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Nuclear and Quantum Physics - Fission

Grade 12IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Nuclear fission is the process by which a heavy nucleus (typically with a high nucleon number A>200A > 200) splits into two or more smaller, more stable nuclei of intermediate mass.

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The process releases a significant amount of energy because the binding energy per nucleon is higher for the daughter nuclei than for the parent nucleus. This difference in energy is released as kinetic energy of the fragments and gamma radiation.

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Induced fission occurs when a heavy nucleus, such as 92235U^{235}_{92}U, captures a thermal neutron (a slow-moving neutron). This creates an unstable isotope 92236U^{236}_{92}U which then undergoes fission.

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A chain reaction occurs when the neutrons released during one fission event go on to trigger further fission events. This can lead to a self-sustaining reaction if at least one neutron per fission causes another fission.

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The critical mass is the minimum mass of fissile material required to maintain a self-sustaining nuclear chain reaction.

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In a nuclear reactor, the moderator (e.g., water or graphite) slows down fast neutrons to thermal speeds to increase the probability of fission, while control rods (e.g., boron or cadmium) absorb excess neutrons to regulate the reaction rate.

📐Formulae

Δm=(∑mreactants)−(∑mproducts)\Delta m = (\sum m_{reactants}) - (\sum m_{products})

E=Δmc2E = \Delta m c^2

EMeV=Δm(u)×931.5 MeV/uE_{MeV} = \Delta m (u) \times 931.5 \text{ MeV/u}

92235U+01n→56141Ba+3692Kr+301n+Energy^{235}_{92}U + ^{1}_{0}n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n + \text{Energy}

💡Examples

Problem 1:

In a specific fission reaction of Uranium-235, the total mass of the reactants is 236.0526 u236.0526 \text{ u} and the total mass of the products is 235.8665 u235.8665 \text{ u}. Calculate the energy released in MeV.

Solution:

Δm=236.0526 u−235.8665 u=0.1861 u\Delta m = 236.0526 \text{ u} - 235.8665 \text{ u} = 0.1861 \text{ u} E=0.1861×931.5 MeVE = 0.1861 \times 931.5 \text{ MeV} E≈173.35 MeVE \approx 173.35 \text{ MeV}

Explanation:

The energy released is calculated by first finding the mass defect (the difference in mass between reactants and products) and then converting that mass into energy using the conversion factor 1 u=931.5 MeV1 \text{ u} = 931.5 \text{ MeV}.

Problem 2:

A nuclear power plant produces 1000 MW1000 \text{ MW} of power. If each fission event releases 200 MeV200 \text{ MeV} of energy, calculate the number of fissions occurring per second.

Solution:

1 MeV=1.6×10−13 J1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} Eper fission=200×1.6×10−13 J=3.2×10−11 JE_{per\ fission} = 200 \times 1.6 \times 10^{-13} \text{ J} = 3.2 \times 10^{-11} \text{ J} Power=Number of fissions×Eper fissiontime\text{Power} = \frac{\text{Number of fissions} \times E_{per\ fission}}{\text{time}} Rate=109 J/s3.2×10−11 J/fission\text{Rate} = \frac{10^9 \text{ J/s}}{3.2 \times 10^{-11} \text{ J/fission}} Rate=3.125×1019 fissions/s\text{Rate} = 3.125 \times 10^{19} \text{ fissions/s}

Explanation:

To find the number of fissions per second, we divide the total power output (Energy per second) by the energy released in a single fission event, ensuring all units are converted to Joules.