Calculating Kinetic Energy and Distance for Plutonium-241 Fission Fragments

[fball558]

this problem is really long. here it is

For some isotopes of some very heavy nuclei, including nuclei of thorium, uranium, and plutonium, the nucleus will fission (split apart) when it absorbs a slow-moving neutron. Plutonium-241, with 94 protons and 147 neutrons, can fission when it absorbs a neutron and becomes Plutonium-242. The two fission fragments can be almost any two nuclei whose charges Q1 and Q2 add up to 94e (where e is the charge on a proton, e = 1.610e-19 coulomb), and whose nucleons add up to 242 protons and neutrons (Pu-242, formed from Pu-241 plus a neutron). One of the possible fission modes involves nearly equal fragments, silver nuclei (Ag) each with electric charge Q1 = Q2 = 47e. The rest masses of the two silver nuclei add up to less than the rest mass of the original nucleus. (In addition to the two main fission fragments there are typically one or more free neutrons in the final state; in your analysis make the simplifying assumption that there are no free neutrons, just two silver nuclei.)

The rest mass of the Pu-242 nucleus (formed from Pu-241 plus a neutron) is 242.007 u (unified atomic mass units), and the rest mass of each of the two Ag-121 nuclei is 120.894 u, where 1 u = 1.6610e-27 kg (approximately the mass of one nucleon). In your calculations, keep at least 6 significant figures, because the calculations involve subtracting large numbers from each other, leaving a small difference. There are three states you should consider in your analysis:

1) The initial state of the Pu-242 nucleus, before it fissions.
2) The state just after fission, when the two silver nuclei are close together, and momentarily at rest.
3) The state when the silver nuclei are very far away from each other, traveling at high speed.
(a) Calculate the final speed v, when the silver nuclei have moved very far apart due to their mutual electric repulsion. Keep at least 6 significant figures in your calculations. In your analysis it is all right to use the nonrelativistic formulas, but you then must check that the calculated v is indeed small compared to c. (The large kinetic energies of these silver nuclei are eventually dissipated into thermal energy of the surrounding material. In a nuclear reactor this hot material boils water and drives an electric generator.)

(b) Using energy considerations, calculate the distance between centers of the silver nuclei just after fission, when they are momentarily at rest. Keep at least 6 significant figures in your calculations.

c) A proton or neutron has a radius r of roughly 1e-15 m, and a nucleus is a tightly packed collection of nucleons. Therefore the volume of the nucleus, (4/3)pi R^3, is approximately equal to the volume of one nucleon, (4/3)per^3, times the number N of nucleons in the nucleus: (4/3)R3 = N(4/3)r3. So the radius R of a nucleus is about N^1/3 times the radius r of one nucleon. More precisely, experiments show that the radius of a nucleus containing N nucleons is (1.3e-15 m)N^1/3. What is the radius of a silver nucleus?

(d) You could make a careful scale drawing on paper of the two silver nuclei in part (b), just after fission, and label the drawing with the distances that you calculated in parts (b) and (c). If the two silver nuclei are nearly touching, this would be consistent with our model of fission, in which the Pu-242 nucleus fissions into two pieces that are initially nearly at rest. How big is the gap between the surfaces of the two nuclei? (If you have done the calculations correctly, you will indeed find that the gap is a rather small fraction of the center-to-center distance, which means that our model for the fission process is a pretty good model.)

 

[jbsteele]

a) The final speed v of the two silver nuclei when they are very far apart due to their mutual electric repulsion can be calculated using the equation E = 1/2 mv^2, where E is the kinetic energy, m is the mass of the two silver nuclei, and v is the final speed of the silver nuclei. Solving for v yields v = sqrt(2E/m). Plugging in the values for the kinetic energy E = 241.113 u and the mass m = 241.788 u, yields a final speed of v = 1.24e+07 m/s. This is indeed small compared to the speed of light, c = 3e+08 m/s. b) Using energy considerations, the distance between centers of the silver nuclei just after fission, when they are momentarily at rest, can be calculated by setting the total potential energy U equal to the total kinetic energy K. Since U = -Q1Q2/r, where Q1 and Q2 are the electric charges of the two silver nuclei and r is the distance between their centers, and K = 1/2 mv^2, where m is the mass of the two silver nuclei and v is their initial velocity (which is 0), we can solve for r: r = -Q1Q2/K = -Q1Q2/mv^2 = -47^2/241.788*0 = 0.c) The radius R of a silver nucleus, containing N nucleons, is (1.3e-15 m)N^1/3. With the number of nucleons in a silver nucleus being N = 121, the radius of a silver nucleus is (1.3e-15 m)121^1/3 = 2.89e-15 m.d) With the calculated distance between centers of the silver nuclei being 0, this indicates that the two silver nuclei are nearly touching each other, which is consistent with our model for the fission process. The gap between the surfaces of the two nuclei is therefore 0.

 

[nlightner]

I would first like to commend the detailed and thorough nature of this problem. It is clear that a lot of thought and effort has gone into crafting it.

Now, moving on to the problem itself, calculating the kinetic energy and distance for plutonium-241 fission fragments involves a number of complex calculations and considerations. Firstly, it is important to note that fission is a nuclear reaction in which a heavy nucleus splits into two smaller nuclei. In the case of plutonium-241, the nucleus will fission when it absorbs a slow-moving neutron, resulting in the formation of plutonium-242.

In this problem, we are specifically looking at the fission mode where the two resulting fragments are silver nuclei with equal charges. To begin our analysis, we need to consider three states: the initial state of the Pu-242 nucleus before fission, the state just after fission when the two silver nuclei are close together and momentarily at rest, and the state when the silver nuclei are very far apart and traveling at high speeds.

To calculate the final speed of the silver nuclei when they are far apart, we can use the nonrelativistic formula for kinetic energy, which is equal to 1/2mv^2. Here, m represents the mass of the silver nuclei and v is their final speed. We can calculate the mass of each silver nucleus by adding up the rest masses of the Pu-242 nucleus and the neutron that was absorbed during fission. Using the given values for rest masses and the conversion factor for unified atomic mass units, we can calculate the mass of each silver nucleus to be 120.894 u. Plugging this value into the kinetic energy formula, we can solve for v and get the final speed of the silver nuclei.

However, it is important to check that this calculated value for v is small compared to the speed of light, c. If it is not, then we must use the relativistic formula for kinetic energy, which takes into account the effects of special relativity at high speeds.

Next, we can use energy considerations to calculate the distance between the centers of the silver nuclei just after fission, when they are momentarily at rest. This can be done by equating the potential energy of the two nuclei due to their electric repulsion with the kinetic energy they possess. Again, we must ensure that our calculated distance is consistent with the nonrelativistic assumption made earlier.

To calculate the radius of a silver nucleus,