In 1910 Rutherford performed a classic experiment in which he directed a beam of
ID: 1437570 • Letter: I
Question
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 2.04 MeV (2.04 106 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.
Explanation / Answer
The Decrease in Kinetic Energy = Increase in Electric potential Energy
Kinetic Energy = KQ1Q2/r = eV
so by substituing
(2.04 *10^6*1.6*10^-19) = (9*10^9 * 79* 1.6*10^-19 * 2*1.6 *10^-19)/r
r = (9*10^9 * 79* 1.6*10^-19 * 2*1.6 *10^-19)/((2.04 *10^6*1.6*10^-19)
So closet distance = 1.11 *10^-13 m