RI 20 Nuclear Physics tutorial solutions

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 Tutorial 20 Nu clear Physics Suggested Solutions DISCUSSION QUESTIONS Nuclear Reactions D1 (a) (i) α-particle comes to a stop at distance of closest approach and then reverses its motion along the same line (zero impact parameter b) (ii) α-particle experiences a large deflection when it passes close to the nucleus (small impact parameter b) (iii) α-particle experiences little deflection when it passes some distance from the nucleus (large impact parameter b) very small, very massive, positive nucleus (not to scale) RafflesInstitution (b) Observation Conclusion Most α-particles experienced little to no deflection Results as expected in Thomson’s plum - pudding model of atom. A small fraction of α -particles experienced large deflection ; a few α- particles are deflected at angles ° 170 . Large deflection possible only if the nucleus is massive (99.99% mass of atom). The very small fraction indicates that the nucleus is very small ( −− − 14 1510 10 m )

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 D2 (a) Neutrons are neutral and cannot be measured directly using electric or magnetic mass spectrometer. Note Experimentally, we can determine the energy or speed of a charged particle using either electric or magnetic mass spectrometer (for low energy) by measuring the amount of deflection, provided the mass of the particle is known. For high energy charged particles, an electromagnetic calorimeter is more appropriate. When a charged particle enters the calorimeter, it initiates a particle showers, i.e., the original charge particle transfers all its energy to other particles (via electromagnetic interaction). The energy of the particle shower is then deposited within the calorimeter and measured. https://www.desy.de/~garutti/LECTURES/ParticleDetectorSS12/L10_Calorimetry.pdf (b) Maximum recoil speed of the target nucleus occurs when the neutron collides head - on with the target nucleus. (c) Consider the head-on collision between a neutron and the target nucleus. mn m mn m vn v un before collision after collision Principle of conservation of momentum: = +nn nnmu mv m v  Since the collision is elastic, = −nnu vv ⇒ = −nnv vu  Substituting  into , ( )= −+nn n nm u m v u mv ⇒ += 2 n n n m v mvu m Since the target nuclei has mass 1m and 2m with recoil speed 1v and 2v , respectively, ++= =1 11 2 22 22 nn n nn mv mv mv mvu mm ⇒ −= − 11 22 21 n mv mvm vv

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 (d) The data for the collision are as follow: p 1.00 mu= ; 71 p 3.30 10 m sv −= × N 14.0 mu= ; 61 N 4.70 10 m sv −= × Substituting the date into the equation × ×− × ×= ×− × −×= =−× 76 67 7 7 1.00 3.30 10 14.00 4.70 10 4.70 10 3.30 10 3.28 10 1.16 2.83 10 n uum u u D3 Nuclear equation: →26 4 13 2H + Li 2 He Decrease in mass ∆ = + −× = 2.013553 6.013476 2 4.001505 0.024019 muu u u Since the total momentum of the system is zero, the α -particles must have equal and opposite momentum and hence have the same energy. Energy of each α-particle ( ) − − = ∆= × ×× × × = × 22 27 811 22 12 0.024019 1.66 10 3.00 10 1.79 10 J E mc D4 (a) Nuclear equation: +→ + +235 1 141 92 1 92 0 56 36 0U n Ba Kr 3 n (b) Nuclear fission (c) Energy released = − =×+×−×= ≈ products reactantsB.E. B.E. 141 8.5 92 8.5 235 7.6 194.5 195 MeV E (d) Kinetic energies of products ( 141 56Ba, 92 36Kr and neutrons) and gamma radiation.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 D5 (a) Decrease in mass ∆= − = ×−= Deuterium-atom Helium-atom2 2 2.014102 4.002603 0.025601 mm m uuu (b) Number of deuterium atoms = ×= × × = × 23 231.0 6.02 10 102.0 2.9914102 A R mNN M Number of nuclear fusion reaction is × 234941. 10 . Energy released from 1.0 g of deuterium ( ) −= ×× ×× × × = × 223 27 8 11 1. 10 0.025601 1.66 10 3.00 10 5.7 10 J 494 2 E Note: Answer given is different from above as it was calculated using approximated mass of each nucleus. (c) Mass of deuterium needed × ×××= ×÷ =× 6 11 82 g100 10 24 60 60 1.0 0.40 37. 5.7 10m Radioactivity D6 (a) (i) A spontaneous decay is one of which the probability of decay of a nucleus is unaffected by any external factors such as temperature, pressure or chemical composition. (ii) Random means that it is impossible to predict when a nucleus will decay. This results in fluctuations in the count rate measured by a detector. (b) Since cobalt -60 nucleus decays into a different nucleus (nicke l-60) and no nuclei escape the box, the number of nuclei in a box remains constant. Correct definition The half-life of a radioactive nuclide is the time taken for the number of undecayed nuclei to decay to half its original number.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 D7 (a) (i) The fluctuating count rate shows that radioactive decay is random. To minimize uncertainty in the determination of half -life, we must draw a line of best fit through these data points. (ii) The relatively constant count rate (subject to random fluctuations) after 400 st= is the background count rate. After a sufficiently long time, the count rate due to the radioactive source becomes negligible. Background count rate must be subtracted from the total count rate to obtain the true count rate due to the radioactive source. Note Background count rate is usually determined in by measuring the count rate with a radiation detector in the absence of a radioactive source. (b) The background count rate is approximately 8.0 Bq. Using the equation 12ln2 00 tttA Ae Aeλ −−= = and rearranging, ( ) 12 0 ln2 ln tt AA=− We read off two points ( )0.0,58.0 and ( )275.0,10.0 from the line of best fit. Accounting for background count rate, we have ( )0.0,50.0 and ( )275.0,2.0 . So ( ) 12 275.0 ln2 59.2 sln 2.0 50.0t ×=−= OR Actual count rate of source 158 8 50 s −= −= For 1 half life, count rate of source should drop to 25 s–1. 0 10 20 30 40 50 60 0 50 100 150 200 250 300 350 400 450 count rate/ s−1 time/ s

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 This should display on the graph as 125 8 33 s −= += Time taken to reach 133 s 60 s− = D8 (a) (i) Decay constant is the probability per unit time of the decay of a nucleus. (ii) Decay constant λ − −−= = = × 1 91ln2 0.131 yr 4.14 10 s5.3 (b) (i) Number of nuclei in 2.0 mg of 60 27 Co −×= ×= × × = × 3 23 19 0 2.0 10 6.02 10 2.01 1060 A R mNN M (ii) Initial activity of 60 27 Co λ −==×××=× 9 19 10 00 4.14 10 2.01 10 8.32 10 BqAN (iii) Initial power output from 2.0 mg of 60 27 Co −−==××=× =× 10 11 1 2 00 8.32 10 2.62 2.18 10 MeV s 3.49 10 WP AE (iv) Activity after 21.2 years ( )= = × 21.2 5.3 91 0 2 5.20 10 BqAA Power output from 2.0 mg of 60 27 Co after 21.2 years −−== ×× = × = ×9 10 1 35.20 10 2.62 1.36 10 MeV s 2.18 10 WP AE (c) Most of the γ rays are not absorbed by internal organs as γ rays have low ionizing power OR γ rays are highly penetrative. (d) 1. The decay of c obalt-60 generates radioactive wastes which must be properly stored for a long time in secure facilities. Linear accelerator do not generate radioactive waste. 2. The radiation from cobalt -60 is emitted in all directions and is thus difficult to focus them on the tumor. The X -rays from accelerators can be focused more easily. From wiki The radiation dose in photon radiation therapy (gamma rays or x -rays) is measured in grays and varies depending on the type and stage of cancer being treated. Radiation dose is defined as the energy absorbed per unit mass of the irradiated matter.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 7 One gray (Gy) is one joule of radiation energy per kilogram of matter. For curative cases, the typical dose for a solid epithelial tumo ur ranges from 60 to 80 Gy, while lympho