# Energy and Mass Equation ## Mass-Energy Equivalence - Einstein's theory of relativity revealed that matter is fundamentally a form of energy - This leads to two remarkable possibilities: - Mass can be converted into energy - Energy can be converted into mass

Mass-energy equivalence: $E = mc^2$ where $E$ is energy (J), $m$ is mass (kg), and $c$ is the speed of light (m s$^{-1}$)

- Where: - $E$ = energy (J) - $m$ = mass (kg) - $c$ = speed of light (m s$^{-1}$) ## Real-World Examples - Mass-energy equivalence isn't just theoretical – it occurs in several observable contexts: - Nuclear fusion powering stars like our Sun - Fission reactions in nuclear power stations generating electricity - Nuclear weapons releasing enormous amounts of energy - Particle accelerators where high-energy collisions create new particles ## Energy Released in Nuclear Reactions - When nuclei undergo reactions (fission or fusion), binding energy is released - The binding energy equals the energy released during nucleus formation - This can be calculated using: $$E = (\Delta m)c^2$$ - Where: - $E$ = Binding energy released (J) - $\Delta m$ = mass defect (kg) - $c$ = speed of light (m s$^{-1}$)

Daughter nuclei from fission and fusion have higher binding energy per nucleon than parent nuclei

### Why Energy Is Released - Both fission and fusion produce daughter nuclei with higher binding energy per nucleon than the parent nuclei - Energy is released due to the mass difference between reactants and products - This mass difference converts directly into energy according to $E = mc^2$ **Binding Energy Per Nucleon Curve:** The graph below shows how binding energy per nucleon varies with mass number. Notice that: - Light nuclei (fusion) move up the curve toward the peak - Heavy nuclei (fission) move down toward the peak - Both directions represent an increase in stability and release energy

## Worked Example: Uranium-235 Fission **Problem:** When Uranium-235 nuclei undergo fission by slow-moving neutrons, two possible reactions occur: Reaction 1: $_{92}^{235}\text{U} + _0^1\text{n} \rightarrow _{54}^{139}\text{Xe} + _{38}^{95}\text{Sr} + 2_0^1\text{n}$ + energy Reaction 2: $_{92}^{235}\text{U} + _0^1\text{n} \rightarrow 2_{46}^{116}\text{Pd} + x\text{c}$ + energy ### Part (a): Identifying Products **Task:** For reaction 2, identify particle c and state the number x of such particles **Step 1: Balance proton numbers (bottom numbers)** $$92 = (2 \times 46) + xn_p$$ where $n_p$ is the number of protons in c $$xn_p = 92 - 92 = 0$$ Therefore, c must be a neutron (zero protons) **Step 2: Balance nucleon numbers (top numbers)** $$235 + 1 = (2 \times 116) + x$$ $$x = 235 + 1 - 232 = 4$$ Therefore, **4 neutrons** are generated in reaction 2 ### Part (b): Calculating Energy Released **Task:** The binding energy per nucleon for each nuclide is given in the table. Show that reaction 1 produces about 210 MeV. | Nuclide | $E$ / MeV | |---------|-----------| | $_{38}^{95}\text{Sr}$ | 8.74 | | $_{54}^{139}\text{Xe}$ | 8.39 | | $_{92}^{235}\text{U}$ | 7.60 | **Step 1: Calculate total binding energy for each nucleus** Total binding energy = Binding energy per nucleon × Mass number - Binding energy of $^{95}\text{Sr}$ = $8.74 \times 95 = 830.3$ MeV - Binding energy of $^{139}\text{Xe}$ = $8.39 \times 139 = 1166.21$ MeV - Binding energy of $^{235}\text{U}$ = $7.60 \times 235 = 1786$ MeV **Step 2: Calculate energy difference** Energy released = Binding energy of products - Binding energy of reactants $$\text{Energy released} = (830.3 + 1166.21) - 1786$$ $$\text{Energy released} = 210.5 \text{ MeV}$$

Energy is the difference between products and reactants, not just the sum of binding energies

### Part (c): Predicting Reaction Probability **Task:** Reaction 2 produces 163 MeV. Suggest which reaction is more likely to occur, with supporting reasoning. **Analysis:** - Reaction 1 releases more energy (210 MeV) than reaction 2 (163 MeV) - Greater energy release means products have higher binding energy per nucleon - Higher binding energy per nucleon indicates greater stability - The products move further up the binding energy per nucleon graph (shown above) - At high nucleon numbers (A), binding energy per nucleon gradually decreases with A - Nuclear reactions favour the more stable route **Conclusion:** Reaction 1 is more likely to happen because it produces more stable products