Kinetic energy fusion of protons

[mshah3]

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.

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?


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

C: Gain of kinetic energy:
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?)
joules

D: Fusion as energy source
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?
joules
(For comparison, around 1e6 joules are obtained from burning a mole of gasoline.)

Homework Equations

Uel = (9E9) (q1q2) / ( r)

The Attempt at a Solution

used formula: Uel = (9E9) (q1q2) / ( r)
as my formula for kinetic energy
was incorrect

:(

 

[learningphysics]

Uel = (9E9) (q1q2) / ( r)

gives potential energy.

use conservation of energy.

initial potential energy + initial kinetic energy = final potential energy + final kinetic energy.

I think you can assume that initial r is infinite... so initial potential energy = 0.

 

[ogb p]

I would like to clarify that the formula you have used, Uel = (9E9) (q1q2) / ( r), is not the correct formula for kinetic energy. The correct formula for kinetic energy is KE = 1/2mv^2, where m is the mass of the particle and v is its velocity.

In the given scenario, the total kinetic energy of both particles an instant before they touch would be the sum of their individual kinetic energies. This can be calculated by using the formula KE = 1/2mv^2, where m is the mass of the particle and v is its velocity.

For the deuterium nucleus, m = 2.0141 u (atomic mass units) and v = 6.2e-14 joules (given in the problem). Converting the mass to kilograms and using the conversion factor 1 u = 1.66054e-27 kg, we get m = 3.3436e-27 kg. Plugging these values into the formula, we get KE = 1/2(3.3436e-27 kg)(6.2e-14 m/s)^2 = 6.155e-41 joules.

Similarly, for the proton, m = 1.0073 u and v = 1.23e-13 joules. Converting the mass to kilograms, we get m = 1.6735e-27 kg. Plugging these values into the formula, we get KE = 1/2(1.6735e-27 kg)(1.23e-13 m/s)^2 = 1.282e-41 joules.

Therefore, the total kinetic energy of both particles an instant before they touch is 6.155e-41 + 1.282e-41 = 7.437e-41 joules.

Moving on to the kinetic energy of the reaction products (helium nucleus plus photon), we need to consider the conservation of energy. In a fusion reaction, the total energy of the reactants (deuterium and proton) should be equal to the total energy of the products (helium nucleus and photon). Therefore, the kinetic energy of the reaction products would be equal to the total kinetic energy of the reactants, which is 7.437e-41 joules.

Finally, to calculate the gain of kinetic energy in