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Review for Final Exam

Review for Final Exam

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Review for Final Exam

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  1. Review for Final Exam

  2. Schedule • Review Session on Monday at 4:00 PM in 101 UPL if it extends beyond 6:00 PM it will move to 707 Keen. • Final exam – Tuesday, April 23 at 7:30 AM in 101 Carroway building. • This Power Point presentation contains 36 slides – most likely I won’t get to more than about 1/3 of them. Make sure you understand the concepts and problems presented here.

  3. B B I I B-Field Due to Currents • Electric currents produce magnetism. • B = oI/(2r) (due to a long straight wire) • Direction from the right-hand rule. Curl your fingers as if following B. Your thumb is in the direction of the current.

  4. q,v R B is out of the page. Example: Mass Spectrometer. F = qvB F = ma ma = qvB a = qvB/m qvB/m = v2/R  R = mv/(qB)

  5. q,v R B is out of the page. Example: Mass Spectrometer. F = qvB F = ma = mv2/R = qvB mv = qBR or p = qBR ½mv2 = (mv)2/(2m) = p2/(2m) Energy = ½mv2 = ½(qBR)2/m q < 0 !!

  6. B2 I1 B2 I2 F12 B2 I1 F21 F12 Forces Between Currents F12/L = I1B2 = I1oI2/(2r) = o I1I2/(2r)

  7. B1 B1 F12 F21 F21 B1 I2 Forces Between Currents I1 I2

  8. Major Concepts • Battery (voltages) • Ohm’s Law: V = IR • Resistance: R = V/I (in ohms “”) • Resistivity: R = L/A ( in /m) • Power: P = I2R (resistors) • Power: P = VI (batteries or resitors)

  9. Major Concepts • Series Circuit • Current must go through all resistors • RT = R1 + R2 + R3 + … • Parallel Circuit • Current is divided between the resistors • 1/RT = 1/R1 + 1/R2 + 1/R3 + … • Terminal Voltage • Accounts for the resistance within the battery. • Vterminal = V – Ir(internal)

  10. A B C Arrow shows the direction that positive charges move. + R = 24  V = 8 V - Current in a Simple Circuit V = IR  I = V/R I = 8 V/24  I = 0.33 Amp Current is the same everywhere in the circuit! Current at A = 0.33 Amps Current at B = 0.33 Amps Current at C = 0.33 Amps

  11. B C D A + R = 24  V = 8 V - Energy in a Simple Circuit Recall: I = 1/3 Amp Consider the energy of a proton moving through the circuit. (recall: q = +1 e) Energy(A) = qVA = 8 eV Energy(B) = qVB = 8 eV Energy(C) = qVC – q(IR) = 8eV – 1e * ( (1/3)*24 V ) = 0 eV Energy(D) = qVD = 0 eV Proton loses energy moving from B to C. It gains energy moving from D to A.

  12. B C D A + R = 24  V = 8 V - Energy in a Simple Circuit Recall: I = 1/3 Amp Consider the energy of a proton moving through the circuit. (recall: q = -1 e) Energy(D) = qVA = -8 eV Energy(C) = qVB = -8 eV Energy(B) = qVC – q(IR) = -8eV – -1e*((1/3)*24 V ) = 0 eV Energy(A) = qVD = 0 eV An electron loses energy moving from C to B. It gains energy moving from A to D.

  13. Circuit Analysis • Kirchhoff’s Rules: • Loop Rule: • The sum of the voltage drops around any closed loop is zero. • Conservation of Energy • V = 0 • Junction Rule: • The net current into and out of any point in a circuit is zero. • Charge conservation. • I = 0

  14. A B C I2 I3 H I1 E D Complex Circuits R1 • I1 goes from A to B and from E to H. • I2 goes from B to F to G to E. • I3 goes from B to C to D to E. F V R2 R3 G R1 = 10 , R2 = 3 , R3 = 6  and V = 24 Volts

  15. Complex Circuits R1 A B C • Junction Rule at B: • I1 - I2 - I3 = 0 • Loop Rule: ABFGEHA • -I1R1 – I2R2 + V = 0 • Loop Rule: ABCDEHA • -I1R1 – I3R3 + V = 0 F V I2 I3 R2 R3 H I1 G E D

  16. A B C F I2 I3 H I1 G E D Complex Circuits R1 • I1 - I2 - I3 = 0 • -I1R1 – I2R2 + V = 0 • So: -10I1 – 3I2 + 24 = 0 • -I1R1 – I3R3 + V = 0 • So: -10I1 – 6I3 + 24 = 0 V R2 R3

  17. Solving the equations. I1 - I2 - I3 = 0 10I1 + 3I2 - 24 = 0 10I1 + 6I3 - 24 = 0 10I1 + 3I2 - 24 = 0 - ( 10I1 + 6I3 - 24 = 0 ) 3I2 – 6I3 = 0 3I2 = 6I3 I2 = 2I3

  18. Solving the equations. I1 - I2 - I3 = 0 I2 = 2I3 I1 - 2I3 - I3 = 0 I1 - 3I3 = 0 I1 = 3I3 10I1 + 6I3 - 24 = 0  10*3I3 + 6I3 - 24 = 0 OR: 36I3 - 24 = 0  I3 = 24/36 = 0.667 A I1 = 3I3  I1 = 3* 0.667 A = 2.0 A

  19. Solving the equations. I1 - I2 - I3 = 0 I2 = 2I3 I3 = 24/36 = 2/3 A I1 = 3I3  I1 = 3* 2/3 A = 2 A I2 = 2I3 = 2 * 2/3 = 4/3 A

  20. Check the Equations • I1 = 2 A, I2 = 4/3 A, I3 = 2/3 A • I1 - I2 - I3 = 0 • 2 – 4/3 – 2/3 = 0  • 10I1 + 3I2 - 24 = 0 • 10*2 + 3*4/3 – 24 = 0  • 10I1 + 6I3 - 24 = 0 • 10*2 + 6*2/3 – 24 = 0 

  21. Balancing the Energy • P(in) = VI1 = 24 * 2 = 48 W • P1 = I12R1 = 22*10 = 40 W • P2 = I22R2 = (4/3)2*3 = 16/3 W • P3 = I32R3 = (2/3)2*6 = 8/3 W • PT = 40 + 16/3 + 8/3 = 48 W 

  22. Reflected Light The angle of incidence equals the angle of reflection. in out

  23. in out Refracted Light The direction changes when the light moves from material to another. The change depends on the material. Snell’s Law determines the angles: N1 sin 1 = N2 sin 2 The index of refraction, N, is the ratio of the speed of light in vacuum to the speed of light in the material. N = c/v c = speed of light in vacuum. v = speed of light in material. glass air

  24. 1 2 image Optic Axis object 3 Lenses 1 – Parallel Ray 2 – Central Ray 3 – Focal Ray

  25. image Optic Axis object Lens Equation 1 1 1 — = — + — F O I F O I

  26. Optic Axis Diverging Lens 1 3 object image 1 – Parallel Ray 2 – Central Ray 3 – Focal Ray 2

  27. Optic Axis Diverging Lens 1 3 object image F = - 20 cm O = 50 cm Use the Lens Equation 2

  28. Diverging Lens w/ Lens Equation 1 1 1 — = — + — F O I F = - 20 cm O = 50 cm 1/(-20) = 1/50 + 1/I  1/I = -1/20 – 1/50 = -5/100 – 2/100 1/I = -7/100  I = -100/7 = - 14.28 cm m = -I/O = - (-14.28 cm)/( 50 cm ) = + 2/7 m > 0  Image is upright. I < 0  image is virtual.

  29. Atomic Model • Electrons moved around nucleus only in certain stable orbits. • Stable orbits are those in which an integral number of wavelengths fit into the diameter of the orbit (2rn = n) • They emitted (absorbed) light only when they changed from one orbital to another. • Orbits have quanta of angular momenta. L = nh/2 • Orbit radius increases with energy rn = n2 r1 (r1 = .529 x 10-10 m)

  30. Ionized atom n = 3 n = 2 - 3.4 eV n = 1 E = - 13.6 eV Atomic Energy Levels • En = Z2/n2 E1 • Hydrogen • En = - 13.6 eV/n2

  31. Emission & Absorption • Energy is conserved. • E = Eatom = Ef - Ei • Photon energy = hf = hc/ • Absorption  photon disappears a electron in the atom changes from a lower energy level to a higher energy level. • Emission  an electron in atom goes from higher energy level to a lower energy level. This change in energy is the energy of the photon.

  32. Heisenberg Uncertainty Principle • Impossible to know both the position and the momentum of a particle precisely. • A restriction (or measurement) of one, affects the other. • x p  h/(2) • Similar constraints apply to energy and time. • E t  h/(2) EXAMPLE: If an electron's position can be measured to an accuracy of 1.96×10-8 m, how accurately can its momentum be known? x p  h/(2)  p = h/(2x) p = 6.63x10-34 Js /(2 1.96x10-8 m) = 5.38 x 10-27 N s

  33. Nuclear Decay Rates • N = -No t • Number of decaying nuclei (in a given time t), depends on: • Number of remaining nuclei, No • Nuclear decay constant for that type of nuclei,  • Number of nuclei remaining after a time t: • N = No e-t

  34. Nuclear Decay Rates At t =  N is 1/e (0.368) of the original amount

  35. ( , ) +  + ( , ) Example • Start with 5 protons  end up with 5 protons. • Start wit 11 baryons  end up with 11 baryons. • Q-value: (Mass Energy of Final State – Mass Energy of Initial State) Mn = 1.008665 u + M(B) = 10.012936 u = 11.021601 u M = 4.002602 u + M(L) = 7.016003 u = 11.018605 u Q = M (in MeV) = Mf – Mi = 11.018605 u - 11.021601 u Q = -0.002996 u = -0.002996 u * 931.5 MeV/u = -2.79 MeV Negative Q-value means we get extra energy out of the reaction.

  36. Example: Nuclear Power • Suppose that the average power consumption, day and night, in a typical house is 340 W. What initial mass of 235U would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fission.) Energy used = rate * time = 340 W * 3.15 x 107 s/yr = 1.07 x 1010 J Energy from each nucleus is: 200 MeV/nuclei = 200 MeV /nuclei * 1.60 x 10-13J/MeV = 3.2 x 10-11 J/nuclei Number of nuclei required = Energy/(Energy per nucleus) Number of nuclei = 1.07 x 1010 J/3.2 x 10-11 J/nuclei = 3.34 x 1020 nuclei Total mass of 235U = number of nuclei * mass/nucleus Mass = 3.34 x 1020 nuclei * 235 amu * 1.66 x 10-27 kg/amu = 1.31 10-4 kg