# Fusion Reaction really long problem

1. Apr 1, 2009

### jchojnac

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.
(a) Consider a situation in which the initial total momentum of the two particles is zero. Draw a diagram showing each particle and its momentum vector before the reaction. Draw another diagram showing each particle and its momentum vector after the reaction. Clearly label the momentum vectors as to which objects they refer to. Refer to your own diagram and identify which diagram shown below is correct for the situation before the fusion reaction and which diagram is correct for the situation after the fusion reaction.

A (before)
B (before)
C (before)
D (before)
E (before)
F (after)
G (after)
H (after)
I (after)

(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 2.4 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

2. Relevant equations
No Idea

3. The attempt at a solution
No idea how to work it or where to start...help?

2. Apr 2, 2009

### hbates3000

Hey, I currently am working on the same problem through WEBASSIGN.
I also do not understand this problem but I do have the answer for the first multiple choice question pertaining to the diagram about the fusion reaction. It should be:

C(before) and I(after)

Good luck!

3. Apr 2, 2009

### dgbull3

I was able to get part b & i'm still working on c, d, & e...
Part b:
First off, you know from the picture that the two momentum of the proton & deuteron are the same

Next, you need to get the basic conservation of energy equation:
Kproton + Kdeuteron = Uelectric
p^2/(2*m(proton)) + p^2/(2*m(deuteron)) = Uelectric --->get this by plugging in for kinetic with momentum
Now what you do is solve for p (note: q for the deuteron is the same as q for the proton since a deuteron is a proton and a neutron)

now that you know p and you know that each has the same momentum, you can do:
p = m(proton)*v(proton) and p = m(deuteron)*v(deuteron)

just solve for each v and put that into the kinetic energy equation 1/2mv^2 and you will have the kinetic energy in Joules, so all you have to do is convert that to eV

4. Apr 2, 2009

### jchojnac

how do you solve for p when you don't have the velocity?

p^2/(2*m(proton)) + p^2/(2*m(deuteron)) = 1/2mv^2 + mv

5. Apr 2, 2009

### dgbull3

you solve for p using the equation:

p^2/(2*m(proton)) + p^2/(2*m(deuteron)) = Uelectric

6. Apr 2, 2009

### jchojnac

right and Uelectric = 1/2mv^2 +mv

which mass and velocity do I use for this part?

7. Apr 2, 2009

### dgbull3

Uelectric = 1/(4*pi*epsilon)*Q*q/r

8. Apr 2, 2009

### jchojnac

what is epsilon, Q, q, and r? They aren't in the problem.

9. Apr 2, 2009

### dgbull3

parts C & D, you have to use the two equations below:

m1*c^2+m2*c^2+Ktot,initial = Khe+m(he)*c^2+E(gamma)

Khe = E^2/(2*m(he)*c^2)

substitute the second one into the first one and you get a quadratic equation of E. in order to solve for E, you have to put it into your calculator and find the "zero" in graphing mode. once you have that, you can substitute it back in to the second equation to get part d

10. Apr 2, 2009

### dgbull3

1/(4*pi*epsilon) = 9e9 N*m^2/C^2

epsilon is a constant
Q is the charge in Coulombs of a proton
q is the charge in Coulombs of a deuteron (in this problem, it's the same as Q)

11. Apr 2, 2009

### jchojnac

I'm still confused how you got b. You find K for each by K = 1/2mv^2. And you find v by p/m. But how do you find p from p^2/(2*m(proton)) + p^2/(2*m(deuteron)) = Uelectric? If Uelectric = 1/(4*pi*epsilon)*Q*q/r is from parts c & d then how are you supposed to get b? I'm really confused.

12. Apr 2, 2009

### dgbull3

Uelectric can still be solved in part a
1/(4*pi*epsilon) = 9e9
Q = 1.602e-19
q = 1.602e-19
and r is given to you in the problem

m(proton) and m(deuteron) are also given

13. Apr 2, 2009

### yoyoz41

ahhh so the r is 1*10^-15 m from problem or 2.4*10^-15 m from part b????

14. Apr 2, 2009

### dgbull3

it's the 2.4e-15

i'm not sure why they even said the 1e-15 because you don't even use it

15. Apr 2, 2009

kk thx lol