Question

A gold nucleus is 466 fm (1 fm = 10-15 m) from a proton, which initially is at rest. When the proton is released, it speeds away because of the repulsion that it experiences due to the charge on the gold nucleus. What is the proton's speed a large distance (assume to be infinity) from the gold nucleus

Answer 1

Answer:

$v=6.8\times10^6m/s$

Explanation:

The sum of the kinetic and electric potential energies of the proton when initially released must be equal to their sum at infinity, so we have:

$K_i+U_i=K_f+U_f$

Which, since $K_i=0J$ because initially the proton is at rest, is:

$\frac{kqQ}{d}=\frac{mv^2}{2}+\frac{kqQ}{r_\infty}$

where $k=9\times10^9Nm^2/C^2$ is Coulomb's constant, $q=1.6\times10^{-19}C$ the charge of the proton, $Q=(79)(1.6\times10^{-19})C$ the charge of the gold nucleus, since it has 79 protons, $d=466\times10^{-15}m$ the initial separation between them, $m=1.67\times10^{-27}kg$ the mass of the proton and v its final velocity. $r_\infty$ is very far away, so the final electric potential will be 0J, and we have:

$v=\sqrt{\frac{2kqQ}{md}}=\sqrt{\frac{2(9\times10^9)(1.6\times10^{-19})^2(79)}{(1.67\times10^{-27})(466\times10^{-15})}}m/s=6839409m/s$