# Review for Final Exam. 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,

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Review for Final Exam

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.

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

Example: Mass Spectrometer. F = qvB F = ma ma = qvB a = qvB/m qvB/m = v 2 /R  R = mv/(qB) R q,v B is out of the page.

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

Forces Between Currents B2B2 B2B2 I1I1 F 21 F 12 I2I2 B2B2 I1I1 F 12 /L = I 1 B 2 = I 1  o I 2 /(2  r) =  o I 1 I 2 /(2  r)

Forces Between Currents I1I1 F 21 F 12 I2I2 B1B1 B1B1 B1B1 F 21 I2I2

Major Concepts Battery (voltages) Ohm’s Law: V = IR Resistance:R = V/I (in ohms “  ”) Resistivity:R =  L/A (  in  /m) Power:P = I 2 R (resistors) Power:P = VI (batteries or resitors)

Major Concepts Series Circuit Current must go through all resistors R T = R 1 + R 2 + R 3 + … Parallel Circuit Current is divided between the resistors 1/R T = 1/R 1 + 1/R 2 + 1/R 3 + … Terminal Voltage Accounts for the resistance within the battery. V terminal = V – Ir (internal)

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

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

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

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

Complex Circuits R1R1 R2R2 V R3R3 I1I1 I2I2 I3I3 F G A B C D E H 1.I 1 goes from A to B and from E to H. 2.I 2 goes from B to F to G to E. 3.I 3 goes from B to C to D to E. R 1 = 10 , R 2 = 3 , R 3 = 6  and V = 24 Volts

Complex Circuits R1R1 R2R2 V R3R3 I1I1 I2I2 I3I3 A B C D E F G H 1.Junction Rule at B: I 1 - I 2 - I 3 = 0 2.Loop Rule: ABFGEHA -I 1 R 1 – I 2 R 2 + V = 0 3.Loop Rule: ABCDEHA -I 1 R 1 – I 3 R 3 + V = 0

Complex Circuits R1R1 R2R2 V R3R3 I1I1 I2I2 I3I3 A B C D E F G H I 1 - I 2 - I 3 = 0 -I 1 R 1 – I 2 R 2 + V = 0 So: -10I 1 – 3I 2 + 24 = 0 -I 1 R 1 – I 3 R 3 + V = 0 So: -10I 1 – 6I 3 + 24 = 0

Solving the equations. I 1 - I 2 - I 3 = 0 10I 1 + 3I 2 - 24 = 0 10I 1 + 6I 3 - 24 = 0 3I 2 – 6I 3 = 0 3I 2 = 6I 3 I 2 = 2I 3 10I 1 + 3I 2 - 24 = 0 - ( 10I 1 + 6I 3 - 24 = 0 )

Solving the equations. I 1 - I 2 - I 3 = 0 I 2 = 2I 3 I 1 - 2I 3 - I 3 = 0 I 1 - 3I 3 = 0 I 1 = 3I 3 10I 1 + 6I 3 - 24 = 0  10*3I 3 + 6I 3 - 24 = 0 OR: 36I 3 - 24 = 0  I 3 = 24/36 = 0.667 A I 1 = 3I 3  I 1 = 3* 0.667 A = 2.0 A

Solving the equations. I 1 - I 2 - I 3 = 0 I 2 = 2I 3 I 3 = 24/36 = 2/3 A I 1 = 3I 3  I 1 = 3* 2/3 A = 2 A I2 = 2I3 = 2 * 2/3 = 4/3 A

Check the Equations I 1 = 2 A, I 2 = 4/3 A, I 3 = 2/3 A I 1 - I 2 - I 3 = 0 2 – 4/3 – 2/3 = 0 10I 1 + 3I 2 - 24 = 0 10*2 + 3*4/3 – 24 = 0 10I 1 + 6I 3 - 24 = 0 10*2 + 6*2/3 – 24 = 0

Balancing the Energy P (in) = VI 1 = 24 * 2 = 48 W P 1 = I 1 2 R 1 = 2 2 *10 = 40 W P 2 = I 2 2 R 2 = (4/3) 2 *3 = 16/3 W P 3 = I 3 2 R 3 = (2/3) 2 *6 = 8/3 W P T = 40 + 16/3 + 8/3 = 48 W

Reflected Light The angle of incidence equals the angle of reflection.  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: N 1 sin  1 = N 2 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.  in  out air glass

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

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

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

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

Diverging Lens w/ Lens Equation 1 1 1 — = — + — F O I 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. F = - 20 cm O = 50 cm

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  r n = 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 r n = n 2 r 1 (r 1 =.529 x 10 -10 m)

Atomic Energy Levels E n = Z 2 /n 2 E 1 Hydrogen E n = - 13.6 eV/n 2 Ionized atom E = - 13.6 eV - 3.4 eV n = 1 n = 2 n = 3

Emission & Absorption Energy is conserved. E  =  E atom =  E f -  E i 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.

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

Nuclear Decay Rates  N = -N o  t Number of decaying nuclei (in a given time  t), depends on: Number of remaining nuclei, N o Nuclear decay constant for that type of nuclei, Number of nuclei remaining after a time t: N = N o e - t

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

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.

Example: Nuclear Power 1. Suppose that the average power consumption, day and night, in a typical house is 340 W. What initial mass of 235 U 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 10 7 s/yr = 1.07 x 10 10 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 10 10 J/3.2 x 10 -11 J/nuclei = 3.34 x 10 20 nuclei Total mass of 235U = number of nuclei * mass/nucleus Mass = 3.34 x 10 20 nuclei * 235 amu * 1.66 x 10 -27 kg/amu = 1.31 10 -4 kg

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