We want to find the number of protons in the particle collider. We know that 1 Kj of energy was released. 1 electron capture releases 511 electron volts. There are ≈6.242 * 1021 electron volts in a kilojoule. Thus there were 1.221 * 1019 reactions. The rate of reactions is proportional to the number of electrons. Let the number of neutrons be N and the electrons E. The increase in electrons is E+, and that of neutrons, N+. E is increased linearly and decreased proportionally to itself. The distance from equilibrium will decrease by a factor of e in every length of time over which it would become zero if it were only being decreased, and at a constant rate equivalent to the instantaneous force of decrease at t=0. We know this from Battle of the Blobs and Spinning Stick. This duration of time equals 1/n. The equilibrium is at nE=E+, so the number of electrons will approach E+/N+. E, at any time, equals E+/N+ * (1-e-t/N+). The number of neutrons is E+t minus all that stuff. We make the equations below for t=1, t=3, t=3.25 and t=3.5 to find our answer. The final equation to get E+ and N+ is rather difficult, and it is necessary to find them to solve the problem. This equation will be a challenge of it’s own, Exquisite Equation, coming next week.
P_{t=1}=5\,000\,000+\frac{E^+}{N^+} (1-e^{\frac{-1}{N^+}})\\[8pt] E_{t=3}=\frac{E^+}{N^+}1-e^{\frac{-3}{N^+}} \\[8pt] E_{t=3.25}=\frac{E_{t=3}}{e^\frac1{N^+}} =1\,000\,000\\[8pt] P_{t=3.5}=P_{t=1}+\frac {E^+}{N^+}(1-e^{\frac {-3}{N^+}})+E_{t=3.25}-3E^+-1\\[8pt] \frac{E^+-E^+e^\frac{-3}{N^+}} {N^+e^\frac1{N^+}}=1\,000\,000