Apply Fission - IB Physics

Consider a chain reaction where the reproduction factor k is exactly 1. If 1000 fission events occur in one generation of the chain reaction, how many fission events will occur ten generations later?
Answer: 1000 A reproduction factor k = 1 signifies a critical state. This means that, on average, each generation of fissions produces exactly the right number of neutrons to initiate the same number of fissions in the next generation. Therefore, the rate of reaction is constant. If there are 1000 fissions in one generation, there will be 1000 fissions in the next generation, and 1000 fissions ten generations later. The power output is stable.
A research reactor uses $^{235}$ U fuel with a thermal fission cross-section of 585 barns and a thermal absorption cross-section of 694 barns. If the average number of neutrons produced per thermal fission is 2.42, what is the reproduction factor η (eta) for this fuel?
Answer: 2.04 The reproduction factor η is defined as η = ν × (σ_f/σ_a), where ν is the average number of neutrons per fission, σ_f is the fission cross-section, and σ_a is the absorption cross-section. Therefore: η = 2.42 × (585/694) = 2.42 × 0.843 = 2.04. Choice A uses the wrong ratio (σ_a/σ_f). Choice C incorrectly assumes all absorptions lead to fission. Choice D adds the cross-sections instead of taking their ratio.
In a fast breeder reactor, $^{238}$ U captures neutrons to eventually produce $^{239}$ Pu through beta decay processes. If the reactor has a breeding ratio of 1.15, and consumes 800 kg of fissile material per year, how much new fissile material is produced annually?
Answer: 920 kg of new fissile material per year Nuclear reactor physics involves understanding how breeding ratios quantify a reactor's ability to produce new fissile material. The breeding ratio tells you how much new fissile material is created relative to what's consumed - it's a fundamental measure of reactor efficiency. A breeding ratio of 1.15 means that for every unit of fissile material consumed, 1.15 units of new fissile material are produced. This is straightforward multiplication: if 800 kg of fissile material is consumed annually, then the reactor produces $800 \text{ kg} \times 1.15 = 920 \text{ kg}$ of new fissile material per year. Looking at the incorrect options: B (680 kg) represents a common error where students might divide instead of multiply ( $800 ÷ 1.15 ≈ 696$ ), misunderstanding what the breeding ratio means. C (1150 kg) occurs if you mistakenly think the breeding ratio is additive rather than multiplicative - incorrectly calculating $800 + (0.15 \times 1000) = 950$ or similar confused arithmetic. D (1380 kg) suggests adding the consumed amount to the produced amount ( $800 + 580$ ), but the question asks specifically for new material produced, not total material handled. The key insight is that breeding ratio is always multiplicative: new fissile material produced = consumed material × breeding ratio. When you see nuclear reactor problems involving breeding ratios, remember this direct relationship - don't overcomplicate with addition or division. The breeding ratio greater than 1.0 indicates the reactor produces more fuel than it consumes, which is the defining characteristic of breeder reactors.
A nuclear reactor is operating with a neutron reproduction factor k = 1.005. Which statement best describes the state of the reactor?
Answer: The reactor is supercritical, and the power output is increasing. The neutron reproduction factor, k, is the average number of neutrons from one fission that cause a subsequent fission. If k > 1, the reactor is supercritical, and the number of fissions (and thus power output) increases with each generation. If k < 1, it is subcritical (power decreases). If k = 1, it is critical (power is stable). A value of k = 1.005 represents a controlled increase in power, not an imminent explosion.
An operator at a nuclear power plant needs to reduce the power output from 80% to 60%. Which control sequence is appropriate?
Answer: Partially insert the control rods to make k < 1, then withdraw them to make k = 1 at the new power level. To decrease power, the rate of fission must be reduced. This is achieved by inserting control rods further into the core, which increases neutron absorption and makes the reactor subcritical (k < 1). The power level will then decrease. Once the desired power level (60%) is reached, the rods are withdrawn slightly to a position where k = 1 again, making the reactor critical and maintaining a stable output at the new, lower level.
A nuclear reactor generates 108 MW of thermal power. What is the approximate rate at which mass is converted to energy in the reactor?
Answer: 1.2 μg s⁻¹ The rate at which mass is converted to energy can be found using Einstein's mass-energy relation E = mc². The rate of mass conversion is Δm/Δt = P/c², where P is the thermal power. Δm/Δt = (1.08 × 10⁸ J/s) / (3.00 × 10⁸ m/s)² = 1.2 × 10⁻⁹ kg/s. Converting to micrograms: (1.2 × 10⁻⁹ kg/s) × (10⁹ μg/kg) = 1.2 μg/s.
Why are the fission products from a nuclear reactor considered hazardous for long periods of time?
Answer: They consist of a mix of neutron-rich radioactive isotopes with a wide distribution of half-lives. Fission breaks a heavy nucleus into two smaller, neutron-rich fragments (e.g., Strontium-90, Caesium-137). These fragments are radioactive and decay towards stability, primarily via beta and gamma emission. The mixture of fission products contains isotopes with half-lives ranging from fractions of a second to hundreds of thousands of years. The long-lived isotopes are what make the nuclear waste a long-term hazard.
Natural uranium consists of over 99% U-238 and less than 1% U-235. What is the primary reason that fuel for most commercial reactors must be enriched to increase the percentage of U-235?
Answer: U-238 undergoes neutron capture much more readily than fission with thermal neutrons, quenching the chain reaction. While U-238 can fission with very fast neutrons, it is not fissile with the thermal neutrons used in most reactors. Instead, it has a high probability of absorbing a thermal neutron without fissioning (a process called neutron capture), turning into U-239. This process removes neutrons from the chain reaction. To achieve a self-sustaining reaction (k ≥ 1), the concentration of fissile U-235 must be high enough to offset this neutron loss to U-238 capture.
A substance is being evaluated for use as a moderator in a thermal nuclear reactor. Which combination of nuclear properties is most desirable?
Answer: Low probability of neutron absorption and small atomic mass. A moderator must slow down neutrons without absorbing them. To slow them down effectively, neutrons should collide with nuclei of similar mass; therefore, a small atomic mass is desirable (e.g., hydrogen in water, carbon in graphite). To ensure neutrons are available for the chain reaction, the moderator must not absorb them, requiring a low neutron absorption cross-section (low probability of absorption).
Cadmium and Boron are common materials used for control rods in nuclear reactors. What specific nuclear property makes them suitable for this purpose?
Answer: They have a very high cross-section for absorbing thermal neutrons without undergoing fission. The function of a control rod is to regulate the neutron population in the reactor core. To do this effectively, the material must remove neutrons from the chain reaction. Cadmium and Boron have a very high probability (a large 'cross-section') of absorbing thermal neutrons. This absorption prevents the neutrons from causing further fission events. While engineering properties like heat resistance (D) are important, the key nuclear property is the high absorption cross-section.
How is a rapid, uncontrolled release of energy in a fission-based nuclear weapon achieved?
Answer: By creating a promptly supercritical mass, leading to an exponential increase in the rate of fission. An explosion requires a very rapid increase in the reaction rate. This is achieved by quickly assembling a mass that is 'promptly supercritical', meaning k is significantly greater than 1 even without considering delayed neutrons. This leads to an exponential growth in the fission rate over nanoseconds, releasing a massive amount of energy before the assembly can physically blow itself apart. Distractor A is incorrect because k itself does not increase exponentially; the power level does.
A Uranium-235 nucleus undergoes induced fission. The resulting fission fragments are typically unstable. What is the primary reason for this instability?
Answer: The fragments have a higher neutron-to-proton ratio than stable nuclei of similar mass. Heavy nuclei like Uranium-235 have a higher neutron-to-proton ratio than lighter, stable nuclei. When fission occurs, the fragments retain a similarly high ratio. For their smaller mass, this ratio is too high for stability. They are 'neutron-rich' and therefore undergo a series of beta-minus decays, where a neutron converts into a proton, to move towards the line of stability.
What is the approximate energy released in this specific fission reaction?
Answer: 185 MeV First, calculate the total mass of the reactants: Mass_reactants = m(U-235) + m(neutron) = 235.0439 u + 1.0087 u = 236.0526 u. Next, calculate the total mass of the products: Mass_products = m(Xe-140) + m(Sr-94) + 2 × m(neutron) = 139.9216 u + 93.9154 u + 2 × (1.0087 u) = 235.8544 u. The mass defect (Δm) is the difference: Δm = 236.0526 u - 235.8544 u = 0.1982 u. Finally, convert the mass defect to energy: E = Δm × 931.5 MeV/u = 0.1982 u × 931.5 MeV/u ≈ 184.6 MeV.
A fast neutron with kinetic energy 2.0 MeV collides elastically with a stationary hydrogen nucleus in a moderator. After the collision, the neutron continues in the same direction but with reduced speed. What is the maximum fraction of its initial kinetic energy that the neutron can lose in this single collision?
Answer: 1.0 (100% energy loss) In an elastic collision between equal masses where one is initially at rest, maximum energy transfer occurs in a head-on collision. For a neutron (mass ≈ proton mass) colliding with a hydrogen nucleus (proton), the neutron can transfer all its kinetic energy to the proton and come to rest. The energy transfer fraction is $\frac{4m_1m_2}{(m_1+m_2)^2}$ , which equals 1 when $m_1 = m_2$ . Choice A applies the formula for a deuteron target. Choice B assumes average energy loss. Choice C uses incorrect collision dynamics.
The core of a nuclear reactor is enclosed in thick shielding material, such as concrete and lead. What is the primary physics reason for this shielding?
Answer: To attenuate the flux of ionizing gamma radiation and neutrons escaping the reactor core. The fission process produces intense, highly penetrating gamma radiation and a large flux of neutrons. These are ionizing and harmful to living tissue and can make surrounding materials radioactive. The primary purpose of the biological shield is to absorb this radiation and reduce the levels outside the reactor to safe limits. While other components handle heat (coolant) and pressure (pressure vessel), the shielding is for radiation protection.
In a nuclear reactor, the reproduction factor k is defined as the ratio of neutrons produced in one generation to neutrons absorbed in the previous generation. If a reactor has an initial neutron population of $10^{12}$ neutrons and k = 1.003, approximately how many generations will it take for the neutron population to double?
Answer: 230 generations The neutron population grows as $N(t) = N_0 k^t$ where t is the number of generations. For doubling: $2N_0 = N_0 k^t$ , so $2 = (1.003)^t$ . Taking natural log: $\ln(2) = t \ln(1.003)$ . Therefore $t = \frac{\ln(2)}{\ln(1.003)} = \frac{0.693}{0.002996} ≈ 231$ generations. Choice A uses $k-1$ instead of $\ln(k)$ . Choice C incorrectly uses $\log_{10}$ . Choice D uses the wrong formula entirely.
A nuclear power plant's reactor core contains fuel assemblies arranged in a critical configuration. If the geometric buckling $B^2$ is $8.5 × 10^{-4} \text{ cm}^{-2}$ and the material buckling $B_m^2$ is $9.2 × 10^{-4} \text{ cm}^{-2}$ , what modification would bring the reactor closer to the critical condition?
Answer: Add neutron reflectors around the core to reduce neutron leakage For criticality, geometric buckling must equal material buckling ( $B^2 = B_m^2$ ). Currently $B^2 < B_m^2$ , indicating the reactor is subcritical due to excessive neutron leakage. Adding reflectors reduces leakage, effectively increasing the geometric buckling toward the material buckling value. Choice A would increase $B_m^2$ further from $B^2$ . Choice B would decrease $B^2$ further. Choice D would decrease reactivity through increased neutron absorption.
Following a fission event, neutrons are emitted. Most are emitted 'promptly' (within ~10⁻¹⁴ s), but a small fraction (~0.65%) are 'delayed', emitted seconds later from decaying fission products. What is the crucial role of these delayed neutrons in a nuclear reactor?
Answer: They significantly slow down the timescale of the chain reaction, allowing it to be mechanically controlled. If the chain reaction depended only on prompt neutrons, the power level would change almost instantaneously, far too quickly for mechanical systems like control rods to respond. The presence of delayed neutrons extends the average time between neutron generations from microseconds to tenths of a second. This slowing of the reaction dynamics is what makes a nuclear reactor controllable.
Which statement best explains why a single, isolated neutron is more likely to induce fission in U-235 than a single, isolated proton of the same kinetic energy?
Answer: The proton is repelled by the electrostatic force of the nucleus, preventing it from getting close enough to be captured. To induce fission, the incident particle must be absorbed by the nucleus. The U-235 nucleus is strongly positive (92 protons). An incoming proton, also being positive, will experience a powerful electrostatic repulsion (Coulomb barrier) that it must overcome to reach the nucleus. A neutron, having no charge, does not experience this repulsion and can approach and be captured by the nucleus much more easily, even at very low (thermal) kinetic energies.
The fission of a single Uranium-235 nucleus releases approximately 200 MeV of energy. A nuclear power plant has a continuous electrical power output of 500 MW. If the plant's overall efficiency is 32%, approximately how many Uranium-235 fissions occur per second? (1 MeV = 1.6 × 10⁻¹³ J)
Answer: 4.9 × 10¹⁹ First, calculate the required thermal power from the reactor. Thermal Power = Electrical Power / Efficiency = (500 × 10⁶ W) / 0.32 = 1.5625 × 10⁹ W. Next, convert the energy per fission to Joules: Energy = 200 MeV × (1.6 × 10⁻¹³ J/MeV) = 3.2 × 10⁻¹¹ J. Finally, the number of fissions per second is the total thermal power divided by the energy per fission: Rate = (1.5625 × 10⁹ J/s) / (3.2 × 10⁻¹¹ J/fission) ≈ 4.9 × 10¹⁹ fissions/s.

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