In 1910 Rutherford performed a classic experiment in which he directed a beam of
ID: 2042544 • 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 2.72 MeV (2.72*10^6 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.THE ANSWER TO THIS SAMPLE PROLEM IS 8.36e-14 m ... HOW DO I GET THIS EXACT ANSWEr. THIS PROBLEM WAS CONFUSING
Explanation / Answer
Initially only kinetic energy finally only potential so KE=PE KE=k q1 q2 /r 2.72E6 eV = 9E9 2e 79 e/r use that e=1.6E-19 r= 9E9 2 * 79* 1.6E-19/2.72E6=8.36 E-14 m there we go