How Is Kinetic Energy Transformed in Nuclear Fusion Reactions?

AI Thread Summary
In nuclear fusion reactions, such as those occurring in the Sun, the nuclei of hydrogen and deuterium combine to form helium and a gamma ray, converting a portion of their rest energy into kinetic energy. The initial kinetic energies of the proton and deuterium nucleus are given as 6.2e-14 joules and 1.23e-13 joules, respectively. As they approach each other, their total kinetic energy before contact can be calculated using the conservation of energy principle. After the fusion occurs, the kinetic energy of the reaction products must be determined, along with the gain in kinetic energy from the reaction. The discussion emphasizes the need to calculate potential energy changes to accurately assess the final kinetic energy of the system.
astros10
Messages
10
Reaction score
0

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?
 
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top