Question

A proton 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?

Answer 1

Answer:

$\frac{4}{5}R$

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 $\frac{1}{2}v_0$ and when the proton is at distance R' from the nucleus with a final velocity equal to $\frac{1}{4}v_0$. Recall that initially the two particles are very far apart, so there is no initial potential energy. For the first case, we have:

$\Delta K=\Delta U\\K_{f}-K_{i}=U_{f}-U_{i}\\\frac{1}{2}m(v_{f})^2-\frac{1}{2}m(v_{0})^2=\frac{ke(Ze)}{R}-0\\\frac{1}{2}m(\frac{1}{2}v_{0})^2-\frac{1}{2}m(v_{0})^2=\frac{kZe^2}{R}\\\frac{1}{2}m\frac{1}{4}(v_{0})^2-\frac{1}{2}m(v_{0})^2=\frac{kZe^2}{R'}\\\frac{1}{8}m(v_{0})^2-\frac{1}{2}m(v_{0})^2=\frac{kZe^2}{R'}\\-\frac{3}{8}m(v_{0})^2=\frac{kZe^2}{R}(1)$

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

$\frac{1}{2}m(\frac{1}{4}v_{0})^2-\frac{1}{2}m(v_{0})^2=\frac{ke(Ze)}{R'}\\\frac{1}{2}m\frac{1}{16}(v_{0})^2-\frac{1}{2}m(v_{0})^2=\frac{kZe^2}{R'}\\\frac{1}{32}m(v_{0})^2-\frac{1}{2}m(v_{0})^2=\frac{kZe^2}{R'}\\-\frac{15}{32}m(v_{0})^2=\frac{kZe^2}{R'}(2)$

Finally, dividing (2) in (1):

$\frac{-\frac{15}{32}m(v_{0})^2}{-\frac{3}{8}m(v_{0})^2}=\frac{\frac{kZe^2}{R'}}{\frac{kZe^2}{R}}\\\\\frac{\frac{15}{32}}{\frac{3}{8}}=\frac{R}{R'}\\R'(\frac{120}{96})=R\\R'(\frac{5}{4})=R\\R'=\frac{4}{5}R$