Bringing the particles together

[shehan1]

1.This is all the given information:

In a fusion reaction, the nuclei of two atoms join to form a single atom of a different element. In such a reaction, a fraction of the rest energy of the original atoms is converted to kinetic energy of the reaction products. A fusion reaction that occurs in the Sun converts hydrogen to helium. Since electrons are not involved in the reaction, we focus on the nuclei.

Hydrogen and deuterium (heavy hydrogen) can react to form helium plus a high-energy photon called a gamma ray:




Objects involved in the reaction: Particle # of protons # of neutrons Charge Rest Mass (atomic mass units)
1H (proton) 1 0 +e 1.0073
2H (deuterium) 1 1 +e 2.0136
3He (helium) 2 1 +2e 3.0155
gamma ray 0 0 0 0

Although in most problems you solve in this course you should use values of constants rounded to 2 or 3 significant figures, in this problem you must keep at least 5 significant figures throughout your calculation. Problems involving mass changes require many significant figures because the changes in mass are small compared to the total mass. In this problem you must use the following values of constants, accurate to 5 significant figures:

Constant Value to 5 significant figures
c (speed of light) 2.9979e8 m/s
e (charge of a proton) 1.6022e-19 coulomb
atomic mass unit 1.6605e-27 kg
8.9875e9 N·m2 /C2

This is the question:

The deuterium nucleus starts out with a kinetic energy of 8.4e-14 joules, and the proton starts out with a kinetic energy of 1.67e-13 joules. The radius of a proton is 0.9e-15 m; assume that if the particles touch, the distance between their centers will be twice that. What will be the total kinetic energy of both particles an instant before they touch?



3. I tried just adding the two kinetic energies it was wrong. My teacher gives homework that's ahead of the classwork so i don't really know what to do or what equation to use.

 

[Apphysicist]

The total energy remains the same. Since the fusion reaction hasn't taken place, I would say the rest energies aren't important...also because the kinetic energy is defined as the total relativistic energy minus the rest energy.

So what kinds of energy changes could there be? Since they've gotten closer, they've lost gravitational potential, but that was minuscule anyway.

There is another conservative force, the coulomb force, that will make these ~point charges have a higher potential energy when they are closer together (because they are oppositely charged, pushing them together requires work, and that means losing kinetic energy but gaining potential).

The proton (which has a higher given KE), is at about 1 MeV, so I would say it's reasonably safe to disregard any relativistic effects anyway (then why supply c at all?).

It looks to me like the question is just K(tot)=K(deu, i) + K(pro, i) - U(elec)

 

[shehan1]

Thanks a lot that gave me the right answer.

 

[shehan1]

How would i then answer this part:

What is the kinetic energy of the reaction products (helium nucleus plus photon)?


I know how to solve the rest of the problem other than those two parts.

 

[Apphysicist]

Well then you need to look at the total energy before and after.

K(i)=K(deu,i)+K(pro,i)
But also there was energy in the masses:
E(i)=E(deu, rest) +E(pro, rest)
I assume they brought the particles together from infinity because there was no electrical potential energy mentioned in the problem initially so your TOTAL INITIAL ENERGY:

K(i)+E(i)

Then your energy after would involve the energy of a photon, the kinetic energy and rest energy of the helium and heat and nuclear binding forces. I assume the problem disregards heat and nuclear binding forces.

I wonder if you'll encounter a problem in solving that second part for this reason: a photon doesn't really have "kinetic energy" as it is massless, but it DOES have energy. Thus, you can't just equate K(i)+E(i)=K(He, f)+E(He, rest)...because really the right-hand-side also has the energy of the photon. The energy of the photon is E=hv (v for frequency), but v=c/λ (λ wavelength)...you know neither the frequency or the wavelength. But maybe the question's fine with you disregarding that??