Atomic Physics Problems Set Solutions
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SPH3U1 - Energy Problems Set 2 Problems 1. Write the nuclear reaction equation for the following nuclei undergoing the indicated type of decay: a) 210 84ππ b) 45 20πΆπ → 45 21ππ c) 46 24πΆπ → 46 23π d) 152 ∗ 66π·π¦ e) 240 94ππ’ → 236 92π + 42π»π g) 214 83π΅π → 214 84π΅π f) 58 ∗ 29πΆπ’ → 58 29πΆπ’ + 00πΎ h) 22 11ππ → 22 10ππ → 206 82ππ → + 42π»π a. b. c. d. e. f. g. h. Nucleus Polonium-210 Calcium-45 Chromium-46 Dysprosium-152 Plutonium-240 Copper-58 Bismuth-214 Sodium-22 + −10π½ + 01π½ 152 66π·π¦ + 00πΎ Decay type alpha decay beta-negative decay beta-positive decay γ decay α decay gamma decay β- decay β+decay + −10π½ + 01π½ 2. The following shows the products of part of a decay series known as the Thorium Series. Write a nuclear reaction equation for each decay stage of the series and identify its type. 232 228 228 228 224 220 90πβ → 88π π → 89π΄π → 90πβ → 88π π → 86π π 232 228 4 90πβ → 88π π + 2π»π 228 88π π → 228 89π΄π + −10π½ 228 89π΄π → 228 90πβ + −10π½ 228 90πβ → 224 88π π + −10π½ 224 88π π → 220 86π π + 42π»π Half-life 3. The half-life of carbon-14 is 5730 a. An ancient piece of bark is estimated to have originally contained 2.00 µg of carbon-14. If its age has been determined to be 1.7x104 a, how much carbon-14 does it now have? 1 π‘⁄ π1/2 π = π0 (2) π= 1.7×104 π⁄ 1 ( 5730 π) (2.00 ππ) ( ) 2 π = 0.2558 ππ π = 0.26 ππ 4. A specimen of peat from an ancient bog contains 800.0 µg of carbon-14. What will be the amount of carbon-14 remaining after 1.0 x104 a? 1 π‘⁄ π1/2 π = π0 (2) 1.0×104 π⁄ 1 ( 5730 π) 2 π = (800.0 ππ) ( ) π = 238.6 ππ π = 240 ππ SPH3U1 - Energy Problems Set 5. Uranium-238 has a half-life of 4.468x109 a. A sample of uranium ore currently has 722 mg of uranium238. If the sample is 5.9x108 a, how much uranium-238 was there when it formed? 1 π‘⁄ π1/2 π = π0 (2) π0 = π0 = π0 = 791.20 ππ π π0 = 790 ππ π‘ 1 ⁄π ( ) 1/2 2 722 ππ 5.9×108 π⁄ ) 1 ( 4.468×109 π ( ) 2 6. Iodine 131 53πΌ is used in diagnostic and therapeutic techniques in the treatment of thyroid disorders. This isotope has a half-life of 8.04 days. What percentage of an initial sample of 131 53πΌ remains after 30.0 days? 1 π‘⁄ π1/2 π π0 π = π0 (2) π π0 π π0 1 π‘⁄ π1/2 = (2) = 1 ( (2) = 0.07529 % πππππππππ = 7.529 % = 7.53 % 30.0 πππ¦π ⁄8.04 πππ¦π ) 7. To make the dial of a watch glow in the dark, 1.000 ×10-9 kg of radium 226 88π π is used. The half-life of this isotope has a value of 1.60 ×103 yr. How many kilograms of radium disappear while the watch is in use for fifty years? 1 2 π‘⁄ π1/2 π = π0 ( ) 50 π⁄ ) 1 ( 1.60×103 π π = (1.000 × 10−9 ππ) (2) π = 9.7857 × 10−10 ππ π ππππ ππππππππ = 1.000 × 10−9 ππ − 9.7857 × 10−10 ππ π ππππ ππππππππ = 2.14 × 10−11 ππ 8. The shroud of Turin is a religious artefact known since the Middle Ages. In 1988 its age was measured using the radiocarbon dating technique, which revealed that the shroud could not have been made before 1200 AD. Of the 146πΆ nuclei that were present in the living matter from which the shroud was made, what percentage remained in 1988? 1 π‘⁄ π1/2 π = π0 (2) 1 π‘⁄ π1/2 π π0 = (2) π π0 = (2) π π0 = 0.9080 1988 π−1200 π⁄ 1 ( 5730 π) % πππππππππ = 90.80 % = 91 % SPH3U1 - Energy Problems Set Nuclear Fission 9. During a nuclear reaction, an unknown particle is absorbed by a copper 63 29πΆπ’ nucleus, and the reaction products are 62 πΆπ’ , a neutron, and a proton. Determine the unknown particle and the nucleus 29 63 temporarily formed when the copper 29πΆπ’ nucleus absorbs the unknown particle. 63 29πΆπ’ ?? + ?? ??? ? → ??? ? → 62 29πΆπ’ 63 29πΆπ’ + 11π+ → → 64 30ππ + 10π + 11π+ 62 29πΆπ’ + 10π + 11π+ 10. A neutron causes 232 90πβ to fission according to the reaction 232 1 π΄ 0 90πβ + 0π → ππ + 0πΎ a. Identify the unknown nucleus , giving its atomic mass number A, its atomic number Z, and the symbol X for the element. 232 1 233 0 90πβ + 0π → 90πβ + 0πΎ b. Nucleus X then undergoes β- decay, and its daughter does too. Identify the final nucleus, giving its atomic mass number, atomic number, and name. 233 233 0 90πβ → 91ππ + −1π½ 233 91ππ → 233 92π + −10π½ 11. A small sample of coal, when completely converted to energy, releases 4.5 × 1014 J of energy. Determine the original mass of coal, assuming the final mass is zero. πΈ = βππ 2 π = 5.0068 × 10−3 ππ πΈ π2 − π1 = π 2 π = 5.0 π 4.5×1014 π = (2.998×108 π½ π/π )2 12. Calculate the mass defect, in kilograms, and the energy released, in joules, of the following nuclear reaction: 140 235 1 93 1 92π + 0π → 55πΆπ + 37π π + 3 0π 93 = 235.044 π’, 10π = 1.009 π’, 140 55πΆπ = 139.909 π’, 37π π = 92.922 π’ βπ = ππππππ’ππ‘π − ππππππ‘πππ‘π βπ = ππΆπ + ππ π + 3ππ − (ππ + ππ ) βπ = 139.909 π’ + 92.922 π’ + 3(1.009 π’) − (235.044 π’ + 1.009 π’) βπ = −0.195 π’ βπ = 0.195 π’ πππ π 235 92π πΈ πΈ πΈ πΈ = βππ 2 = (0.195 π’)(1.660538782 × 10−27 ππ/π’)(2.998 × 108 π/π )2 = 2.91036 × 10−11 π½ = 2.91 × 10−11 π½ SPH3U1 - Energy Problems Set 13. How much energy, in MeV, is released by the following fission reaction? 235 92π 235 92π = 235.044 π’, 10π = 1.009 π’, 96 38ππ + 10π → 94 38ππ 1 + 140 54ππ + 2 0π = 93.915 π’, 140 54ππ = 139.922 π’ βπ = ππππππ’ππ‘π − ππππππ‘πππ‘π βπ = πππ + πππ + 2ππ − (ππ + ππ ) βπ = 93.915 π’ + 139.922 π’ + 2(1.009 π’) − (235.044 π’ + 1.009 π’) βπ = −0.198 π’ βπ = 0.198 π’ πππ π πΈ πΈ πΈ πΈ = βππ 2 = (0.198 π’)(930 πππ/π’) = 184.14 πππ = 184 πππ 2 14. The energy released during each 235 92π fission is 2.0 × 10 MeV. How many kilograms of coal must be 7 burned to produce the same energy as 1.0 kg of 235 92π if 3.0 × 10 J of energy is released when 1.0 kg of 24 coal is burned? (n.b. 2.6 x10 U nuclei are found in a 1 kg sample of uranium) ππ ππ’ππππ ππ ππ’ππππ ππ ππ’ππππ ππ ππ’ππππ = (πππ π ππ π235 ) ÷ (πππππ πππ π ππ π235 ) × (6.02 × 1023 ππ‘πππ /πππ) = (1.0 ππ) ÷ (0.235 ππ/πππ) × (6.02 × 1023 ππ‘πππ /πππ) = 2.56170 × 1024 = 2.6 × 1024 πΈ = (ππππππ‘ππππ )(πΈπππππ¦/πππππ‘πππ) πΈ = (2.6 × 1024 )(2.0 × 102 πππ)(1.602 × 10−13 π½/πππ) πΈ = 8.3304 × 1013 π½ πΈπππππ¦ ππππ’ππππ π’πππππ€π πππ π ππππ πΈπππππ¦ = 1.0 ππ π’πππππ€π πππ π ππππ π’πππππ€π πππ π ππππ π’πππππ€π πππ π ππππ π’πππππ€π πππ π ππππ ππππ = (πΈπππππ¦ ππππ’ππππ)/(πΈπππππ¦/ππ) = (8.3304 × 1013 π½)/(3.0 × 107 π½/ππ) = 2776800 ππ = 2.8 × 106 ππ SPH3U1 - Energy Problems Set Nuclear Fusion 15. One reaction to produce carbon in stars is as follows: 4 4 4 12 2π»π + 2π»π + 2π»π → 6πΆ Calculate the mass defect, in atomic mass units, and energy released, in electronvolts, of this reaction. 12 ( 42π»π m=4.002603 u, 6πΆ m = 12.00000 u) βπ = ππππππ’ππ‘π − ππππππ‘πππ‘π βπ = ππΆ − 3(ππ»π ) βπ = 12.00000 π’ − 3(4.002603 π’) βπ = −7.809 × 10−3 π’ βπ = 7.8 × 10−3 π’ πππ π πΈ πΈ πΈ πΈ = βππ 2 = (7.8 × 10−3 π’ )(930 πππ/π’) = 7.254 × 106 ππ = 7.3 × 106 ππ 16. Consider the induced nuclear reaction 21π» + 147π → 126πΆ + 42π»π . The atomic masses are 21π» (2.014 102 u), 147π (14.003 074 u), 126πΆ (12.000 000 u), and 42π»π (4.002 603 u). Determine the energy released, in joules, when the 126πΆ and 42π»π nuclei are formed in this way. βπ = ππππππ’ππ‘π − ππππππ‘πππ‘π βπ = ππΆ + ππ»π − (ππ + ππ» ) βπ = 12.00000 π’ + 4.002 603 π’ − (14.003 074 π’ + 2.014 102 π’) βπ = −0.014573 π’ πΈ πΈ πΈ πΈ = βππ 2 = (0.014573 π’ )(930 πππ/π’)(1.602 × 10−13 π½/πππ) = 2.17117 × 10−12 π½ = 2.2 × 10−12 π½ 17. Two deuterium atoms ( 21π»= 2.014102 u) react to produce tritium ( 31π»= 3.016050 u) and hydrogen ( 11π»= 1.007825 u) according to the reaction: 2 2 3 1 1π» + 1π» → 1 π» + 1π» What is the energy released by this deuterium-deuterium reaction? βπ = ππππππ’ππ‘π − ππππππ‘πππ‘π βπ = ππ»−3 + ππ» − 2(ππ»−2 ) βπ = 3.016050 π’ + 1.007825 π’ − 2(2.014102 π’) βπ = −4.329 × 10−3 π’ πΈ πΈ πΈ πΈ πΈ = βππ 2 = (4.329 × 10−3 π’ )(930 πππ/π’) = 4.02597 πππ = 4.0 πππ = 6.4 × 10−3 π½ SPH3U1 - Energy Problems Set 18. One proposed fusion reaction combines lithium 63πΏπ (6.015 u) with deuterium 21π» (2.014 u) to give helium 4 2π»π (4.003 u): 2 1π» + 63πΏπ → 2 42π»π a) How many of the above reactions would need to occur to supply the energy needs of one household for a year (estimated to be 3.8 × 1010 J)? βπ = ππππππ’ππ‘π − ππππππ‘πππ‘π βπ = 2ππ»π − (ππΏπ + ππ»−2 ) βπ = 2(4.003 π’) − (6.015 π’ + 2.014 π’) βπ = −0.023 π’ πΈ = βππ 2 πΈ = (0.023 π’)(930 πππ/π’)(1.602 × 10−13 π½/πππ) πΈ = 3.426678 × 10−12 π½ ππππππ‘ππππ = ππππππ‘ππππ = π‘ππ‘ππ ππππππ¦ ππππππ¦ πππ πππππ‘πππ 3.8×1010 π½ 3.426678×10−12 π½/πππππ‘πππ 22 ππππππ‘ππππ = 1.10895 × 10 πππππ‘ππππ ππππππ‘ππππ = 1.1 × 1022 πππππ‘ππππ b) How many grams of lithium would be needed if there are 8.673x1025 Li nuclei in 1 kg of a lithium sample? π₯ ππ 1 ππ = 1.1×1022 ππ’ππππ 8.673×1025 ππ’ππππ (1 ππ)(1.1×1022 ππ’ππππ) π₯= 8.673×1025 ππ’ππππ −4 π₯ = 1.268304 × 10 πππ π πΏπ = 0.13 π ππ 19. The following shows the carbon-nitrogen-oxygen (CNO) cycle that occurs inside stars. Write a nuclear reaction equation for each stage in the cycle and identify its type. 12 13 13 14 15 15 12 4 6πΆ → 7π → 6πΆ → 7π → 8π → 7π → 6πΆ + 2π»π 12 1 13 nuclear fusion ( + gamma particle released) 6πΆ + 1π» → 7π 13 7π → 13 6πΆ 13 6πΆ + 11π» → 14 7π nuclear fusion ( + gamma particle released) 14 7π + 11π» → 15 8π nuclear fusion ( + gamma particle released) 15 8π → 15 7π + 11π» → 15 7π + 01π½ beta+ decay + + 01π½ 12 6πΆ + 42π»π beta+ decay nuclear fusion ( + alpha particle released) SPH3U1 - Energy Problems Set Conceptual Questions 1. To which of the following objects, each about 1000 yr old, can the radiocarbon dating technique not be applied: a wooden box, a gold statue, and some plant seeds? Explain. Wooden box and plant seeds. Both of these objects contain carbon, so can be used for radiocarbon dating, whereas a gold statue does not. 2. Some preserved seed were found and their age, through the technique of carbon dating, was determined to be 5000 years old. However, new evidence suggests that the amount of carbon in the air at that time is higher than previously thought. Are the seeds older or younger than 5000 years? Explain. Older. If there were more C-14 in the beginning but that same amount currently, this would mean that the seeds have lost more C-14 than previously thought. In order to do so, they must have existed for a longer amount of time, so are older than 5000 years. 3. Compare and contrast nuclear fission and fusion. How are they alike? How are they different? Fission and fusion are similar because both processes result in more stable nuclei. They both achieve this by releasing energy which can be used as a power source. They differ in that a large nucleus breaks into smaller fragments when undergoing fission and smaller nuclei combine to make a larger one when undergoing fusion. Fusion also, in general, releases more energy per reaction than fission. 4. Why are neutrons such good projectiles for producing nuclear reactions? Neutrons are good projectiles for producing nuclear reactions because they are neutral and they are massive. If you want a particle to hit the nucleus with a lot of energy, a more massive particle is the better choice. A light electron would not be as effective. Using a positively charged projectile like an alpha or a proton means that the projectile will have to overcome the large electrical repulsion from the positively charged nucleus. Neutrons can penetrate directly to the nucleus and cause nuclear reactions. 5. Explain why nuclear fusion is much more difficult to obtain than nuclear fission To fuse two nuclei together they must have enough energy (travelling fast enough) to overcome their mutual repulsion. To break apart a large nucleus, a neutron with relatively low energy can be used to strike and destabilize it as the neutron has no charge and is not repelled. 6. Why is nuclear fusion more desirable than nuclear fission for power generation? Explain. Fusion releases more energy per reaction than does fission, and so it is more efficient. Also, the products of fusion are stable nuclei, and have minimal environmental impact, whereas fission products are radioactive (and can be for thousands of years) and so storing them pose serious environmental (and technological) challenges. 7. The energy from nuclear fission appears in the form of thermal energy – but the thermal energy of what? Explain. The thermal energy from nuclear fission appears in the kinetic energy of the fission products (daughter nuclei and neutrons). In other words, the fission products are moving very fast (especially the neutrons).
