How Is Kinetic Energy Transformed in Nuclear Fusion Reactions?

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SUMMARY

Nuclear fusion reactions, such as the conversion of hydrogen to helium in the Sun, involve the transformation of rest energy into kinetic energy of reaction products. In the specific scenario discussed, a proton (1H) and a deuteron (2H) collide, resulting in the formation of a helium-3 nucleus and a gamma ray. The initial kinetic energies of the proton and deuterium are 1.23e-13 joules and 6.2e-14 joules, respectively. The total kinetic energy of the reaction products is greater than the initial kinetic energies, indicating a gain in kinetic energy during the fusion process.

PREREQUISITES
  • Understanding of nuclear fusion principles
  • Familiarity with kinetic energy calculations
  • Knowledge of atomic structure, specifically protons and deuterons
  • Basic grasp of energy conservation laws in physics
NEXT STEPS
  • Calculate the total kinetic energy of fusion products in nuclear reactions
  • Explore the role of gamma rays in nuclear fusion
  • Investigate the energy yield of fusion reactions involving hydrogen and deuterium
  • Learn about the conditions necessary for nuclear fusion to occur
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Students studying nuclear physics, physicists interested in fusion energy, and educators teaching concepts of energy transformation in nuclear reactions.

astros10
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Homework Statement


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:

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

A proton (1H nucleus) and a deuteron (2H nucleus) start out far apart. An experimental apparatus shoots them toward each other (with equal and opposite momenta). If they get close enough to make actual contact with each other, they can react to form a helium-3 nucleus and a gamma ray (a high energy photon, which has kinetic energy but zero rest energy). Consider the system containing all particles.

A)The deuterium nucleus starts out with a kinetic energy of 6.2e-14 joules, and the proton starts out with a kinetic energy of 1.23e-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?



B) Now that the proton and the deuterium nucleus are touching, the reaction can occur. 1.What is the kinetic energy of the reaction products (helium nucleus plus photon)?

2.What was the gain of kinetic energy in this reaction? (The products have more kinetic energy than the original particles did when they were far apart. How much more?)

3. If a mole of hydrogen and a mole of deuterium underwent this fusion reaction, how much kinetic energy would be generated?


Homework Equations


c (speed of light)= 2.9979e8 m/s
e (charge of a proton) =1.6022e-19 coulomb
atomic mass unit =1.6605e-27 kg
coulomb's constant=8.9875e9 N·m2 /C2
Kf + Uf = Ki+Ui
U=-(GMm)/R



The Attempt at a Solution


The only one I have attempted thus far is A because I believe I need that answer to solve the three questions in part B. What I did for the first part was take:

Kf + Uf = Ki+Ui
Ui=0, Ki=6.2e-14j+1.23e-13j
Could the answer to the first part be 0 for the final kinetic energy, since all the energy has been transferred to potential energy?
 
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astros10 said:

Homework Equations


Kf + Uf = Ki+Ui
U=-(GMm)/R
The last formula is for the gravitational potential energy of two point masses (or spherical masses). Gravity is not relevant here.

astros10 said:

The Attempt at a Solution


The only one I have attempted thus far is A because I believe I need that answer to solve the three questions in part B. What I did for the first part was take:

Kf + Uf = Ki+Ui
Ui=0, Ki=6.2e-14j+1.23e-13j
Could the answer to the first part be 0 for the final kinetic energy, since all the energy has been transferred to potential energy?
No. You need to calculate Uf and then substitute into your equation
Kf + Uf = Ki+Ui
 

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