-1 points seroP10 16 P.023 w. 0/10 Submissions Used My Notes Ask Your Teacher In
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-1 points seroP10 16 P.023 w. 0/10 Submissions Used My Notes Ask Your Teacher In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha particles (having charges of +2e and masses of 6.64 x 10 27 kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 1.41 x 107 m/s directly toward the nucleus, as in the figure below. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. 79e 2e Need Help? ReadExplanation / Answer
In essence, this is a conservation of energy problem. The two energies of interest are the kinetic energy of the alpha particle, and the potential energy of the alpha particle-gold nucleus pair. Since the two are both positively charged, they will repel each other, and their electrical potential energy will be positive at any finite distance. If the alpha particle is initially very, very far away, we can approximate their starting separation as infinite, meaning their initial electrical potential energy is zero, and the total energy of the system is just the alpha particle’s kinetic energy.
How close does the alpha particle get? When it used all of its initial kinetic energy up as electrical potential energy. The closer it gets, the larger the electrical potential energy, and the more kinetic energy it must spend. At some point, it is all gone, and the particle instantaneously stops and then turns around. At that point of closest approach, the alpha particle’s kinetic energy is zero. Comparing the energy in the initial and final cases will allow us to find the distance of closest approach R
KE of the approaching alpha = ½mv²
½(6.64*10^-27)(1.41*10^7)² = 6.60*10^-13 J
Alpha comes to rest (about to be turned) when all it's KE is transferred into electrical potential energy (due to the electric field created by gold nucleus)
Elec. pot. energy = elec potential (V as J/C) x Q2(C) =1/4(Q1)/R x Q2 .. in J
Q1 = gold charge =79e Coulomb
Q2 = alpha charge = 2e Coulomb
e = 1.6*10^-19 C
1/4 = 9*10^9Nm²/C²
R = separation of the two (point) charges (m)
E.pot.energy = (9*10^9)(79e*2e) / R = (9*10^9)*158(1.6*10^-19)² /R = (3.64*10^-26) /R
6.60*10^-13 Jke = (3.64*10^-26) /R
R = (3.64*10^-26) / 6.60*10^-13 = 5.52*10^-14 m
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