Question

4. + ㅢ1 points My Notes Ask Your Teacher In 1910 Rutherford performed a classic experiment in which he directed a beam of alpha particles at a thin gold foil. He unexpectedly observed a few of the particles scattered almost directly backward. This result was not consistent with then current models of atomic structure and led Rutherford to propose the existence of a very dense concentration of positive charge at the center of an atom-the atomic nucleus. The alpha particle has a charge of +2e and the gold nucleus a charge of +79e. Suppose that an alpha particle is initially a great distance from the gold, has a kinetic energy of 3.41 MeV (3.41 × 106 eV), and is headed directly at a gold nucleus. How close will the particle come to the center of the nucleus? Treat the nucleus and the alpha particle as point charges

Answer 1

The alpha is stipulated to be initially far away so you can set its electric PE=0 . Then its total energy "Ei" initially is just its kinetic energy;

Ei = 3.41 Mev

When the alpha is stopped by the Gold nucleus a distance "r" from the nucleus all of its energy at that point is PE .So its final energy, at that point is;

Ef = kqQ/r = k(158)e^2/r

Since energy is conserved just set the energies equal and solve for "r"

with

k = 9 x 10^9 J-m/C^2

e = 1.6 x 10^-19 C

So,

(3.41*10^6*1.6*10^-19 J) = (9*10^9)(158)(1.6*10^-19)^2/r

=> r = 6.67*10^-14 m