A plutonium- 239 nucleus at rest spontaneously undergoes fission via the following reaction: $$ { }_{94}^{239} \mathrm{Pu} \rightarrow{ }_{46}^{112} \mathrm{Pd}+{ }_{48}^{124} \mathrm{Cd}+3 \mathrm{n} . $$ The mass of plutonium-239 in atomic mass units is \({ }^{239} \mathrm{Pu}=239.0521\). The mass of palladium-112 is 111.9073. The mass of cadmium-124 is \(123.9176\). The mass of a neutron is \(1.0087\). a) What is the Q value of the reaction? b) If we ignore the neutrons, what should be the speeds of the palladium and cadmium nuclei? (Assume Newtonian physics.) c) If the palladium and cadmium nuclei can be treated as spheres, and are initially far apart, how close can they get to each other at these velocities? d) \({ }^{112} \mathrm{Pd}\) decays via beta emission to \({ }^{112}\) At with a half-life of 21 hrs. If a sample contains \(97 \%\) argon, how long has the palladium been decaying?

We can determine the Q value of the reaction by finding the difference in mass between the reactants and products. The Q value can then be converted into energy using the Einstein mass-energy equivalence formula (\(E = mc^2\)). The mass difference is given by: $$ \Delta m = m_{Pu} - (m_{Pd} + m_{Cd} + 3m_n), $$ where \(m_{Pu}\) is the mass of plutonium-239, \(m_{Pd}\) is the mass of palladium-112, \(m_{Cd}\) is the mass of cadmium-124, and \(m_n\) is the mass of a neutron. Plug in the given values and calculate the mass difference (in atomic mass units): $$ \Delta m = 239.0521 - (111.9073 + 123.9176 + 3 \times 1.0087) = 0.2011 \, \text{amu}. $$ Now, convert this to energy using the Einstein mass-energy equivalence formula: $$ Q = \Delta m \times c^2, $$ where \(c\) is the speed of light (\(2.998 \times 10^8 \, \text{m/s}\)). Since we are dealing with atomic mass units, the conversion factor between energy and mass is \(1 \, \text{amu} = 931.5 \, \text{MeV}\). Therefore, $$ Q = 0.2011 \cdot 931.5 = 187.21 \, \text{MeV}. $$

Ignoring the neutrons, the conservation of linear momentum implies that the two resulting nuclei (Palladium and Cadmium) must have equal and opposite momenta. We can use the formula for kinetic energy and momentum conservation to find the velocities of both nuclei. Using kinetic energy formular, we have: $$ Q = K_{Pd} + K_{Cd} = \frac{1}{2}m_{Pd}v_{Pd}^2 + \frac{1}{2}m_{Cd}v_{Cd}^2, $$ where \(K_{Pd}\) and \(K_{Cd}\) represent the kinetic energy of palladium and cadmium, respectively. Since the momenta are equal and opposite, we have: $$ m_{Pd}v_{Pd} = m_{Cd}v_{Cd}. $$ We can solve for \(v_{Pd}\) and \(v_{Cd}\) using these equations and the given masses. Our first equation can be rearranged as: $$ v_{Cd}^2 = \frac{2(Q - \frac{1}{2}m_{Pd}v_{Pd}^2)}{m_{Cd}}. $$ And by substituting the momentum conservation equation into the above equation: $$ v_{Pd} = \frac{m_{Cd}v_{Cd}}{m_{Pd}}. $$ Solve for \(v_{Pd}\) and \(v_{Cd}\) to get their velocities.

We can determine the closest approach distance by using the concept of stopping distance when considering the repulsive electrostatic force between the two nuclei, which have charges equal to their atomic numbers. The stopping distance can be found using the conservation of energy, considering the kinetic energy of the two nuclei is converted into potential energy due to electrostatic repulsion. The electrostatic potential energy can be calculated as: $$ U = \frac{kZe^2}{r}, $$ where \(k\) is Coulomb's constant, \(Z\) is the atomic number of each nucleus, \(e\) is the elementary charge, and \(r\) is their separation. The total initial kinetic energy of the two nuclei (ignoring the neutrons) can be calculated as: $$ K = \frac{1}{2}m_{Pd}v_{Pd}^2 + \frac{1}{2}m_{Cd}v_{Cd}^2 = Q. $$ By equating the initial kinetic energy with the potential energy, we can solve for the closest approach distance, \(r\): $$ \frac{k(Z_{Pd}Z_{Cd})e^2}{r} = Q. $$

Given the half-life of the palladium-112 decaying into argon-112 and the fact that our sample contains \(97\%\) argon, we can find the time elapsed since the start of the decay using the half-life formula: $$ N = N_0 \times (1/2)^{t/T}, $$ where \(N\) is the remaining amount of palladium-112, \(N_0\) is the initial amount, \(t\) is the elapsed time, and \(T\) is the half-life. Since we have \(97 \%\) argon, it means we have \(3 \%\) of palladium remaining, implying \(N = 0.03N_0\). Replacing these values and the given half-life, we can find the time elapsed since the start of the palladium decay.