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Early Models of the Atom Contents: J. J. Thomson The electron Plum pudding model of the atom

Early Models of the Atom Contents: J. J. Thomson The electron Plum pudding model of the atom Rutherford Gold foil experiment Nucleus model Solving closest approach problems Whiteboard. J.J. Thomson 1856 - 1940. “Plum Pudding” model.

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Early Models of the Atom Contents: J. J. Thomson The electron Plum pudding model of the atom

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  1. Early Models of the Atom • Contents: • J. J. Thomson • The electron • Plum pudding model of the atom • Rutherford • Gold foil experiment • Nucleus model • Solving closest approach problems • Whiteboard

  2. J.J. Thomson 1856 - 1940 “Plum Pudding” model Discovers the electron (charge/mass ratio Straight/curved) Electrons are negatively charged Electrons are surrounded in atom with positive “stuff” TOC

  3. Ernest Rutherford(1871-1937) Sends alpha particles into gold foil Measures the angle of scatter Should go right through plum pudding… TOC

  4. Ernest Rutherford(1837-1937) Rutherford’s atom: (It has a nucleus) • But is also has problems: • Why doesn’t the electron radiate energy? • How does this explain the spectral lines they had been observing? • I won’t Bohr you with the solution to this right now… TOC

  5. Most Alphas are not deflected much More deflection closer to nuclei

  6. Solving closest approach Ek = 1/2mv2 V = W/Q, W = VQ V = kQ/r VQp = 1/2mv2 kinetic = potential 1/2mv2 = (kQN/r)Qp r QN Qp Example 1: What is the closest approach of a proton going 2.6 x 106 m/s if it approaches a carbon nucleus head on? TOC

  7. Whiteboards: Closest Approach 1 | 2 TOC

  8. What is the closest approach in nm of a proton going 15,000 m/s to a hydrogen nucleus. Mp = 1.673 x 10-27 kg A hydrogen nucleus has 1 proton in it kinetic energy = electric potential energy Ek = 1/2mv2 = 1/2(1.673 x 10-27 kg)(15000 m/s)2 = 1.88213 x 10-19 J Ek = electric potential energy = VQ = Ve Ek = (kQ/r)e = (ke/r)e = ke2/r, r = ke2/Ek = (9.0 x 109 Nm2/C2)(1.602 x 10-19 C)2/(1.88213 x 10-19 J) = 1.22721 x 10-9m = 1.2 nm W 1.2 nm

  9. An Alpha particle’s closest approach brings it to within 61 nm of a Gold nucleus. What is its velocity in m/s? (3) Mp = 1.673 x 10-27 kg, Mn = 1.675 x 10-27 kg,  = 2n+2p M = 6.696 x 10-27 kg A gold nucleus has 79 protons in it, and an alpha 2 kinetic energy = electric potential energy electric potential energy = VQ = V2e (alpha) V = kQ/r = (9.0 x 109 Nm2/C2)(79*1.602 x 10-19 C)/(61 x 10-9 m) = 1.867 V elec. pot. = V2e = (1.867 V)(2)(1.602 x 10-19 C) = 5.983 x 10-19 J 5.983 x 10-19 J = 1/2mv2 = 1/2(6.696 x 10-27 kg)v2 v = 13367 m/s = 13,000 m/s W 13,000 m/s

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