The liquid-drop model of the nucleus suggests that high-energy...

Question

The liquid-drop model of the nucleus suggests that high-energy oscillations of certain nuclei can split (“fission”) a large nucleus into two unequal fragments plus a few neutrons. Using this model, consider the case of a uranium nucleus fissioning into two spherical fragments, one with a charge q₁ = +38e and radius r₁ = 5.5 x 10⁻¹⁵ m , the other with q₂ = + 54e and r₂ = 6.2 x 10⁻¹⁵ m. Calculate the electric potential energy (MeV) of these fragments, assuming that the charge is uniformly distributed throughout the volume of each spherical nucleus and that their surfaces are initially in contact at rest. The electrons surrounding the nuclei can be neglected. This electric potential energy will then be entirely converted to kinetic energy as the fragments repel each other. How does your predicted kinetic energy of the fragments agree with the observed value associated with uranium fission (approximately 200 MeV total)? [ 1 MeV = 10⁶ eV.]

Accepted Answer

Welcome back. Everyone. In a nuclear physics lab, researchers are studying the fission of thorium nuclei which split into two spherical fragments. One with a charge of plus 40 E and radius 4.8 multiplied by 10 to the negative 15 m and another with plus 60 E and radius 5.7 multiplied by 10 to the negative 15 m. Initially in contact at rest, calculate the electric potential energy U between these fragments and then determine the predicted kinetic energy K after fission. Based on this potential energy, compare your predicted kinetic energy with the observed value of approximately 190 mega electron volts associated with thorium fission where the charge of an electron E is 1.6 multiplied by 10 to the negative 19 coulombs and one joule equals 6.24 multiplied by 10 to the 12 mega electron volts. A says U equals 3.3 multiplied by 10 squared mega electron volts K equals 3.3 multiplied by 10 squared mega electron volts. And the predicted kinetic energy is 73% higher than the observed value. B says U is 3.3 multiplied by 10 squared mega electron volts K is 1.9 multiplied by 10 squared mega electron volts. And it's 79% higher than the observed value. C says there are 3.3 multiplied by 10 squared mega electron volts, 3.3 multiplied by 10 squared mega electron volts. And the predicted kinetic energy is 79% lower than the observed value. And D says they are 3.3 multiplied by 10 squared mega electron volts, 1.9 multiplied by 10 squared mega electron volts. And that the predicted kinetic energy is 73% lower than the observed value. Now, let's start with the first part of this problem. We want to calculate the electric potential energy. What do we already know? Well, so far, OK. We know that one of our fragments has a charge of plus 40 E. So we can call that Q one, OK. The charge of fragment one plus 40 E. And now we know E is 1.6 multiplied by 10 to the negative 19 coulombs. So it's 40 multiplied by that value which equals 6.4 multiplied by 10 to the negative 18 coulombs. So that's the actual charge of fragment. One likewise fragment two has a charge of 60 E. So that's 60 multiplied by 1.6 multiplied by 10 or raise, multiplied by 10 to the negative 19 Coombs sorry, which is equal to 9.6 multiplied by 10 to the negative 18 Coombs. Outside of that, we know that the radius of fragment one R one is 4.8 multiplied by 10 to the negative 15 m. While the radius of fragment two is 5.7 multiplied by 10 to the negative 15 m. So first of all, how can we use all of this or how can we relate it to the electric potential energy? Well, recall, so that's the first part of our problem. You should put it in red actually. OK. So in for you, so first recall that the electric potential energy U between two point charges is equal to the product of the charges. Q one Q two divided by four pi epsilon knot multiplied by D the distance between the center of the two fragments. Now, if we think about it here, we have our two fragments. OK. So let's say this is fragment one and this is fragment two and we know the radii for both fragments. So the distance between the center is going to be R one plus R two. So really, then we could rewrite this as Q one, Q two divided by four pi epsilon knott multiplied by R one plus R two. We know all of those values. So we can go ahead and solve because by substituting those values, OK, where we know that epsilon knot is the permittivity of free space and that values are constant, then it's going to be equal to Q 16.4 multiplied by 10 to the negative 18 Coombs multiplied by Q 29.6 multiplied by 10 to the negative 18 Coombs. Ok. Let me fix that properly here. All divided by four pi epsilon knot which is 8.85 multiplied by 10 to the negative 12 fats per meter multiplied by R one plus R two. So that's 4.8 where both of them are multiplied by 10 to the negative 15 m. So that's going to be really, let me set this properly here. That's really going to be 4.8 plus 5.7 multiplied by 10 to the negative 15 m. Now, when we go ahead and solve here, OK, then we should get our electric potential energy to be equal to 5.25 multiplied by 10 to the negative 11th joules. But now remember all of our answers were in mega electron volts. So we need to convert and we already know our conversion factor. OK. So converting two mega electron volts, then U is going to be equal to 5.25 multiplied by 10 to the negative 11th. Jules multiplied by 6.24 multiplied by 10 to the 12 mega electron volts per joule. And now when we go ahead and multiply, we get you to be 327.6 mega electron volts or to two significant figures, approximately 3.3 multiplied by 10 squared mega electron volts. Thus, this is the value for our electric potential energy. Next, our problem tells us to determine the predicted kinetic energy. OK. The predicted kinetic energy predicted. Wow, I hope you guys are not seeing that I can't spell predicted, but this is the predicted kinetic energy, right? And now assuming that all potential energy converts into kinetic energy after the fission. OK, then K is going to be equal to you, which is also going to be approximately equal to 3.3. Rather, let me write the exact value here. Three, 327 0.6 mega electron volts or approximately 3.3 multiplied by 10 squared mega electron volts. Now let's think about the final part of our problem. How does this predicted kinetic energy compare to the actual observed value? Well, the percentage difference, OK, let me put that in black here. The percentage difference between the predicted and observed kinetic energies, OK is going to be equal to the predicted kinetic energy minus the observed kinetic energy divided by the observed kinetic energy. That's what we're comparing it with multiplied by 100 to change it to a percent. So let's try to figure that out. So that's going to be 327.6 mega electron volts minus 190 mega electron volts divided by 190 mega electron volts multiplied by 100. I know when we go ahead and divide here, then we should get it to be equal to 72.6% of our percentage difference or approximately 73% 2 significant figures. And again, that makes sense because our, our predicted kinetic energy is almost double the actually observed kinetic energy. Thus, that tells us that the calculated electric potential energy resulted in a predicted kinetic energy significantly higher than the observed value. This discrepancy could be attributed to factors such as energy losses and the fission process or just related to the complexities of nuclear interactions. Therefore, the values of U sorry, the values are U equals 3.3 multiplied by 10 squared mega electron volts K is 3.3 multiplied by 10 squared mega electron volts. And the predicted kinetic energy is 73% higher than the observed value. If we look back at our answer choices that tells us that a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Electric Potential Energy

Uniform Charge Distribution

Kinetic Energy from Potential Energy

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