Aproton is projected toward a fixed nucleus of charge +ze with velocity vo. initially the two particles are very far apart. when the proton is a distance r from the nucleus its velocity has decreased to 1/2 vo. how far from the nucleus will the proton be when its velocity has dropped to 1/4 vo?

Explanation:

Using the law of conservation of energy for both cases, when the proton is at distance R from the nucleus with a final velocity equal to and when the proton is at distance R' from the nucleus with a final velocity equal to . Recall that initially the two particles are very far apart, so there is no initial potential energy. For the first case, we have:

In the same way, for the second case, we have:

Finally, dividing (2) in (1):

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