**Nuclear fission**, the opposite of fusion. Here, instead of combining small atoms, we split a large one apart. This process releases a huge amount of energy and can create a powerful chain reaction. Your task is to apply the principles we've learned to calculate the exact amount of energy released in a specific fission reaction, similar to the worked example. ## Nuclear Fission * **Definition**: A process where the nucleus of an atom splits into two or more smaller nuclei, with the release of a large amount of energy. * **Discovery**: Discovered by Otto Hahn and Fritz Strassmann. * **Initiation**: Typically initiated by bombarding a heavy nucleus with a **neutron**, causing it to become unstable and split. * **Products**: The reaction produces daughter nuclei (which can be different elements or isotopes), additional neutrons, and a significant amount of energy (Q). * **Examples of Fission Reactions**: * ${ }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{92}^{236} \mathrm{U}^{*} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n}+\mathrm{Q}$ * ${ }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{92}^{236} \mathrm{U}^{*} \rightarrow{ }_{50}^{132} \mathrm{Sn}+{ }_{42}^{101} \mathrm{Mo}+3{ }_{0}^{1} \mathrm{n}+\mathrm{Q}$ * ${ }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{92}^{236} \mathrm{U}^{*} \rightarrow{ }_{56}^{139} \mathrm{Ba}+{ }_{36}^{95} \mathrm{Kr}+2{ }_{0}^{1} \mathrm{n}+\mathrm{Q}$ * The number of emitted neutrons and the daughter nuclei can vary. ## Fission Chain Reaction * **Mechanism**: A single fission event releases multiple neutrons (typically 2 or 3) that can then cause further fission reactions, leading to a succession of fissions. * **Requirement for Sustaining**: * For fission, **slow neutrons** are required. Fast neutrons produced by fission may escape without causing further reactions. * At least one released neutron must produce another fission reaction to sustain the chain. * **Critical Mass**: The **minimum mass** of the fissile material required to sustain a fission chain reaction. * **Subcritical Mass**: If mass is less than critical mass, the fission process soon ends. * **Critical Mass (Controlled)**: If mass is equal to critical mass, or conditions are suitable for producing one more fission reaction per event, controlled nuclear energy is produced (e.g., in nuclear reactors). * **Supercritical Mass (Uncontrolled)**: If mass is greater than critical mass, the fission process becomes uncontrolled, leading to a rapid and explosive release of energy (e.g., in nuclear weapons). * **Critical Volume**: The volume occupied by the critical mass of a material. ## Energy in Fission Reaction * **Magnitude**: Nuclear fission releases a huge amount of energy, significantly greater than chemical reactions. * A single fission of uranium-235 releases about **200 MeV** of energy. * Chemical reactions release energy on the order of a few electron volts (eV) per reaction (e.g., methane combustion $\approx 9.2 \mathrm{eV}$). * **Comparison**: Fission of 1 kg of uranium-235 can produce about **2.5 million times more energy** than burning 1 kg of coal. * **Applications**: * **Nuclear Power Plants**: Controlled chain reactions generate electricity. * **Nuclear Weapons**: Uncontrolled chain reactions lead to explosive energy release. ## Stability and Binding Energy * Heavier nuclei have a smaller binding energy per nucleon, making them less stable. * Fission allows unstable heavy nuclei to split into smaller fragments, which are generally more stable (have larger binding energy per nucleon). ## Assignment 22.7 Find the amount of energy released by the following fission reaction if the masses of uranium, Xenon, Strontium and neutron are 235.12142 u, 139.9055 u, 93.9064 u and 1.008665 u respectively constant. $$ { }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_{0}^{1} \mathrm{n}+\mathrm{Q} $$ ### Worked Example 22.7 Q&A - **Q:** What is the first step to calculate energy released in a fission reaction using masses? **A:** Calculate the total mass of reactants (L.H.S.) and products (R.H.S.). - **Q:** For the reaction ${ }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n}+\mathrm{Q}$, given: $m(\mathrm{U}-235) = 235.12142 \mathrm{u}$, $m(\mathrm{Ba}-141) = 140.883 \mathrm{u}$, $m(\mathrm{Kr}-92) = 91.9262 \mathrm{u}$, $m(\mathrm{n}) = 1.008665 \mathrm{u}$, calculate the total mass on the L.H.S. **A:** Total mass on L.H.S. $= m(\mathrm{n}) + m(\mathrm{U}-235) = 1.008665 \mathrm{u} + 235.12142 \mathrm{u} = 236.130085 \mathrm{u}$. - **Q:** For the same reaction and given masses, calculate the total mass on the R.H.S. **A:** Total mass on R.H.S. $= m(\mathrm{Ba}-141) + m(\mathrm{Kr}-92) + 3 \times m(\mathrm{n}) = 140.883 \mathrm{u} + 91.9262 \mathrm{u} + 3 \times 1.008665 \mathrm{u} = 235.835195 \mathrm{u}$. - **Q:** What is the mass defect ($\Delta m$) for the reaction in Example 22.7? **A:** $\Delta m = (\text{Total mass L.H.S.}) - (\text{Total mass R.H.S.}) = 236.1301 \mathrm{u} - 235.8352 \mathrm{u} = 0.2949 \mathrm{u}$. - **Q:** Calculate the energy released (Q) in MeV for the reaction in Example 22.7 using the calculated mass defect and the conversion factor $931.5 \mathrm{MeV}/\mathrm{u}$. **A:** $Q = \Delta m \times 931.5 \mathrm{MeV}/\mathrm{u} = 0.2949 \mathrm{u} \times 931.5 \mathrm{MeV}/\mathrm{u} = 274.7 \mathrm{MeV}$. | Concept | Description | Formula / Value | | :------------------ | :-------------------------------------------------------------------------- | :---------------------------------------------------------------------------------------- | | Nuclear Fission | Splitting of heavy nucleus into lighter ones + energy. | ${ }_{0}^{1} \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow \text{products} + \text{neutrons} + \mathrm{Q}$ | | Energy per U-235 Fission | Energy released per single event. | $\approx 200 \mathrm{MeV}$ | | Critical Mass | Minimum mass to sustain chain reaction. | Varies by material and conditions | | Energy-Mass Equivalence | Relates mass defect to energy released. | $E = \Delta m c^2$ (often $Q = \Delta m \times 931.5 \mathrm{MeV}/\mathrm{u}$) |