Question 24.100: An alpha particle (which has two protons) is sent directly t......

An alpha particle (which has two protons) is sent directly toward a target nucleus containing 92 protons. The alpha particle has an initial kinetic energy of 0.48 pJ. What is the least center-to-center distance the alpha particle will be from the target nucleus, assuming the nucleus does not move?

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The distance r being looked for is that where the alpha particle has (momentarily) zero kinetic energy. Thus, energy conservation leads to

K_0+U_0=K+U \Rightarrow\left(0.48 \times 10^{-12} J \right)+\frac{(2 e)(92 e)}{4 \pi \varepsilon_0 r_0}=0+\frac{(2 e)(92 e)}{4 \pi \varepsilon_0 r} .

If we set r_0=\infty\left(\text { so } U_0=0\right) then we obtain r=8.8 \times 10^{-14} \,m .

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