Answer
See answers.
Work Step by Step
Energy is conserved. At the beginning, when the particles are very far apart, there is KE but no PE. At the end there is PE but no kinetic energy.
$$KE_i=PE_f$$
The initial KE is given as 7.7 MeV, which is $1.232\times10^{-12}J$. Find the separation of the 2 particles at the end. Use equation 30–1.
$$1.232\times10^{-12}J =PE_f$$
$$1.232\times10^{-12}J =k\frac{q_{\alpha}q_{Fm}}{ r_{separation}}$$
$$ r_{separation}=k\frac{q_{\alpha}q_{Fm}}{1.232\times10^{-12}J }$$
$$ r_{separation}=(8.99\times10^9 Nm^2/C^2)\frac{(2)(79)(1.60\times10^{-19}C)^2}{1.232\times10^{-12}J }$$
$$=3.0\times10^{-14}m$$
Find the ratio of this distance to a gold atom’s nuclear radius. Use equation 30–1 to calculate the size of the gold nucleus.
$$\frac{ r_{separation}}{r_{Au}}=\frac{2.952\times10^{-14}m }{197^{1/3}(1.2\times 10^{-15}m)}=4.2$$
