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Nuclear and Quantum Physics - Fusion and stars

Grade 11IBPhysics

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

🔑Concepts

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Nuclear fusion is the process in which two light nuclei combine to form a more massive nucleus. For fusion to occur, the nuclei must have sufficient kinetic energy to overcome the electrostatic repulsion (the Coulomb barrier). This requires extremely high temperatures (around 107 K10^7 \text{ K}) and high pressures, typically found in stellar cores.

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The energy released in fusion is due to the mass defect Δm\Delta m. The mass of the resulting nucleus is less than the sum of the masses of the individual reactant nuclei. This 'lost' mass is converted into energy according to Einstein's mass-energy equivalence principle E=Δmc2E = \Delta mc^2.

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Binding energy per nucleon is a measure of nuclear stability. In the binding energy curve, fusion occurs for light elements (where mass number A<62A < 62) because the product nucleus has a higher binding energy per nucleon than the reactants, resulting in a more stable state and the release of energy.

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Main sequence stars, like our Sun, primarily generate energy through the proton-proton (p-p) chain. The net reaction involves four hydrogen nuclei (protons) fusing to form one helium-4 nucleus, two positrons, and two neutrinos: 411H→24He+2+10e+2ν+γ4^1_1H \rightarrow ^4_2He + 2^0_{+1}e + 2\nu + \gamma.

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A star maintains its size through hydrostatic equilibrium, which is the balance between the inward pull of gravity and the outward radiation pressure generated by nuclear fusion in the core.

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The Hertzsprung-Russell (H-R) diagram plots stars according to their luminosity (LL) on the y-axis and surface temperature (TT) on the x-axis (with temperature increasing to the left). Most stars fall along the Main Sequence, where they fuse hydrogen into helium.

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Stellar evolution depends on the initial mass of the star. Low-mass stars (like the Sun) evolve into Red Giants and eventually shed their outer layers to leave a White Dwarf. High-mass stars evolve into Red Supergiants and end their lives in a Supernova, leaving behind a Neutron Star or a Black Hole.

📐Formulae

E=Δmc2E = \Delta mc^2

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

λmaxT=2.90×10−3 m K\lambda_{max} T = 2.90 \times 10^{-3} \text{ m K}

L=σAT4=4πR2σT4L = \sigma A T^4 = 4\pi R^2 \sigma T^4

L∝M3.5L \propto M^{3.5}

E (in MeV)=Δm (in u)×931.5 MeV u−1E \text{ (in MeV)} = \Delta m \text{ (in u)} \times 931.5 \text{ MeV u}^{-1}

💡Examples

Problem 1:

Calculate the energy released (in MeV\text{MeV}) in the fusion reaction: 12H+13H→24He+01n^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n. Use the following atomic masses: m(12H)=2.014102 um(^2_1H) = 2.014102 \text{ u}, m(13H)=3.016049 um(^3_1H) = 3.016049 \text{ u}, m(24He)=4.002603 um(^4_2He) = 4.002603 \text{ u}, m(01n)=1.008665 um(^1_0n) = 1.008665 \text{ u}. Use 1 u=931.5 MeV1 \text{ u} = 931.5 \text{ MeV}.

Solution:

  1. Calculate the total mass of reactants: 2.014102+3.016049=5.030151 u2.014102 + 3.016049 = 5.030151 \text{ u}

  2. Calculate the total mass of products: 4.002603+1.008665=5.011268 u4.002603 + 1.008665 = 5.011268 \text{ u}

  3. Calculate the mass defect Δm\Delta m: 5.030151−5.0112680.018883\begin{array}{r} 5.030151 \\ -5.011268 \\ \hline 0.018883 \end{array} Δm=0.018883 u\Delta m = 0.018883 \text{ u}

  4. Convert mass defect to energy: E=0.018883×931.5≈17.59 MeVE = 0.018883 \times 931.5 \approx 17.59 \text{ MeV}

Explanation:

The energy released is found by calculating the difference in mass between the reactants and the products (mass defect) and then using the conversion factor to turn atomic mass units into Mega-electronvolts.

Problem 2:

A star is observed to have a peak emission wavelength of λmax=500 nm\lambda_{max} = 500 \text{ nm}. Estimate the surface temperature of the star.

Solution:

  1. Convert wavelength to meters: λmax=500×10−9 m\lambda_{max} = 500 \times 10^{-9} \text{ m}

  2. Use Wien's Displacement Law: λmaxT=2.90×10−3 m K\lambda_{max} T = 2.90 \times 10^{-3} \text{ m K}

  3. Rearrange to solve for TT: T=2.90×10−3500×10−9T = \frac{2.90 \times 10^{-3}}{500 \times 10^{-9}}

  4. Calculate the value: T=2.90×10−35.0×10−7=5800 KT = \frac{2.90 \times 10^{-3}}{5.0 \times 10^{-7}} = 5800 \text{ K}

Explanation:

Wien's Displacement Law relates the blackbody temperature of a star to the wavelength at which it emits the most radiation. By measuring the peak wavelength, we can determine the star's surface temperature.