Alpha particles are very ionising So alpha particles have very low penetrating power Or so alpha particles will be absorbed/stopped by the skin

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1 (a)(i) (a)(ii) (a)(iii) Alpha particles are very ionising So alpha particles have very low penetrating power Or so alpha particles will be absorbed/stopped by the skin Gamma rays are very penetrating Or Gamma rays will pass through the skin Handled using (long) tongs Or never handled directly Or (closed source) pointed away from people (b) Kept in a lead-lined box (when not being used) We cannot be sure which nuclei will decay next/when Or All nuclei will (eventually) decay We know that the activity halves in a fixed period of time Or We can calculate the activity using A Ae t Or We know that the activity decreases exponentially Or Probability of decay is constant for a source Total for Question 7

2 (a) (b)(i) (b)(ii) (b)(iii) 4 7 N n C 3 H Top line correct Bottom line correct 6 Background radiation would increase the count rate (by a constant amount) Or Background count rate has to be subtracted (from the activity) Record the count for a long period of time Or Record the count more than once and find an average value Use of t/ ln Use of A A e t Correct time identified (65 years) A = 4 Bq Or Use of Correct time identified (65 years) Use of A = 4 Bq / Example of calculation ln year t.3year / A A e t (c)(i).8bq A 563year 65year e.8bq A 4.Bq.57 Mass difference calculation 4 Conversion to kg Use of E c m ΔE =.8 - (J) Example of calculation Δm = ( ) u ( ) u =.89 u Δm =.89 u.66 7 kg u - = kg ΔE = c Δm = (3 8 m s - ) kg =.8 - J 4

3 (c)(ii) MAX Very high temperatures [accept T~ 7 K] so that nuclei have sufficient energy to come close enough to overcome electrostatic repulsion [accept reference to strong interaction] A collision rate large enough to sustain fusion (from a very high density) Total for Question 4

4 3(a) Activity is the rate of decay of (unstable) nuclei Or activity is the number of (unstable) nuclei that decay in unit time 3(b)(i) 3(b)(ii) Background radiation/count will increase the recorded count Or background count must be subtracted from the recorded count Or background radiation contributes systematic error to the count [Do not accept to correct for background radiation ] Radioactive decay is a random process (so count for a fixed period will vary) [Ignore references to spontaneous, accurate, reliable] Idea that repeating enables a mean/average value to be calculated 3(b)(iii) ln Use of t Use of A A e t [allow.5 Bq for A here; allow use of N N e t ] A =.47 Bq [Allow calculation of number of half lives elapsed 3 and use of A A t t / for mp and mp] Example of calculation: ln d t 8.d A A e t 6.38 e 866d 3d 6.38Bq.74.47Bq 3(b)(iv) Idea that people have to be close to or ingest seaweed for any degree of risk Or β particles are moderately ionising Or β particles can enter body through the skin The half-life is short Or after a month the activity has decayed to negligible levels Or the radioisotope doesn t remain in the seaweed for very long Total for Question 9

5 4(a)(i) Ionising radiation removes electrons from atoms/molecules 4(a)(ii) Least ionising most ionising γ β α 4(b)(i) Paper.5 cm aluminium.5 cm lead α radiation stopped stopped stopped β radiation passes through stopped stopped γ radiation passes through passes through passes through 3 4(b)(ii) (There is the possibility of) exposure to neutrons Uncharged particles are not (directly) ionising Total for question 7

6 5(a)(i) 5(a)(ii) Top line correct Bottom line correct 3 4 H H He n Attempt at mass deficit calculation E =.75 GeV (accept.8 - J) Example of calculation: m = ( ) GeV/c =.75 GeV/c ΔE =.75 GeV 5(a)(iii) Momentum is conserved Mass of neutron is smaller, so speed is greater E k = ½ mv, so E k is larger Or Momentum is conserved E k = p /m m of neutron is smaller, so E k is larger 3 5(b) ln Use of t Use of A A e t t = 4 (years) Example of calculation: ln year t.3 year A A e t ln A A ln. t.563 year 4.9 years 3

7 *5(c) QWC Work must be clear and organised in a logical manner using technical wording where appropriate There is little possibility of a runaway fusion reaction (unlike fission) There would not be any radioactive waste produced in the fusion process Or the flux of neutrons would produce radioactive isotopes when absorbed by materials in the reactor A very/extremely high temperature (plasma) is required Plasma must not touch reactor walls, so strong magnetic fields are required If plasma touches the walls of the reactor its temperature falls (and fusion stops) 5 Total for question 5