The actual size of a gold nucleus is approximately 7 fm. Knowing this, calculate the kinetic energy (KE, or KEa, however you may choose to label the measurement) that an alpha particle would need to just touch the outside of the nucleus. Does this seem like a reasonable number?
Hint - the LHC is currently the most powerful particle accelerator on Earth, and i?t operates at 13 TeV (TeV = 1012 eV).
