The deuterium nucleus starts out with a kinetic energy of 1.88e-13 joules, and the proton starts out with a kinetic energy of 3.77e-13 joules. The radius of a proton is 0.9e-15 m; assume that if the particles touch, the distance between their centers will be twice that. What will be the total kinetic energy of both particles an instant before they touch?
Remember that the q values of a deuterium and proton is 1.6e-19. You can apply this to the (9e9)*(q1q2/r) formula where Ki + Kf = (9e9)*(q1q2/r)(1.88e-13 joules + 3.77e-13 joules). Advice for my son who's planning his curriculum ::
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