PHY20004 Nuclear Physics Tutorial 2 Nuclear Fission & Fusion. Week 10 Quiz review A few concepts from the Assignment! Assignment discussion time

PHY20004 Nuclear Physics Tutorial 2

Nuclear Fission & Fusion

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Week 10 Quiz review

Week 10

Week 10 Quiz Review

A – The energy released is equivalent to the mass difference ∆7 = 89.907737 − 89.907151 = 0.000586 C, which is converted to F-G as 0.000586 × 931.5 = 0.546 F-G (see Class 3 slide 6).

Quiz 2

1) In the decay scheme J_,- _{→ L} ,.!- + N + O, what are the particles N and O? A. -/ _{and P} B. -. _{and Q̅} 0 C. -/ _{and Q} 0 D. -/ _{and Q̅} 0 C – This decay is representing D/ _{decay, in which a proton has transformed into a} neutron, reducing the atomic number " by 1 and keeping the mass number % fixed. D/ decay produces a positron and a neutrino (see Class 3 slide 7). 2) In a particular radioactive decay chain, a nuclide emits an .-particle, followed by two D. _{particles, followed by a S-particle.} A. The final nucleus is an isotope of the original nucleus B. The final nucleus is an isotone of the original nucleus C. The final nucleus is an isobar of the original nucleus D. The final nucleus is the same as the original nucleus A – The .-particle reduces the number of protons by 2 and the number of neutrons by 2. The two D. _{decays then convert 2 neutrons to protons. The S decay does not change} the neutron or proton numbers. Hence overall, the number of protons is unchanged and the number of neutrinos decreases by 4. The final nucleus is hence an isotope of the original nucleus (see Class 3 slide 11). Class 4 3) 4 × 10!' _{atoms decay with a half-life of 2.3 +A. How many are remaining after 3.7 +A?} A. 2.5 × 10!' B. 1.7 × 10!' C. 1.3 × 10!' D. 1.1 × 10!'

C – The radioactive decay law is ((U) = ((0) -.12_{, where W = 0.693/Y}

!/". Hence

((3.7 +A) = 4 × 10!' _{× -}.$.%+&×(!.#_$.!) _{= 1.3 × 10}!' _{(see Class 4 slide 3).}

Week 10 Quiz Review

A – The energy released is equivalent to the mass difference ∆7 = 89.907737 − 89.907151 = 0.000586 C, which is converted to F-G as 0.000586 × 931.5 = 0.546 F-G (see Class 3 slide 6).

Quiz 2

1) In the decay scheme J_,- _{→ L} ,.!- + N + O, what are the particles N and O? A. -/ _{and P} B. -. _{and Q̅} 0 C. -/ _{and Q} 0 D. -/ _{and Q̅} 0 C – This decay is representing D/ _{decay, in which a proton has transformed into a} neutron, reducing the atomic number " by 1 and keeping the mass number % fixed. D/ decay produces a positron and a neutrino (see Class 3 slide 7). 2) In a particular radioactive decay chain, a nuclide emits an .-particle, followed by two D. _{particles, followed by a S-particle.} A. The final nucleus is an isotope of the original nucleus B. The final nucleus is an isotone of the original nucleus C. The final nucleus is an isobar of the original nucleus D. The final nucleus is the same as the original nucleus A – The .-particle reduces the number of protons by 2 and the number of neutrons by 2. The two D. _{decays then convert 2 neutrons to protons. The S decay does not change} the neutron or proton numbers. Hence overall, the number of protons is unchanged and the number of neutrinos decreases by 4. The final nucleus is hence an isotope of the original nucleus (see Class 3 slide 11). Class 4 3) 4 × 10!' _{atoms decay with a half-life of 2.3 +A. How many are remaining after 3.7 +A?} A. 2.5 × 10!' B. 1.7 × 10!' C. 1.3 × 10!' D. 1.1 × 10!'

C – The radioactive decay law is ((U) = ((0) -.12_{, where W = 0.693/Y}

!/". Hence

((3.7 +A) = 4 × 10!' _{× -}.$.%+&×(!.#_$.!) _{= 1.3 × 10}!' _{(see Class 4 slide 3).}

This decay is representing 𝛽! decay, in which a proton has

Week 10 Quiz Review

A – The energy released is equivalent to the mass difference ∆7 = 89.907737 − 89.907151 = 0.000586 C, which is converted to F-G as 0.000586 × 931.5 = 0.546 F-G (see Class 3 slide 6).

Quiz 2

1) In the decay scheme J-_, → _,.!-L + N + O, what are the particles N and O? A. -/ _{and P} B. -. _{and Q̅} 0 C. -/ _{and Q} 0 D. -/ _{and Q̅} 0 C – This decay is representing D/ _{decay, in which a proton has transformed into a} neutron, reducing the atomic number " by 1 and keeping the mass number % fixed. D/ decay produces a positron and a neutrino (see Class 3 slide 7). 2) In a particular radioactive decay chain, a nuclide emits an .-particle, followed by two D. _{particles, followed by a S-particle.} A. The final nucleus is an isotope of the original nucleus B. The final nucleus is an isotone of the original nucleus C. The final nucleus is an isobar of the original nucleus D. The final nucleus is the same as the original nucleus A – The .-particle reduces the number of protons by 2 and the number of neutrons by 2. The two D. _{decays then convert 2 neutrons to protons. The S decay does not change} the neutron or proton numbers. Hence overall, the number of protons is unchanged and the number of neutrinos decreases by 4. The final nucleus is hence an isotope of the original nucleus (see Class 3 slide 11). Class 4 3) 4 × 10!' _{atoms decay with a half-life of 2.3 +A. How many are remaining after 3.7 +A?} A. 2.5 × 10!' B. 1.7 × 10!' C. 1.3 × 10!' D. 1.1 × 10!'

C – The radioactive decay law is ((U) = ((0) -.12_{, where W = 0.693/Y}

!/". Hence

((3.7 +A) = 4 × 10!' _{× -}.$.%+&×(!.#_$.!) _{= 1.3 × 10}!' _{(see Class 4 slide 3).}

Week 10 Quiz Review

A – The energy released is equivalent to the mass difference ∆7 = 89.907737 − 89.907151 = 0.000586 C, which is converted to F-G as 0.000586 × 931.5 = 0.546 F-G (see Class 3 slide 6).

Quiz 2

1) In the decay scheme J-_, → _,.!-L + N + O, what are the particles N and O? A. -/ _{and P} B. -. _{and Q̅} 0 C. -/ _{and Q} 0 D. -/ _{and Q̅} 0 C – This decay is representing D/ _{decay, in which a proton has transformed into a} neutron, reducing the atomic number " by 1 and keeping the mass number % fixed. D/ decay produces a positron and a neutrino (see Class 3 slide 7). 2) In a particular radioactive decay chain, a nuclide emits an .-particle, followed by two D. _{particles, followed by a S-particle.} A. The final nucleus is an isotope of the original nucleus B. The final nucleus is an isotone of the original nucleus C. The final nucleus is an isobar of the original nucleus D. The final nucleus is the same as the original nucleus A – The .-particle reduces the number of protons by 2 and the number of neutrons by 2. The two D. _{decays then convert 2 neutrons to protons. The S decay does not change} the neutron or proton numbers. Hence overall, the number of protons is unchanged and the number of neutrinos decreases by 4. The final nucleus is hence an isotope of the original nucleus (see Class 3 slide 11). Class 4 3) 4 × 10!' _{atoms decay with a half-life of 2.3 +A. How many are remaining after 3.7 +A?} A. 2.5 × 10!' B. 1.7 × 10!' C. 1.3 × 10!' D. 1.1 × 10!'

C – The radioactive decay law is ((U) = ((0) -.12_{, where W = 0.693/Y}

!/". Hence

((3.7 +A) = 4 × 10!' _{× -}.$.%+&×(!.#_$.!) _{= 1.3 × 10}!' _{(see Class 4 slide 3).}

The 𝛼-particle reduces the number of protons by 2 and the number of neutrons by 2. The two 𝛽" decays then convert 2 neutrons to protons. The 𝛾 decay does not change the neutron or proton numbers. Hence overall, the number of protons is unchanged and the number of

Week 10 Quiz Review

A – The energy released is equivalent to the mass difference ∆7 = 89.907737 − 89.907151 = 0.000586 C, which is converted to F-G as 0.000586 × 931.5 = 0.546 F-G (see Class 3 slide 6).

Quiz 2

1) In the decay scheme J-_, → _,.!-L + N + O, what are the particles N and O? A. -/ _{and P} B. -. _{and Q̅} 0 C. -/ and Q₀ D. -/ _{and Q̅} 0 C – This decay is representing D/ _{decay, in which a proton has transformed into a} neutron, reducing the atomic number " by 1 and keeping the mass number % fixed. D/ decay produces a positron and a neutrino (see Class 3 slide 7). 2) In a particular radioactive decay chain, a nuclide emits an .-particle, followed by two D. _{particles, followed by a S-particle.} A. The final nucleus is an isotope of the original nucleus B. The final nucleus is an isotone of the original nucleus C. The final nucleus is an isobar of the original nucleus D. The final nucleus is the same as the original nucleus A – The .-particle reduces the number of protons by 2 and the number of neutrons by 2. The two D. _{decays then convert 2 neutrons to protons. The S decay does not change} the neutron or proton numbers. Hence overall, the number of protons is unchanged and the number of neutrinos decreases by 4. The final nucleus is hence an isotope of the original nucleus (see Class 3 slide 11). Class 4 3) 4 × 10!' _{atoms decay with a half-life of 2.3 +A. How many are remaining after 3.7 +A?} A. 2.5 × 10!' B. 1.7 × 10!' C. 1.3 × 10!' D. 1.1 × 10!'

C – The radioactive decay law is ((U) = ((0) -.12, where W = 0.693/Y_!/". Hence ((3.7 +A) = 4 × 10!' _{× -}.$.%+&×(!.#_$.!) _{= 1.3 × 10}!' _{(see Class 4 slide 3).}

Week 10 Quiz Review

A – The energy released is equivalent to the mass difference ∆7 = 89.907737 − 89.907151 = 0.000586 C, which is converted to F-G as 0.000586 × 931.5 = 0.546 F-G (see Class 3 slide 6).

Quiz 2

1) In the decay scheme J-_, → _,.!-L + N + O, what are the particles N and O? A. -/ _{and P} B. -. _{and Q̅} 0 C. -/ and Q₀ D. -/ _{and Q̅} 0 C – This decay is representing D/ _{decay, in which a proton has transformed into a} neutron, reducing the atomic number " by 1 and keeping the mass number % fixed. D/ decay produces a positron and a neutrino (see Class 3 slide 7). 2) In a particular radioactive decay chain, a nuclide emits an .-particle, followed by two D. _{particles, followed by a S-particle.} A. The final nucleus is an isotope of the original nucleus B. The final nucleus is an isotone of the original nucleus C. The final nucleus is an isobar of the original nucleus D. The final nucleus is the same as the original nucleus A – The .-particle reduces the number of protons by 2 and the number of neutrons by 2. The two D. _{decays then convert 2 neutrons to protons. The S decay does not change} the neutron or proton numbers. Hence overall, the number of protons is unchanged and the number of neutrinos decreases by 4. The final nucleus is hence an isotope of the original nucleus (see Class 3 slide 11). Class 4 3) 4 × 10!' _{atoms decay with a half-life of 2.3 +A. How many are remaining after 3.7 +A?} A. 2.5 × 10!' B. 1.7 × 10!' C. 1.3 × 10!' D. 1.1 × 10!'

C – The radioactive decay law is ((U) = ((0) -.12, where W = 0.693/Y_!/". Hence ((3.7 +A) = 4 × 10!' _{× -}.$.%+&×(!.#_$.!) _{= 1.3 × 10}!' _{(see Class 4 slide 3).}

_{The radioactive decay law is 𝑁 𝑡 = 𝑁 0 𝑒}"#$_{, where 𝜆 = 0.693/𝑇}

%/'. Hence 𝑁 3.7 𝑦𝑟 = 4×10%(×𝑒").+,-×(!.#$.!) = 1.3×10%(.

Week 10 Quiz Review

4) The half-life of radioactive iodine-137 is 8.0 days. How many iodine nuclei are necessary to produce an activity of 1.0 Z![? A. 2.9 × 10+ B. 4.6 × 10+ C. 3.7 × 10!$ D. 7.6 × 10!" C – The radioactivity in decays per second is given by 78 72 = W( where Y!/" = 0.693/W. Here, 78 72 = 10 .% _{× 3.7 × 10}!$ _\.! _{= 3.7 × 10}# _\.!_{. Hence the number of nuclei is ( =} ! 1 78 72 = '×"#×&%$$ $.%+& × 3.7 × 10 # _{= 3.7 × 10}!$ _{(see Class 4 slide 6).} 5) A certain radioactive element has a half-life of 20 days. How long would it take for 7/8 of the atoms originally present to disintegrate? A. 20 days B. 40 days C. 60 days D. 80 days C – The fraction of atoms remaining after 20, 40, 60, 80 days is 1/2, 1/4, 1/8, 1/16, hence the fraction of atoms that have disintegrated at these times is 1/2, 3/4, 7/8, 15/16. Hence, 7/8 of the atoms have disintegrated after 60 days (see Class 4 slide 4). Class 5 6) What are the number of protons " and neutrons ( in the missing nuclide J in the following fission reaction? P + "&(0 → J + !#$!\ + 4P A. " = 55, ( = 37 B. " = 37, ( = 55 C. " = 92, ( = 37 D. " = 37, ( = 58 B – The total number of protons (") and neutrons (() on either side of the equation must match. The atomic number of 0 is " = 92, so the left-hand side corresponds to " = 92 and ( = 144. The atomic number of !\ is " = 55, so the right-hand side

Week 10 Quiz Review

4) The half-life of radioactive iodine-137 is 8.0 days. How many iodine nuclei are necessary to produce an activity of 1.0 Z![? A. 2.9 × 10+ B. 4.6 × 10+ C. 3.7 × 10!$ D. 7.6 × 10!" C – The radioactivity in decays per second is given by 78 72 = W( where Y!/" = 0.693/W. Here, 78 72 = 10 .% _{× 3.7 × 10}!$ _\.! _{= 3.7 × 10}# _\.!_{. Hence the number of nuclei is ( =} ! 1 78 72 = '×"#×&%$$ $.%+& × 3.7 × 10 # _{= 3.7 × 10}!$ _{(see Class 4 slide 6).} 5) A certain radioactive element has a half-life of 20 days. How long would it take for 7/8 of the atoms originally present to disintegrate? A. 20 days B. 40 days C. 60 days D. 80 days C – The fraction of atoms remaining after 20, 40, 60, 80 days is 1/2, 1/4, 1/8, 1/16, hence the fraction of atoms that have disintegrated at these times is 1/2, 3/4, 7/8, 15/16. Hence, 7/8 of the atoms have disintegrated after 60 days (see Class 4 slide 4). Class 5 6) What are the number of protons " and neutrons ( in the missing nuclide J in the following fission reaction? P + "&(0 → J + !#$!\ + 4P A. " = 55, ( = 37 B. " = 37, ( = 55 C. " = 92, ( = 37 D. " = 37, ( = 58 B – The total number of protons (") and neutrons (() on either side of the equation must match. The atomic number of 0 is " = 92, so the left-hand side corresponds to " = 92 and ( = 144. The atomic number of !\ is " = 55, so the right-hand side

excluding nuclide J contains " = 55 and ( = 89. Therefore, J contains " = 37 and ( = 55 (see Class 5 slide 4).

The radioactivity in decays per second is given by 12_1$ = 𝜆𝑁 where 𝑇_%/' = 0.693/𝜆. Here, 12_1$ = 10"+×3.7×10%) 𝑠"% = 3.7×103 𝑠"%. Hence the number of nuclei is 𝑁 = %_#12_1$ = (×'3×-+))_).+,- ×3.7×103 = 3.7×10%).

Week 10 Quiz Review

4) The half-life of radioactive iodine-137 is 8.0 days. How many iodine nuclei are necessary to produce an activity of 1.0 Z![? A. 2.9 × 10+ B. 4.6 × 10+ C. 3.7 × 10!$ D. 7.6 × 10!" C – The radioactivity in decays per second is given by 78₇₂ = W( where Y_!/" = 0.693/W. Here, 78₇₂ = 10.% _{× 3.7 × 10}!$ _\.! _{= 3.7 × 10}# _\.!_{. Hence the number of nuclei is ( =} ! 1 78 72 = '×"#×&%$$ $.%+& × 3.7 × 10 # _{= 3.7 × 10}!$ _{(see Class 4 slide 6).} 5) A certain radioactive element has a half-life of 20 days. How long would it take for 7/8 of the atoms originally present to disintegrate? A. 20 days B. 40 days C. 60 days D. 80 days C – The fraction of atoms remaining after 20, 40, 60, 80 days is 1/2, 1/4, 1/8, 1/16, hence the fraction of atoms that have disintegrated at these times is 1/2, 3/4, 7/8, 15/16. Hence, 7/8 of the atoms have disintegrated after 60 days (see Class 4 slide 4). Class 5 6) What are the number of protons " and neutrons ( in the missing nuclide J in the following fission reaction? P + "&(0 → J + !#$!\ + 4P A. " = 55, ( = 37 B. " = 37, ( = 55 C. " = 92, ( = 37 D. " = 37, ( = 58 B – The total number of protons (") and neutrons (() on either side of the equation must match. The atomic number of 0 is " = 92, so the left-hand side corresponds to " = 92 and ( = 144. The atomic number of !\ is " = 55, so the right-hand side

Week 10 Quiz Review

4) The half-life of radioactive iodine-137 is 8.0 days. How many iodine nuclei are necessary to produce an activity of 1.0 Z![? A. 2.9 × 10+ B. 4.6 × 10+ C. 3.7 × 10!$ D. 7.6 × 10!" C – The radioactivity in decays per second is given by 78₇₂ = W( where Y_!/" = 0.693/W. Here, 78₇₂ = 10.% _{× 3.7 × 10}!$ _\.! _{= 3.7 × 10}# _\.!_{. Hence the number of nuclei is ( =} ! 1 78 72 = '×"#×&%$$ $.%+& × 3.7 × 10 # _{= 3.7 × 10}!$ _{(see Class 4 slide 6).} 5) A certain radioactive element has a half-life of 20 days. How long would it take for 7/8 of the atoms originally present to disintegrate? A. 20 days B. 40 days C. 60 days D. 80 days C – The fraction of atoms remaining after 20, 40, 60, 80 days is 1/2, 1/4, 1/8, 1/16, hence the fraction of atoms that have disintegrated at these times is 1/2, 3/4, 7/8, 15/16. Hence, 7/8 of the atoms have disintegrated after 60 days (see Class 4 slide 4). Class 5 6) What are the number of protons " and neutrons ( in the missing nuclide J in the following fission reaction? P + "&(0 → J + !#$!\ + 4P A. " = 55, ( = 37 B. " = 37, ( = 55 C. " = 92, ( = 37 D. " = 37, ( = 58 B – The total number of protons (") and neutrons (() on either side of the equation must match. The atomic number of 0 is " = 92, so the left-hand side corresponds to " = 92 and ( = 144. The atomic number of !\ is " = 55, so the right-hand side

excluding nuclide J contains " = 55 and ( = 89. Therefore, J contains " = 37 and ( = 55 (see Class 5 slide 4).

The fraction of atoms remaining after 20, 40, 60, 80 days is 1/2, 1/4, 1/8, 1/16, hence the fraction of atoms that have disintegrated at these times is 1/2, 3/4, 7/8, 15/16. Hence, 7/8 of the atoms have

Week 10 Quiz Review

4) The half-life of radioactive iodine-137 is 8.0 days. How many iodine nuclei are necessary to produce an activity of 1.0 Z![? A. 2.9 × 10+ B. 4.6 × 10+ C. 3.7 × 10!$ D. 7.6 × 10!" C – The radioactivity in decays per second is given by 78₇₂ = W( where Y_!/" = 0.693/W. Here, 78₇₂ = 10.% _{× 3.7 × 10}!$ _\.! _{= 3.7 × 10}# _\.!_{. Hence the number of nuclei is ( =} ! 1 78 72 = '×"#×&%$$ $.%+& × 3.7 × 10 # _{= 3.7 × 10}!$ _{(see Class 4 slide 6).} 5) A certain radioactive element has a half-life of 20 days. How long would it take for 7/8 of the atoms originally present to disintegrate? A. 20 days B. 40 days C. 60 days D. 80 days C – The fraction of atoms remaining after 20, 40, 60, 80 days is 1/2, 1/4, 1/8, 1/16, hence the fraction of atoms that have disintegrated at these times is 1/2, 3/4, 7/8, 15/16. Hence, 7/8 of the atoms have disintegrated after 60 days (see Class 4 slide 4). Class 5 6) What are the number of protons " and neutrons ( in the missing nuclide J in the following fission reaction? P + "&(0 → J + !#$!\ + 4P A. " = 55, ( = 37 B. " = 37, ( = 55 C. " = 92, ( = 37 D. " = 37, ( = 58 B – The total number of protons (") and neutrons (() on either side of the equation must match. The atomic number of 0 is " = 92, so the left-hand side corresponds to " = 92 and ( = 144. The atomic number of !\ is " = 55, so the right-hand side

Week 10 Quiz Review

4) The half-life of radioactive iodine-137 is 8.0 days. How many iodine nuclei are necessary to produce an activity of 1.0 Z![? A. 2.9 × 10+ B. 4.6 × 10+ C. 3.7 × 10!$ D. 7.6 × 10!" C – The radioactivity in decays per second is given by 78₇₂ = W( where Y_!/" = 0.693/W. Here, 78₇₂ = 10.% _{× 3.7 × 10}!$ _\.! _{= 3.7 × 10}# _\.!_{. Hence the number of nuclei is ( =} ! 1 78 72 = '×"#×&%$$ $.%+& × 3.7 × 10 # _{= 3.7 × 10}!$ _{(see Class 4 slide 6).} 5) A certain radioactive element has a half-life of 20 days. How long would it take for 7/8 of the atoms originally present to disintegrate? A. 20 days B. 40 days C. 60 days D. 80 days C – The fraction of atoms remaining after 20, 40, 60, 80 days is 1/2, 1/4, 1/8, 1/16, hence the fraction of atoms that have disintegrated at these times is 1/2, 3/4, 7/8, 15/16. Hence, 7/8 of the atoms have disintegrated after 60 days (see Class 4 slide 4). Class 5 6) What are the number of protons " and neutrons ( in the missing nuclide J in the following fission reaction? P + "&(0 → J + !#$!\ + 4P A. " = 55, ( = 37 B. " = 37, ( = 55 C. " = 92, ( = 37 D. " = 37, ( = 58 B – The total number of protons (") and neutrons (() on either side of the equation must match. The atomic number of 0 is " = 92, so the left-hand side corresponds to " = 92 and ( = 144. The atomic number of !\ is " = 55, so the right-hand side

excluding nuclide J contains " = 55 and ( = 89. Therefore, J contains " = 37 and ( = 55 (see Class 5 slide 4).

Week 10 Quiz Review

7) Consider the fission reaction: 0 "&( _{+ P →} !#!_{=? + ^A}+" _{+ P-CUA_P\} How many excess neutrons are produced after the reaction? A. Zero B. 1 C. 2 D. 3 D – Uranium 0, barium =? and krypton ^A have atomic numbers " = 92, 56, 36. The number of neutrons entering the reaction is (235 − 92) + 1 = 144. The number leaving in the =? and ^A is (141 − 56) + (92 − 36) = 141. Therefore, 3 excess neutrons are produced (see Class 5 slide 4). 8) In the following nuclear reaction, what particle is missing? / !% <sub>+ `-</sub># <sub>→ (-</sub>!+ <sub>+ J</sub> A. a B. P C. -. D. S B – Oxygen /, helium `- and neon (- have atomic numbers " = 8, 2, 10. Hence the number of protons already balance, and J must be a neutron (see Class 5 slide 4). 9) What is the main purpose of a moderator in a fission reactor? A. It protects the uranium fuel B. It slows down neutrons to thermal velocities C. It absorbs the released energy D. It transports the released energy out of the reactor B – The moderator slows down the neutrons to increase the cross-section of their subsequent fission reactions (see Class 5 slide 9).

Uranium 𝑈, barium 𝐵𝑎 and krypton 𝐾𝑟 have atomic numbers 𝑍 = 92, 56, 36. The number of neutrons entering the reaction is (

) 235 −

Week 10 Quiz Review

7) Consider the fission reaction: 0 "&( _{+ P →} !#!_{=? + ^A}+" _{+ P-CUA_P\} How many excess neutrons are produced after the reaction? A. Zero B. 1 C. 2 D. 3 D – Uranium 0, barium =? and krypton ^A have atomic numbers " = 92, 56, 36. The number of neutrons entering the reaction is (235 − 92) + 1 = 144. The number leaving in the =? and ^A is (141 − 56) + (92 − 36) = 141. Therefore, 3 excess neutrons are produced (see Class 5 slide 4). 8) In the following nuclear reaction, what particle is missing? / !% <sub>+ `-</sub># <sub>→ (-</sub>!+ <sub>+ J</sub> A. a B. P C. -. D. S B – Oxygen /, helium `- and neon (- have atomic numbers " = 8, 2, 10. Hence the number of protons already balance, and J must be a neutron (see Class 5 slide 4). 9) What is the main purpose of a moderator in a fission reactor? A. It protects the uranium fuel B. It slows down neutrons to thermal velocities C. It absorbs the released energy D. It transports the released energy out of the reactor B – The moderator slows down the neutrons to increase the cross-section of their subsequent fission reactions (see Class 5 slide 9).

Week 10 Quiz Review

7) Consider the fission reaction: 0 "&( _{+ P →} !#!_{=? + ^A}+" _{+ P-CUA_P\} How many excess neutrons are produced after the reaction? A. Zero B. 1 C. 2 D. 3 D – Uranium 0, barium =? and krypton ^A have atomic numbers " = 92, 56, 36. The number of neutrons entering the reaction is (235 − 92) + 1 = 144. The number leaving in the =? and ^A is (141 − 56) + (92 − 36) = 141. Therefore, 3 excess neutrons are produced (see Class 5 slide 4). 8) In the following nuclear reaction, what particle is missing? / !% <sub>+ `-</sub># <sub>→ (-</sub>!+ <sub>+ J</sub> A. a B. P C. -. D. S B – Oxygen /, helium `- and neon (- have atomic numbers " = 8, 2, 10. Hence the number of protons already balance, and J must be a neutron (see Class 5 slide 4). 9) What is the main purpose of a moderator in a fission reactor? A. It protects the uranium fuel B. It slows down neutrons to thermal velocities C. It absorbs the released energy D. It transports the released energy out of the reactor B – The moderator slows down the neutrons to increase the cross-section of their subsequent fission reactions (see Class 5 slide 9).

Week 10 Quiz Review

Class 6 10) The following fusion reaction occurs in the Sun: `-" & <sub>+</sub> `-" # _→ =-# )

Week 10 Quiz Review

Class 6 10) The following fusion reaction occurs in the Sun: `-" & <sub>+</sub> `-" # _→ =-# )

where the masses of the nuclei are, 7b `-& c = 3.01605 C, 7b `-# c = 4.00260 C and 7b =-) c = 7.01693 C. How much energy is absorbed or released by the reaction? A. 920 F-G, absorbed B. 1.6 F-G, absorbed C. 920 F-G, released D. 1.6 F-G, released B – The energy absorbed or released is equivalent to the mass difference, Δ7 = 3.01605 + 4.00260 − 7.01693 = 0.00172 C, which is converted to F-G as 0.00172 × 931.5 = 1.6 F-G. This energy is absorbed because the binding energy is decreased after the reaction (see Class 6 slide 5).

The energy absorbed or released is equivalent to the mass difference, Δ𝑚 = 3.01605 + 4.00260 − 7.01693 = 0.00172 𝑢, which is

Assignment concept summary

Let’s briefly summarise a few concepts used in Assignment Q1:

•

Producing nuclear energy

•

_{Chain reactions}

•

Reaction cross-section

•

Mean free path

•

Critical mass & moderation

•

Design of a fission reactor

Producing nuclear energy

•

Let’s consider again the variation of the binding energy per

nucleon, 𝐵/𝐴, with the number of nucleons, 𝐴:

Energy may be released if nuclei with large 𝐴 are split

up (known as fission) Energy may be released if

nuclei with small 𝐴 are joined together (known as fusion)

Increasing binding energy is equivalent to

releasing energy

Reaction cross-section

•

When particles are bombarding a target, the probability of a

nuclear reaction occurring is given by its cross-section

(symbol 𝜎), defined as:

Number of reactions

per second per nucleus

Cross-section

(𝜎)

Flux of incident

particles (𝜙)

=

_×

Units: s"% Units: m' Units: m"' s"%

Beam of particles (number passing

unit area per second = 𝜙)

Geometrical interpretation: each

target nucleus is associated with a little

projected area = 𝜎

Mean free path

•

The reaction cross-section also determines the mean free

path of the bombarding particles (how far they travel,

before initiating a reaction, symbol 𝜆

– yes, another one!

)

• Let the number density of the target nuclei be 𝑛

• Consider a cylinder of cross-sectional area 𝜎 and length 𝜆

• The number of reactions that happen in this

cylinder is 𝑛×𝜆𝜎 = 1 for the mean free path

• Hence,

𝜆 = 1/𝑛𝜎

Critical mass & moderation

Critical mass

The fissile material must be larger than

the mean free path of the bombarding neutrons, otherwise they’ll pass

straight through – this minimum size is known as the critical mass

Moderation

Fission produces neutrons with far too

Design of a fission reactor

•

A few more design points …

•

"$%

U

is a highly fissile nuclide, but only comprises 0.7% of

natural uranium, whereas

"$#

U

is 99.3%. Fission will

proceed more efficiently if the

"$%

U

fraction is

enriched

•

"$#

U

in the fuel will absorb neutrons to produce

"$&

U

,

which 𝛽-decays into

"$&

Pu

, another very useful nuclear fuel

(reactors producing

"$&

Pu

are known as “

breeder reactors

”)

–

See Assignment Q1 e)

Assignment