Subject: Nuclear and Particle Physics, Code: 2825/04

Size: px

Start display at page:

Download "Subject: Nuclear and Particle Physics, Code: 2825/04"

Erick Bates
5 years ago
Views:

1 Subject: Nuclear and Particle Physics, Code: 2825/04 Session: Jan... Year: Mark Scheme (sixth draft, operational) MAXIMUM MARK (including common question) 90 43

2 ADVICE TO EXAMINERS ON THE ANNOTATION OF SCRIPTS. Please ensure that you use the final version of the Mark Scheme. You are advised to destroy all draft versions. 2. Please mark all post-standardisation scripts in red ink. A tick ( ) should be used for each answer judged worthy of a mark. Ticks should be placed as close as possible to the point in the answer where the mark has been awarded. The number of ticks should be the same as the number of marks awarded. If two (or more) responses are required for one mark, use only one tick. Half marks (½) should never be used. 3. The following annotations may be used when marking. No comments should be written on scripts unless they relate directly to the mark scheme. Remember that scripts may be returned to Centres. x ^ bod ecf con sf = incorrect response (errors may also be underlined) = omission mark = benefit of the doubt (where professional judgement has been used) = error carried forward (in consequential marking) = contradiction (in cases where candidates contradict themselves in the same response) = error in the number of significant figures 4. The marks awarded for each part question should be indicated in the margin provided on the right hand side of the page. The mark total for each question should be ringed at the end of the question, on the right hand side. These totals should be added up to give the final total on the front of the paper. 5. In cases where candidates are required to give a specific number of answers, (e.g. give three reasons ), mark the first answer(s) given up to the total number required. Strike through the remainder. In specific cases where this rule cannot be applied, the exact procedure to be used is given in the mark scheme. 6. Correct answers to calculations should gain full credit even if no working is shown, unless otherwise indicated in the mark scheme. (An instruction on the paper to Show your working is to help candidates, who may then gain partial credit even if their final answer is not correct.) 7. Strike through all blank spaces and/or pages in order to give a clear indication that the whole of the script has been considered. 8. An element of professional judgement is required in the marking of any written paper, and candidates may not use the exact words that appear in the mark scheme. If the science is correct and answers the question, then the mark(s) should normally be credited. If you are in doubt about the validity of any answer, contact your Team Leader/Principal Examiner for guidance. 44

3 / = alternative and acceptable answers for the same marking point Abbreviations, ; = separates marking points annotations and NOT = answers which are not worthy of credit conventions used in the ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit Mark Scheme ecf = error carried forward AW = alternative wording ora = or reverse argument Question Expected Answers Marks (a) repulsion/attraction correctly labelled on axis; [] correct point N - where strong line crosses distance axis; at N (resultant) force is zero; () so neutrons must be at equilibrium; () any not just 'forces equal' correct point P; at P electrostatic and strong forces balance (or AW); [2] [2] (c) crosses axis at P; allow P on either curve if forces equal crosses e/s force line at point vertically above N; generally correct shape, entirely above strong line; [3] (d)(i) (F =) Q 2 /[4πε 0 (x) 2 ] allow (F =) Q Q 2 /[4πε 0 (x) 2 ] 25 = (.6 x 0-9 ) 2 / (4π x 8.85 x 0-2 [d] 2 ) subs. d = 3.0(3) x 0-5 m allow 3 x 0-5 m [] [2] 2(a) either produced in a nuclear (fission) reactor or bombard (natural) uranium with neutrons () uranium 238 (nucleus) absorbs / captures a neutron () product (uranium 239) undergoes β-decay () any 2 2 [2] alpha particle Pu -> 4 2He U each correct product nucleus gets () [] 2 [2] (c)(i) years / 7.57 x 0 s [] either λ = 0.693/24000 or N = N 0 (½) 9000/24000 equation(s) = 2.89 x 0-5 y - = 5 x 0 20 (½) subs. N = N 0 e -λt (= 3.85 x 0 20 ) = 5 x 0 20 exp(-2.89 x 0-5 x 9000 ) ( = 3.85 x 0 20 ) [2] 45

4 (d)(i) ratio = 4.0 original ratio N 240 /N 239 = 40 x 0 20 /(5 x 0 20 ) = 8 (ratio after 9000 years = 4 ) equal numbers after another = 8000 years so total time = = years [] [3] 3(a) plasma is electron-nuclei/ion mixture; give nuclei enough energy to overcome Coulomb barrier; nuclei get close enough to fuse; plasma: matter at high temperature / fourth state of matter; () electrons stripped off atom / pulled away from nucleus; () high temperatures: give electrons energy to escape from nucleus; () explains Coulomb barrier - repulsion between positive nuclei; () confinement: increases plasma density / brings nuclei closer together; () methods: gravitational (eg in Sun) detail: strong gravity field pulls plasma together; () inertial; detail: (inward-acting) lasers push plasma together / compress plasma; () magnetic; detail: nuclei / ions / electrons spiral along B lines; () any 3 3 [9] (b) either number of atoms of 4 2He in.0 kg = ( / 0.004) x 6.02 x 0 23 ( =.5 x 0 26 ) so total energy =.5 x 0 26 x 28.4 x.6 x 0-9 x 0 6 = 6.9 x 0 4 J (accept 6.8 x 0 4 J) or mass of 4 2He = 4 x.67 x 0-27 (= 6.68 x 0-27 kg) so number of 4 2He in.0 kg = / (6.68 x 0-27 ) (=.50 x 0 26 ) () and energy generated =.50 x 0 26 x 28.4 MeV (= 4.25 x 0 33 ev) () = 4.25 x 0 33 x.6 x 0-9 J = 6.8 x 0 4 J () [3] 4(a) equation H + 2 H -> 3 2He (+ energy) [] ke of nuclei converted to (electric) potential energy all ke is converted [2] pe = (.6 x 0-9 ) x (.6 x 0-9 ) / (4π x 8.85 x 0-2 x 3.07 x 0-3 ) correct charge subs. () correct remaining subs. () (= 7.5 x 0-6 J) 2 [2] (iii) initial momentum = final momentum or equivalent m u - (2m) v = 0 (so u = 2v ) [2] 46

5 (iv) ke = ½ m v 2 ke of deuterium = ½ (2 m) v 2 (= m v 2 ) ke of hydrogen = ½ m (2v) 2 (= 2m v 2 ) so deuterium has / 3 of total ke ) hydrogen has 2 / 3 ) working so ke of hydrogen = 2 / 3 x 7.5 x 0-6 = 0.50 x 0-5 J ) ke of deuterium = / 3 x 7.5 x 0-6 = 0.25 x 0-5 J ) 2 [4] 5(a) principle of acceleration: charged particles move between electrodes at different voltages / charged electrodes / through electric field; magnetic fields to bend beam(s) / exert motor force; either magnetic field strength is increased as particles accelerate / gain. energy or frequency of accelerating voltage changes to synchronise with. particles; linear accelerator / linac for injecting particles into synchrotron; () accelerators on circular path to increase / maintain speed / energy of particles; () (magnetic force) provides centripetal force / force perpendicular to path; () magnetic field for focusing (particle beam); () any 2 2 [5] (b) energy of 0 0 Z = m c 2 = (.63 x 0-25 ) x (3.0 x 0 8 ) 2 (=.46 x 0-8 J ) =.46 x 0-8 / (.6 x 0-9 x 0 9 ) (= 9.7) GeV subs. [2] (c)(i) minimum particle energy = (9.7)/2 = 45.8 GeV or 92/2 = 46 GeV because both particle energies are used / available and 0 0Z (particle) can (in theory) be at rest / have zero energy after collision product particles must have some ke because initial mtm. (of system) not zero, so final mtm. not zero [3] [2] 47

6 6(a) hadron baryon lepton neutron proton electron neutrino 4 lines correct 2/2: 3 lines correct /2: 2 or line correct 0/2 2 [2] (iii) (iv) 0-5 minutes - any value within range weak force / interaction d -> u + e - + ν (-bar) omits e - or ν loses each (u) (u) (d) (d) charge: - / 3 (+ 2 / 3 - / 3 ) -> 2 / 3 (+ 2 / 3 - / 3 ) - (+0) baryon number: / 3 (+ / 3 + / 3 ) -> / 3 (+ / 3 + / 3 ) + 0 (+0) nuclear values: charge 0 = - (+ 0) and baryon no. = + 0 gets /2 [] [] 2 [2] [2] (c)(i) arrowed line plus 'resultant' / p r label anti- () neutrino () is emitted carries away some momentum () shows neutrino momentum vector () any 3 [] 3 [3] 7. As for synoptic question in Telecommunications unit. [20] 48

2825/04 Nuclear and Particle Physics. June Mark Scheme

2825/04 Nuclear and Particle Physics June 2003 Mark Scheme The following annotations may be used when marking: X = incorrect response (errors may also be underlined) ^ = omission mark bod = benefit of