1. Tardigrade
  2. Question
  3. Physics
  4. What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus (Z=79) ?

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. What is the distance of closest approach when a $5 MeV$ proton approaches a gold nucleus $(Z=79) ?$

Atoms

Solution:

$E =\frac{( Ze )( e )}{4 \pi \varepsilon_{ o } r }=\frac{9 \times 10^{9} \times 79 \times\left(1.6 \times 10^{-19}\right)^{2}}{ r \times 1.6 \times 10^{-13}}$
$r=2.3 \times 10^{-14} m$