Semi difficult fusion problem, thanks

[rob36]

Homework Statement

a) 1H + 2H 3He + gamma
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 2.7 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 = ??eV
Kdeuteron = ??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 = ??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 =?? 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) = ??eV

Homework Equations

Ef = Ei
Uf = Ki = 1/2 m v1^2 + 1/2 m v2^2
Ptot = 0
P^2*(1/2m1 + 1/2m2)

The Attempt at a Solution

I think initial total kinetic is 4.2666E-14 and beyond that I'm obviously doing something wrong...

 

[rob36]

ok
Kproton = 355414 eV
Kdeuteron = 177920eV

still trying to find the rest though
i think c involves a quadratic

 

[m2003]

I would like to offer some guidance and clarification on the problem at hand.

Firstly, the given reaction is a fusion reaction, where two particles (proton and deuteron) combine to form a heavier particle (helium-3) and release energy in the form of a gamma ray. This reaction is a type of nuclear reaction that occurs in the core of stars, including our sun. It is also a potential source of energy for future power generation.

Now, let's break down the problem and approach it step by step. In part (a), we are given the reaction equation and the rest masses of the particles involved. The key concept here is conservation of energy, where the total energy of the reactants (proton and deuteron) must equal the total energy of the products (helium-3 and gamma ray). This means that the sum of the kinetic energies and the rest energies of the reactants must equal the sum of the kinetic energies and the rest energies of the products.

Moving on to part (b), we are asked to find the minimum kinetic energies of the proton and deuteron for the reaction to take place. To do this, we need to consider the momentum principle, which states that the initial total momentum of the system is equal to the final total momentum of the system. In this case, the initial total momentum is zero, since the particles are at rest, and the final total momentum is also zero, since the products are also at rest. This means that the momentum of the proton and deuteron must be equal in magnitude but opposite in direction, so that they cancel each other out. We can use this information to solve for the minimum kinetic energies of the particles.

In part (c), we are asked to find the energy of the gamma ray. This can be done using the energy principle, which states that the initial total energy of the system is equal to the final total energy of the system. We know the initial total energy, which is the sum of the kinetic energies and rest energies of the reactants, and we know the final total energy, which is the energy of the gamma ray. Using the given equation and the fact that the photon has zero rest mass, we can solve for the energy of the gamma ray.

In part (d), we are asked to find the kinetic energy of the helium-3 nucleus. This can be done using the same approach as in part (c), where we use