Chapter J 1 a Partiles are produed in ollisions one example out of many is: a ollision of an eletron with a positron in a synhrotron. If we produe a pair of a partile and its antipartile eah of mass M we will need an energy of at least M. This energy will e provided y the eletron positron pair. Their total energy will e ( me + E K ). Hene ( me + EK) = M EK = M me. Thus, if M is to e large the kineti energy must large as well. Using the result in a, EK = M me = 175GeV 0.511MeV 175GeV. This is the kineti energy of the eletron and we need an equal amount for the positron for a total of 350 GeV. To see something one must use a wavelength that is at least of the same order of magnitude as the size that is eing investigated. If the wavelength is too large the ojet annot e seen. Therefore photons and other partiles used to proe a struture must have a de Broglie wavelength that is of the same size as the struture itself. For elementary partiles of sizes smaller than nulear diameters ( 15 m ) we therefore h need very small wavelengths. These neessarily imply high energies sine λ = for E h photons and λ = for partiles with mass. me 34 8 h 6.63 3 15 3 a The de Broglie wavelength is λ = = =.5 m. 9 19 E 50 1.6 i Yes, sine it is omparale to a nuleus diameter. ii No, sine it is larger than a nuleon size. 4 The de Broglie wavelength is given y λ = h p = h. me K The alpha partile has the larger mass and therefore the smaller de Broglie wavelength, so it would e more suitale. 5 a Synhrotron radiation is eletromagneti radiation emitted y a harged partile that is aelerating. This radiation is a onstraint on the aelerator eause a fration of the energy provided for aeleration is atually lost as radiation. Synhrotron radiation is more pronouned in irular mahines. 6 At the same energy the lighter partile would radiate more. In a lina, the length would e muh larger to ahieve aeleration of a heavy partile suh as the proton to energies omparale to those ahieved y eletrons. 7 Neutrons eing eletrially neutral annot e aelerated y eletri fields. Copyright Camridge University Press 011. All rights reserved. Page 1 of 8
8 The question has a typo. It should say Why an magneti fields not e used to aelerate partiles. a Magneti fields produe a fore that is at right angles to the veloity of the partile and so this fore does zero work. Hene it annot hange the kineti energy and hene speed of the partile. They are used to deflet the partiles into irular paths. 9 One reason is to provide shielding for the synhrotron radiation and another is to e far from virations from aove. With a onstant alternating soure frequeny, the time taken for the voltage in the next tue to hange sign is onstant. Hene the partile spends the same time travelling in any one tue. However, as the partile moves down the line it is moving faster eause it is eing aelerated and so the length of the tues must e inreasing. Eventually the partiles are moving lose to the speed of light, and so the speed is pratially onstant and therefore so is the length of the tues. 11 In a ylotron, the partile to e aelerated is plaed at the entre of an area in etween opposite poles of a magnet. The partiles are emitted with a small initial speed from the entre and then egin to follow a irular path inside the region of magneti field. The lower magnet is split into two parts (the Ds) and an alternating potential differene is applied etween the two split Ds. Every time a partile rosses the Ds it is aelerated. For this to happen the frequeny of the applied potential differene must equal the frequeny of revolution of the partile in the field. This frequeny is onstant and does not depend on the speed of the partile. With inreasing speed, the partile follows a path of inreasing radius (a spiral path) and eventually emerges out of the region of the magneti field with the help of additional magneti fields. 1 The proton inreases its energy y 30 kev every time it rosses the D s. To ahieve a 5 3 kineti energy of 5 MeV it must ross 833 times, and this orresponds to 30 aout 833 417 revolutions. Copyright Camridge University Press 011. All rights reserved. Page of 8
13 a From page 735 in Physis for the IB Diploma, we know that the ylotron 19 qb 1.6 1. 7 frequeny has to e f = = = 1.8 Hz. 7 πm π 1.67 The maximum kineti energy is E K = q B R m = (1.6 19 1. 0.0) 1.67 7 = 4.41 13 J =.8MeV. The proton inreases its energy y 35 kev every time it rosses the Ds. To.8 3 ahieve a kineti energy of.8 MeV it must ross = 80 times, and this 35 orresponds to 80 40 = revolutions. 14 The potential differene affets the numer of revolutions the partile will make in the region of the magneti field. As shown in the text (see page 735 in Physis for the IB Diploma), the final energy of the partile is determined y how large the radius of the ylotron is. 15 The magnets are used to end the harged partiles in the eam into a irular path. The radius of the irular path in a magneti field is given y R = mv qb. As the speed inreases, the radius would inrease as well. However, in a synhrotron the partiles follow a irular path of fixed radius. This an happen only if the strength of the magnets inreases in suh a way so as to keep the radius fixed. The partiles are aelerated y eletri fields in etween the magnets. 16 From R = mv, we see that dereasing the B field y a fator of 4 would inrease the qb radius y a fator of 4, i.e. the irumferene would have to e 4 7 = 8km. 17 a The separation of two onseutive unhes is 7 3 3000 = 4.5m. The unhes have a relative speed of twie the speed of light and so the time for 4.5 two to meet is 3 = 7.5 8 9 s. The frequeny is just the inverse of the meeting time, i.e. 1 f = 7.5 = 1.33 9 8 Hz. d The numer of ollisions per seond is N = 0 f =.7 9 s 1. Copyright Camridge University Press 011. All rights reserved. Page 3 of 8
18 a mv mv p E qvb = R = = =. R qb qb qb 1 19 E 7.0 1.6 B = = = 5.5T. qr 19 3 8 1.6 4.6 3 19 The great advantage of the synhrotron is that muh higher energies an e ahieved. The linear aelerator would require a very great length to ahieve a omparale energy. In a synhrotron, all of the total energy of the olliding partiles is availale energy to produe the rest energy of new partiles. In a lina the target is stationary and some of the energy must e provided to give kineti energy to the produts in addition to rest energy. Another advantage is that the time of the ollision an e ontrolled (unlike the situation in a lina). 0 In a olliding eam experiment all the energy of the olliding partiles an go into produing new partiles. In a stationary target experiment, the energy must go into produing the partiles as well as providing them with suffiient kineti energy to satisfy momentum onservation. In the olliding eam, no kineti energy needs to e provided to the reated partiles eause the total momentum efore the ollision is zero. 1 The disadvantage of olliding eam experiments is the low proaility of ollision and the very large perentage of energy lost as synhrotron radiation. The total energy is 938 = 1876MeV. Eah photon must arry away half of this energy i.e. 938 MeV. h 6 19 Hene, = 938MeV = 938 1.6 J, λ i.e. λ = 6.63 34 3 8 938 6 1.6 = 1.3 19 15 m. 3 We use EA = m E+ ( m ), where m is the mass of the target (the proton). Here, E A = 938 + 940 + 140 = 018 MeV. Hene, 018 = 938 E + 938 E = 133MeV. The kineti energy of the proton is then K = 133 938 = 95 MeV. (The answer on page 817 in Physis for the IB Diploma gives the total energy of the proton.) 4 We use EA = m E+ ( m ). The availale energy must e and so 16( m ) = m E+ ( m ) m E = 14( m ), i.e. E EA = 4m, = 7m. 5 The energy required to produe the partiles is E A = 938 + 135 = 011MeV. Hene the total energy of eah of the protons must e E = 011 = 05.5 MeV. The kineti energy is then E K = 05.5 938 = 67.5 MeV. Copyright Camridge University Press 011. All rights reserved. Page 4 of 8
6 We use EA = M E+ ( M ) + ( m ), where M is the mass of the target (the proton). Here, E A = 1193 + 498 = 1691MeV. Hene 1691 = 938 E + 938 + 140 E = 45 MeV. 7 Detetors are plaed around the ollision point in a layered fashion. They egin with a drift hamer, whih will determine the paths of partiles. This is followed y alorimeters that measure the energy of the partiles and end with muon detetors. The first thing that the experimenter wants to know is the diretion of motion of the produed partiles. This is done in the drift hamers, whih is why they are put losest to the interation point. It is important that in the drift hamer (see Q30) the eletri fields do not hange the kineti energy and the diretion of the produed partiles. 8 They measure the energy of a partile. They are typially sheets of lead or iron with gas in etween. The partile will ause the emission of other partiles as it enters the alorimeters. In the eletromagneti alorimeter, ionization aused y eletrons, positrons and photons in the gas inside the alorimeter is deteted. The amount of ionization is related to the energy of the partile. Hadrons pass through the eletromagneti alorimeter relatively unaffeted and enter the thiker hadroni alorimeter that surrounds the eletromagneti alorimeter. Its onstrution is similar to the eletromagneti alorimeter the differene is that in the eletromagneti alorimeter it is the eletromagneti fore that is responsile for reating the seondary partiles, whereas in the hadroni alorimeter it is the strong interation. Muons, eing very penetrating, pass through the eletromagneti and hadroni alorimeters relatively unaffeted and so the muon detetors are plaed in the outer layers. 9 In a ule hamer, a liquid (usually this is liquid hydrogen) is kept near its oiling point. A harged partile entering the hamer will produe ions along its path and ules will e formed along the path. This is photographed giving an image of the path. A magneti field ends the paths of the partiles, and the radius of the path gives information on the momentum of the partile. Copyright Camridge University Press 011. All rights reserved. Page 5 of 8
30 Consider the diagram elow, whih shows six fine wires (represented y the irles). The wires are into the plane of the page. The wires on the left are kept at a negative potential and those on the right at a positive potential. Therefore there is an eletri field etween the wires. These wires are in a hamer and the hamer is filled with a gas. _ + eletrons drifting _ + _ + path of harged partile A harged partile enters the hamer along the path shown. Assume for simpliity that the path is along the plane of the page. The partile will ionize the gas, and eletrons will drift towards the wires with the positive potential. By suitale hoie of gas and wire eletri potentials the eletron drift speed an e made onstant (or almost onstant). The arrival of the eletrons at eah wire is registered as a small urrent and the arrival time is reorded (the unertainty in the arrival time is less than 1 ns). Sine the drift speed is onstant and known, we an dedue the distanes the eletrons travelled to get to the wire. To fully reonstrut the path, we need an additional set of wires at right angles to the first set. The two sets of wires together (a grid) an e used to find the position from whih the eletrons were emitted, i.e. the position of the harged partile on a plane. A seond grid of wires parallel to the first an then e used to loate the position of the partile in three dimensions. 31 Drift hamers have two main advantages over the ule hamer. The first is that no photographs are taken. The images are digital and so are analysed y a omputer. With ule hamer photographs thousands of hours were spent analysing the images. The seond advantage is that there is no dead time: with the ule hamer a ertain amount of time had to pass from taking one photograph to the next. Copyright Camridge University Press 011. All rights reserved. Page 6 of 8
3 The question is admittedly vague apologies! The idea is to use the priniples of energy and momentum onservation to the proess that leads to the reation of a partile. We an make the question more onrete and only for those taking the relativity option y asking the following question. You will need to use the relativisti formula E = (m ) + p. This is a homework exerise for the good physis student something like this will not appear on an IB exam. It also assumes knowledge of hapter J3 (see pages 746 757 in Physis for the IB Diploma). New question 3 3 The diagram is a opy of part of a ule hamer photograph showing a pion ( π ) olliding with a stationary proton and produing a neutral kaon and an unknown 0 partile X: p + π K + X. The kaon susequently deays into two pions and partile X deays in a proton and a pion. The dashed lines showing the paths of the K and the X are not part of the photograph and are inferred y the experimenter. pions inoming pion K X proton pion a Explain why the proton that was omarded y the pion does not show in the photograph. Explain why the paths of the K and the X also did not show in the photograph. Explain whether the deay of the kaon is given y 0 0 0 K π + π. + or 0 + K π π d The inoming pion had a momentum of 90 MeV 1. The kaon is produed with momentum 40 MeV 1 at an angle of 1 o to the diretion of the pion. Calulate the mass of the unknown partile X. Copyright Camridge University Press 011. All rights reserved. Page 7 of 8
e Partile X is a hadron. Is it a meson or a aryon? (You are given the masses: proton: 938 MeV, pion: 135 MeV kaon: 498MeV.) and Answer a d e Beause it is stationary and so auses no ionization. Beause they are neutral and so ause no ionization. 0 + It is a deay into harged pions, K π + π. The traks of the neutral pions would not show up in the photograph. Or, we see that the pions are turning in opposite diretions in the magneti field of the ule hamer. This implies that they are oppositely harged. Applying momentum onservation. You should find the per apita expense on this type of researh and ompare it to other expenditures, for example defene. 90 = 40 os1 o + p x p x = 685MeV 1 0 = 40sin1 o p y p x = 50MeV 1 The momentum of the X is therefore p X = p x + p y = 685 + 50 = 687 MeV 1. The total energy efore the ollision is mp + ( mπ ) + pπ = 938 + 135 + 90 = 1868 MeV. The total energy of the produed kaon is ( mk ) + pk = 498 + 40 = 553MeV and so the total energy of the X is E = 1868 553 = 1315 MeV. Hene the rest energy of the X is mx = E px = 1315 687 = 110MeV. The mass is m X = 110MeV. e To onserve aryon numer, X has to e a aryon. (The X is in fat the Λ 0 aryon). 33 This is for you! You may want to onsider the atual ost per apita for this researh and then ompare it to the ost per apita for military expenditures of various ountries. You may also want to refer to the importane of nulear magneti resonane imaging and PET sans in mediine and how these tehniques originated in nulear and partile physis. Copyright Camridge University Press 011. All rights reserved. Page 8 of 8