The question asks: Uranium nuclei are unstable and decay by emitting a helium nucleus, which contains two protons and two neutrons. Suppose the original nucleus has 92 protons and 146 neutrons before it emits the helium nucleus, and that the helium nucleus has an knknown speed as it departs from the surface of the remaining nucleus, which is spherical with a radius of about 7.4 fm. What is the helium nucleus's minimum kinetic energy far from the emitting nucleus and why is this the lower limit? My attempt at a solution: First, note that m_p = 1.673 * 10^-27kg, m_n = 1.675 * 10^-27kg. also, e = 1.602 * 10^-19C Initially the energy of the system is 0. We can ignore gravitational effects because they are very small compared to the potential energy from the electrical attraction. Therefore, KE_f = (kq_1q_2/r) + (1/2)m_1v_1^2 + (1/2)m_2v_2^2. To calculate q_1 and q_2 I took q_2 = 2 * e, and q_1 = -q_2. By conservation of momentum, m_1v_1 = m_2v_2 so that v_1 = (m_2/m_1)*v_2. If you let r = 7.4 fm, and plug all of this in, and set KE_f = 0, you will find that (1/2)m_2v_2^2 = .12 pJ. The answer in the book is 5.6 pJ. I don't see how they get this, and I'm confused about the question asking why the energy is minimal. Based on the equations, the minimal kinetic energy occurs when r --> infinity, in which case the kinetic energy goes to 0.