A high-energy beam of carbon nuclei (612C^{12}_6\text{C}612C) is directed at a target consisting of a thin gold foil. Some of these carbon nuclei pass extremely close to a gold nucleus (79197Au^{197}_{79}\text{Au}79197Au) and undergo electrostatic deflection.
At the position of closest approach, the distance between the carbon nucleus and the gold nucleus is 4.8 × 10-14 m. Calculate the magnitude of the electrostatic force F F\,F between the carbon nucleus and the gold nucleus at this position. (Take ε0=8.85×10−12 F m−1\varepsilon_0 = 8.85 \times 10^{-12}\text{ F m}^{-1}ε0=8.85×10−12 F m−1 and e=1.60×10−19 Ce = 1.60 \times 10^{-19}\text{ C}e=1.60×10−19 C)
The initial kinetic energy of each carbon nucleus in the beam is 18.0 MeV18.0\text{ MeV}18.0 MeV. Calculate the magnitude of the initial momentum p p\,p of each carbon nucleus. (Take the mass of the carbon nucleus to be 2.0 × 10-26 kg)
