Question

Consider the following fusion reaction, which allows stars to produce progressively heavier elements: \({ }_{2}^{3} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{4}^{7} \mathrm{Be}+\gamma\). The mass of \({ }_{2}^{3} \mathrm{He}\) is \(3.016029 \mathrm{u}\), the mass of \({ }_{2}^{4}\) He is \(4.002603 \mathrm{u}\), and the mass of \({ }_{4}^{7} \mathrm{Be}\) is \(7.0169298 \mathrm{u}\). The atomic mass unit is \(u=1.66 \cdot 10^{-27} \mathrm{~kg} .\) Assuming the Be atom is at rest after the reaction and neglecting any potential energy between the atoms and kinetic energy of the He nuclei, calculate the minimum possible energy and maximum possible wavelength of the photon \(\gamma\) that is released in this reaction.

Short answer

Answer: The minimum possible energy of the photon is 2.5430195 x 10^-13 J, and the maximum possible wavelength is 7.817 x 10^-7 m.

Step-by-step solution

Understand the conservation of mass-energy in the fusion reaction

In the fusion reaction, the total mass-energy of the initial particles (\({ }_{2}^{3} \mathrm{He}\) and \({ }_{2}^{4}\) He) should be equal to the total mass-energy of the final particles (\({ }_{4}^{7} \mathrm{Be}\) and the photon). We will be neglecting the potential energy between the atoms and the kinetic energy of the helium nuclei, so our focus will be on the masses of the particles involved.

Find the energy-equivalent of the initial and final masses

Use the formula for energy-equivalent mass \(E=mc^2\). In this case, \(m\) is the mass difference between the initial particles and the final particle: \(E = \Delta m \times c^2\) Where \(\Delta m\) is the change in mass, and c is the speed of light (\(3\times10^8\) m/s). Calculate \(\Delta m\) using the given masses: \(\Delta m = (m_{^3He} + m_{^4He}) - m_{^7Be}\) \(\Delta m = (3.016029 u + 4.002603 u) - 7.0169298 u\) \(\Delta m = 0.0017022 u\) Convert \(\Delta m\) from atomic mass units to kilograms using the conversion given: \(\Delta m = 0.0017022 u \times 1.66 \cdot 10^{-27} \mathrm{~kg/u}\) \(\Delta m = 2.826355 \times 10^{-30} \mathrm{~kg}\) Now, use that mass to find energy: \(E = \Delta m \times c^2\) \(E = 2.826355 \times 10^{-30} \mathrm{~kg} \times (3 \times 10^8 \mathrm{~m} / \mathrm{s})^2\) \(E = 2.5430195 \times 10^{-13} \mathrm{~J}\)

Calculate the minimum energy and maximum wavelength of the photon

To find the maximum wavelength, we need to first find the minimum energy of the photon. We assume that the Be nucleus is at rest after the reaction, and all the energy is carried away by the photon. The minimum energy of the photon is the energy difference from step 2. So, the minimum energy of the photon is: \(E_\gamma = 2.5430195 \times 10^{-13} \mathrm{~J}\) Now, we find the maximum wavelength of the photon by using the Planck-Einstein relation: \(E_\gamma = h \times f\) and \(f=c/\lambda\) Where \(E_\gamma\) is the minimum energy of the photon, \(h\) is the Planck constant (\(6.626 \times 10^{-34} \mathrm{~Js}\)), \(f\) is the frequency of the photon, and \(\lambda\) is the wavelength. Rearranging and substituting \(f\) with the speed of light (\(c\)) and the wavelength (\(\lambda\)) we get: \(\lambda_\mathrm{max} = \frac{hc}{E_\gamma}\) Plug in the values and solve for \(\lambda_\mathrm{max}\): \(\lambda_\mathrm{max} = \frac{(6.626 \times 10^{-34} \mathrm{~Js})(3 \times 10^8 \mathrm{~m} /\mathrm{s})}{2.5430195 \times 10^{-13} \mathrm{~J}}\) \(\lambda_\mathrm{max} = 7.817 \times10^{-7} \mathrm{~m}\)

Write down the final results

So, the minimum possible energy of the photon is \(2.5430195 \times 10^{-13} \mathrm{~J}\), and the maximum possible wavelength is \(7.817 \times10^{-7} \mathrm{~m}\).

Key concepts

Conservation of Mass-Energy

The principle of conservation of mass-energy is central to understanding nuclear reactions such as fusion in stars. This concept, upheld by Einstein's theory of relativity, states that the total amount of mass and energy in an isolated system remains constant. In the context of a fusion reaction, the mass-energy before and after the reaction should be equivalent, even though the form of that mass-energy can change from pure mass to pure energy or vice versa.

During fusion, the mass of the initial reactants is not entirely converted into the mass of the products; a small fraction is converted into energy, typically in the form of photons or kinetic energy of the particles. This is because the binding energy of the resulting nucleus is different from that of the initial reactants, leading to a mass discrepancy known as the 'mass defect'. The energy released in this process is the energy that powers stars and is calculated by the energy-equivalent mass formula derived from Einstein's famous equation, E=mc².

Energy-Equivalent Mass

Einstein's renowned equation, E=mc², introduces the concept of energy-equivalent mass, revealing that mass can be converted into energy and vice versa. The 'energy-equivalent' of mass is the amount of energy that would be released (or required) if mass were converted entirely into energy. This critical relationship shines a spotlight on the fusion process within stars, where the slight mass lost during fusion reactions is what provides the incredible amounts of energy radiated by the star.

In our exercise, the calculated energy release comes from the mass difference (Δm) between the initial helium nuclei ({_{2}^{3} He} and {_{2}^{4} He}) and the final beryllium nucleus ({_{4}^{7} Be}). This energy is the 'missing' mass multiplied by the square of the speed of light. Thus, even a tiny amount of mass can be converted into a significant amount of energy, underscoring the profound impact of mass-energy conservation in nuclear processes.

Planck-Einstein Relation

The Planck-Einstein relation is a fundamental equation in quantum mechanics bridging the wave and particle nature of energy. It states that the energy of a photon (E) is proportional to its frequency (f), with Planck's constant (h) as the proportionality factor. Mathematically, this is expressed as E = hf. Since the frequency and wavelength (λ) of light are inversely related (f = c/λ, where c is the speed of light), the energy of a photon can also determine its wavelength.

In our stellar fusion exercise, after calculating the minimum energy of the photon released during the reaction, we use the Planck-Einstein relation to find the maximum wavelength of the photon. It's at this point that quantum mechanics intersect with classical physics, allowing us to understand the energy and associated properties of radiation emitted during nuclear reactions such as the fusion inside stars. The magnificent range of wavelengths produced in such processes span the electromagnetic spectrum and are a direct result of energy conversions at the atomic level.