Energy in a fusion reaktion (Have I done it right?)

[Ante]

Homework Statement

The deuterium nucleus starts out with a kinetic energy of 1.09e-13 joules, and the proton starts out with a kinetic energy of 2.19e-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.

A: What will be the total kinetic energy of both particles an instant before they touch?

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


C: 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?)

D: Kinetic energy can be used to drive motors and do other useful things. If a mole of hydrogen and a mole of deuterium underwent this fusion reaction, how much kinetic energy would be generated?

Homework Equations

E_U= (1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r

E_F=E_i

The Attempt at a Solution

A: Used the law of energy conservation
E=(1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r -(k_i)

B: Potential energy becomes kinetic energy: E=(1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r = K

C: (1/(4(Pi)(Epsilon_0)))*(q_1*q_2)/r - (k_i)

D:


One mole deuterium = 6.02*10^2p
One mol protons = 6.02*10^2p
One mole gives: 7.326*10^10 J

 

[anathema]

or 73.26 GJ

A: The total kinetic energy of both particles an instant before they touch would be the sum of their individual kinetic energies, so it would be 1.09e-13 J + 2.19e-13 J = 3.28e-13 J.

B: The kinetic energy of the reaction products can be found by subtracting the total kinetic energy of the initial particles from the total kinetic energy before the reaction, so it would be 3.28e-13 J - (1.09e-13 J + 2.19e-13 J) = 0 J.

C: The gain of kinetic energy in this reaction would be the difference between the kinetic energy of the reaction products and the initial particles, so it would be 0 J - (1.09e-13 J + 2.19e-13 J) = -3.28e-13 J. This means that the products have 3.28e-13 J less kinetic energy than the initial particles.

D: To calculate the kinetic energy generated by one mole of hydrogen and deuterium undergoing this fusion reaction, we can use the formula E = mc^2, where m is the mass of the reactants and c is the speed of light. The mass of one mole of hydrogen and deuterium is 2 grams, or 0.002 kg. The speed of light is approximately 3 x 10^8 m/s. Plugging these values into the equation, we get E = (0.002 kg)(3 x 10^8 m/s)^2 = 1.8 x 10^14 J. This is equivalent to 180 TJ (terajoules) or 180,000 GJ (gigajoules).