I can't figure this out!?

Question

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?
K1H + K2H ? joules

Answer 1

First of all, it may be convenient to convert from joules to electron volts, (where 1 eV = 1.602(10^-19) J )

In this case, the relative motion of the particles is not as important as the the *total distance the particles have to move through* in order to get to their current position. This problem can be greatly simplified, by assuming instead, that the deuterium nucleus remains stationary, and the proton has all of the 5.65(10^-13) J of kinetic energy. As long as *energy is conserved*, we need not worry about altering our frame of reference in such a manner.

The work done by a charged particle, as it moves through a point electric field is:

E = kc*q1*q2* { (1 / r_final) - (1/ r_initial) },

where E is in electron volts. Since the particles start out at infinite distance, the second term is zero, reducing to:

E = kc*q1*q2 / r_final.

Note: kc ≈ 8.988(10^9) N m^2 / C^2

All that remains now is to calculate, using

q1 = q2 = e ≈ 1.602(10^-19) C,

and subtract it from the original kinetic energy.

Hope that's not too confusing,
~W.O.M.B.A.T.

Answer 2

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Answer 3

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