# Finding Speed with the Energy Principle - Nuclear Fission

The question goes along the lines:

Uranium-$236$ fissions when it absorbs a slow-moving neutron. The two fission fragments can be almost any two nuclei whose charges $Q_1$ and $Q_2$ add up to $92e$ ($e$ is the charge of a proton, $1.6 \times 10^{-19}C$), and whose nucleons add up to $236$ protons and neutrons. One of the possible fission modes involves nearly equal fragments of the two palladium nuclei with $Q_1 = Q_2 = 46e$. The rest masses of the two palladium nuclei add up to less than the rest of the mass of the original nucleus. Make the assumption that there are no free neutrons, jus the palladium nuclei. The rest mass of the U-$236$ is 235.996 u (unified atomic mass units), and the rest mass of each $Pd-118$ nucleus is 117.894 u, where $1$ $u = 1.7 \times 10^{-27} kg$.

(a) Calculate the final speed $v$, when the Pd nuclei have moved far apart (due to the mutual electric repulsion). Is this speed small enough that $\frac{p^2}{2m}$ is an adequate ($p$ is momentum) approximation for the kinetic energy of one of the palladium nuclei? (make the non relativistic assumption first, then compare $v$ is indeed small enough to $c$)

(b) Using energy considerations, calculate the distance between centers of the palladium nuclei just after fission, when they are starting from rest.

So the beginning step is to chose the system, for which I include both the Pd nuclei. So the energy principle is now: $$\Delta E_{sys} = W_{ext}$$ We know that $W_{ext} = 0$ since my system is defined not including any external work done on the system. So now instead the system is $\Delta E_{sys} = 0 = \Delta K_{m_1} + \Delta K_{m_2} + \Delta U_{elec_1} + \Delta U_{elec_2}$ We can ignore the earth and the gravitational potential energy sine the masses are very small. If we expand the equation, we get: $$K_{f_1} - K_{i_1} + K_{f_2} + K_{i_2} + U_{f_1} - U_{i_1} + U_{f_2} - U_{i_2} = 0$$ getting rid of terms that are $0$, we get: $$K_{f_1} + K_{f_2} - U_{i_1} - U_{i_2} = 0$$ furthermore: $$K_{f_1} + K_{f_2} = U_{i_1} + U_{i_2}$$ We know that $K = \frac{1}{2}mv^2$ and that $K_{f_1} = K_{f_2}$ are equal since the masses are the same. So we can simplify that to $2K_f = mv^2$. Same for $U = \frac{1}{2 \pi \epsilon_0} \frac{Q_1 Q_2}{r}$. So $U_{i_1} = U_{i_2} = 2U_i = \frac{2}{2 \pi \epsilon_0} \frac{Q_1 Q_2}{r}$. So we can essentially write: $$mv^2 = 2 \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r}$$ $$v = \sqrt{\frac{2}{m} \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r}}$$ I then get lost n finding $r$ since it is not given, furthermore, (b) asks us to find the distance before they move apart. So I am approaching the problem incorrectly. I would need to figure out the error here and how to commence part b.

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Welcome to Physics.SE! Hmmm...you are being too meticulous about this and at the same time ignoring a important aspect of the problem. Make your life easier...consider the situation when the fragments are far enough apart that you can ignore the electrical potential (if any, there are enough electrons around after all) between them. And remember that mass is a form of energy (very important when we start talking about nuclear reactions, that). –  dmckee Oct 14 '11 at 0:58
Note to potential answerers: solving the problem would render this into a "particular homework exercise" and run afoul the FAQ, but addressing the whys of the solution looks like a "principles at work" question to me and is good to go. –  dmckee Oct 14 '11 at 1:01
I just remembered that the initial mass energy is not equal to the final mass energy. There is a different of .208 u, or $.3536 \times 10^{-27} kg$. So the first part is covered since $E = mc^2$ for the rest energy. I can use what I have now essentially for part b. Gracias @dmckee –  Salazar Oct 14 '11 at 1:05