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2 protons are fired at a nucleus, the second one with 2 times the velocityof the...

2 protons are fired at a nucleus, the second one with 2 times the velocityof the first. how would the elctrostatic energy of the second protonat its turning radius compare with the electrostatic energy of the first at its turning radius?

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Answer #1

when a proton encounters a nucleus, the nucleus is assumed to be at rest ,to a good approximation.

the electrostatic force between any two charged particles is given by

where, k is a constant

in the given problem, two protons are fired at a nucleus with the velocity of 2nd one twice the first.

when the proton approaches nucleus with velocity v, it slows down and comes to stop at distance called turning radius.

since, we see that the electrostatic energy is independent of velocity, so the electrostatic energy of both the protons at turning radius will be the same.

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