Nov6-09, 05:47 PM
1. The problem statement, all variables and given/known data
One of the thermonuclear or fusion reactions that takes place inside a star such as our Sun is the production of helium-3 (3He, with two protons and one neutron) and a gamma ray (high-energy photon, denoted by the lowercase Greek letter gamma, ) in a collision between a proton (1H) and a deuteron (2H, the nucleus of "heavy" hydrogen, consisting of a proton and a neutron):
1H + 2H 3He +
The rest mass of the proton is 1.0073 u (unified atomic mass unit, 1.66 10-27 kg), the rest mass of the deuteron is 2.0136 u, the rest mass of the helium-3 nucleus is 3.0155 u, and the gamma ray is a high-energy photon, whose mass is zero. The strong interaction has a very short range and is essentially a contact interaction. For this fusion reaction to take place, the proton and deuteron have to come close enough together to touch. The approximate radius of a proton or neutron is about 110-15 m.
(b) In this situation where the initial total momentum is zero, what minimum kinetic energy must the proton have, and what minimum kinetic energy must the deuteron have, in order for the reaction to take place? Express your results in eV. Assume that the center to center distance at collision is 2.3 10-15 m. You will find that the proton and deuteron have speeds much smaller than the speed of light (which you can verify if you like after calculating their kinetic energies). Keep in mind what you see in the diagram you drew in part (a). You may find it useful to remember that kinetic energy can be expressed either in terms of speed or in terms of the magnitude of momentum. It is very important to do the analysis symbolically; don't plug in numbers until the very end. If you try to do the problem numerically, and/or ignore part (a), you will probably not be able to complete the analysis.
Kproton = 2Your answer is incorrect. eV
Kdeuteron = 3 eV
(c) Becaus the helium-3 nucleus is massive, its kinetic energy is very small compared to the energy of the massless photon. Therefore, what will be the energy of the gamma ray in eV? The relationship E2 - (pc)2 = (mc2)2 is valid for any particle, including a massless photon, so the momentum of a photon is p = E/c, where E is the photon energy. You may need to consider the momentum principle as well as the energy principle in your analysis.
Egamma ray = 4Your answer is incorrect. eV
(d) Now that you know the energy of the gamma ray, calculate the (small) kinetic energy of the helium-3 nucleus. Hint: You will find that the speed of the helium-3 nucleus is very small compared to the speed of light.
KHe-3 nucleus = 5 eV
(e) You see that there is a lot of energy in the final products of the fusion reaction, which is why scientists and engineers are working hard to try to build a practical fusion reactor. The problem is the difficulty and energy cost in getting the electrically charged reactants close enough to fuse (the proton and deuteron in this reaction). If these problems can be overcome, what is the gain in available energy in this reaction?
(KHe-3 nucleus+Egamma ray) - (Kproton+Kdeuteron) = 6 eV
2. Relevant equations
3. The attempt at a solution
I tried using the rest energies to convert to kinetic energy, but it didn't work.
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