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TIME OF COMPLETION_______________ NAME______SOLUTION_______________________ DEPARTMENT OF NATURAL SCIENCES PHYS 3650, Exam 2 Version 1 Total Weight: 100 points Section 1 October 31, 2005 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 10:30 a.m. 11:45 a.m PROBLEM POINTS 1-6 30 7 20 8 20 9 15 10 15 TOTAL 100 PERCENTAGE CREDIT CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The p state of an electronic configuration corresponds to a. n = 2. b. l = 2. (5) c. l = 1. d. n = 0. 2. Bohr’s quantum condition on electron orbits requires a. That the angular momentum of the electron about the hydrogen nucleus equal nh/(2). b. That no more than one electron occupy a given stationary state. (5) c. That the electrons spiral into the nucleus while radiating electromagnetic waves . d. None of the above . 3. The total number of quantum states of hydrogen with quantum number n = 4 is a. 4. b. 16. (5) c. 32. d. 36. 4. An electron in the L shell means that a. l = 1. b. n = 1. (5) c. n = 2. d. m = 2. 5. The restriction that no more than one electron can occupy a given quantum state in an atom was first stated by which of the following scientists? a. Bohr. b. De Broglie. (5) c. Heisenberg. d. Pauli. 6. If the principal quantum number for hydrogen is 5, which one of the following is not a permitted orbital magnetic quantum number for that atom? a. 6. b. -2. (5) c. 0. d. 3. 7. The optical spectra of atoms with two electrons in the same outer shell are similar, but they are quite different from the spectra of atoms with just one outer electron because of the interaction of the two electrons. Separate the following elements into two groups such that those in each group have similar spectra: lithium (Z = 3), beryllium (Z = 4), sodium (Z = 11), magnesium (Z = 12), potassium (Z = 19), calcium (Z = 20), chromium (Z = 24), nickel (Z = 28). Please list the electronic configurations of all of these elements. Li: 1s22s1 Be: 1s22s2 Na: 1s22s22p63s1 Mg: 1s22s22p63s2 K: 1s22s22p63s23p64s1 Ca: 1s22s22p63s23p64s2 Cr: 1s22s22p63s23p64s23d4 Ni: 1s22s22p63s23p64s23d8 1 electron in the outer shell: Li, Na, K 2 electrons in the outer shell: Be, Mg, Ca, Cr, Ni 8. Calculate the wavelength of the K line of rhodium (Z = 45). K: transition from L shell to K shell EK = - (Z-1)2 E0/12 = -26,330 eV EL = - (Z-1)2 E0/22 = -6,582 eV E = EL – EK = 19,748 eV = hc/E = 0.0629 nm Calculate the wavelength of the L line of rhodium. L: transition from M shell to L shell EM = - (Z-9)2 E0/32 = -1,958 eV EL = - (Z-1)2 E0/22 = -6,582 eV E = EM – EL = 4,624 eV = hc/E = 0.269 nm 9. Spectral lines of the following wavelengths are emitted by singly ionized helium: 164 nm, 230.6 nm, and 541 nm. Identify the transitions that result in these spectral lines. He: Z = 2 En = - Z2 E0/n2 E1 = -54.4 eV E2 = -13.6 eV E3 = -6.04 eV E4 = -3.40 eV E5 = -2.18 eV E6 = -1.51 eV E7 = -1.11 eV E8 = -0.850 eV E9 = -0.671 eV E10 = -0.544 eV = hc/E E= hc/ E1= hc/ (1243 eVnm)/(164 nm) = 7.58 eV E2= hc/ (1243 eVnm)/(230.6 nm) = 5.39 eV E3 E2 transition E9 E3 transition E3= hc/ (1243 eVnm)/(541 nm) = 2.30 eV E7 E4 transition 10. A hydrogen atom is in its tenth excited state according to the Borh model (n = 11). a. What is the radius of the Bohr orbit? rn = n2a0 r11 = (11)2(0.0529 nm) = 6.40 nm b. What is the angular momentum of the electron? Ln = n h/(2) L11 =1 h/(2) = 1.16 x 10-33 J-s c. What is the electron’s kinetic energy? L=pr p = L/r = 1.78 x 10-25 kg-m/s KE = p2/(2me) = (1.78 x 10-25 kg-m/s)2/(2(9.11 x 10-31 kg)) = 1.73 x 10-20 J = 0.112 eV d. What is the electron’s potential energy? PE = - ke e2/r PE = - (8.99 x 109 N-m2/kg2) (1.60 x 10-19 C)2/(6.40 x 10-9 m) = -0.224 eV e. What is the electron’s total energy? E = KE + PE = 0.112 eV + (-0.224 eV) = -0.112 eV TIME OF COMPLETION_______________ NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 3650, Exam 2 Version 1 Total Weight: 100 points Section 1 November 4, 2006 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple choice and three (3) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 11:30 a.m. 12:20 p.m PROBLEM POINTS 1-6 25 7 25 8 25 9 25 10 25 TOTAL 100 CREDIT PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. According to the Bohr model of the atom, the angular momentum of an electron around the nucleus a. Could equal any positive value. b. Must equal an integral multiple of h. (5) c. Must equal an integral multiple of h/2π. d. Decreases with time, eventually becoming zero. 2. The reason the position of a particle cannot be specified with infinite precision is the a. Exclusion principle. b. Uncertainty principle. (5) c. Photoelectric effect. d. Principle of relativity. 3. The principal quantum number can have any integer value ranging from a. -∞ to +∞. b. 0 to ∞. (5) c. 1 to ∞. d. 1 to 100. 4. The spin quantum number can have values of a. -1/2, -1, 0, +1, +1/2. b. -1/2, -1, +1, +1/2. (5) c. -1/2, 0, +1/2. d. -1/2, +1/2. 5. In the ground state, the quantum numbers (n, l, ml, ms) for hydrogen are, respectively, a. 1, 1, 1, 1. b. 1, 0, 0, 0. (5) c. 1, 0, 0, ±1/2. d. 1, 1, 1, ±1/2. 6. In terms of an atom's electron configuration, the letters K, L, M, and N refer to a. Different shells with n equal to 1, 2, 3, or 4 respectively. b. Different sub shells with l equal to 1, 2, 3, or 4 respectively. (5) c. The four possible levels for the magnetic quantum number. d. The four possible quantum numbers. 7. What is the full electron configuration in the ground state for elements with Z equal to (a) 10, 1s22s22p6 (b) 16, 1s22s22p63s23p4 (c) 28? 1s22s22p63s23p64s23d8 8. Use the Bohr theory to estimate the wavelength for an n 3 to n 1 transition in molybdenum (Z 42). The measured value is 0.063 nm. Why do we not expect perfect agreement? EK = - (Z-1)2 E0/12 = -22,862 eV EM = - (Z-9)2 E0/32 = -1,64 eV E = EM – EK = 21,216 eV = hc/E = 0.0584 nm 9. (a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the Paschen series (nf = 3). 1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/inf2) = 0.12222 x 107 m-1 = 8.18 x 10-7 m = 818 nm (b) Calculate the wavelengths for the three longest wavelengths in this series. 1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/42) = 0.05347 x 107 m-1 = 18.70 x 10-7 m = 1870 nm 1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/52) = 0.07111 x 107 m-1 = 14.06 x 10-7 m = 1406 nm 1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/62) = 0.08333 x 107 m-1 = 12.00 x 10-7 m = 1200 nm 10. Make a table of all of the allowed four quantum numbers for the first three shells of the hydrogen atom. How many electrons can each shell accommodate? K (2 electrons): n = 1 l=0 (1 0 0 -½ ) 2 1s electrons ml = 0 ms = +- 1/2 (1 0 0 ½ ) L (8 electrons): n = 2 l=0 (2 0 0 -½ ) 2 2s electrons ml = 0 ms = +- 1/2 (2 0 0 ½ ) l=1 ml = -1 ms = +- 1/2 (2 1 -1 ½ ) (2 1 -1 -½ ) 1 0 -½ ) l=1 6 2p electrons ml = 0 ms = +- 1/2 (2 1 0 ½ ) (2 l=1 ml = 1 ms = +- 1/2 (2 1 1 ½ ) (2 1 1 -½ ) M (18 electrons): n = 3 l=0 (3 0 0 -½ ) 2 3s electrons ml = 0 ms = +- 1/2 (3 0 0 ½ ) l=1 ml = -1 ms = +- 1/2 (3 1 -1 ½ ) (3 1 -1 -½ ) (3 1 0 -½ ) l=1 6 3p electrons ml = 0 ms = +- 1/2 (3 1 0 ½ ) l=1 ml = 1 ms = +- 1/2 (3 1 1 ½ ) l=2 ml = -2 (3 1 1 -½ ) (3 2 -2 -½ ) ms = +- 1/2 (3 2 -2 ½ ) l=2 ml = -1 ms = +- 1/2 (3 2 -1 ½ ) (3 2 -1 -½ ) (3 2 0 -½ ) l=2 10 3d electrons ml = 0 ms = +- 1/2 (3 2 0 ½ ) l=2 ml = 1 ms = +- 1/2 (3 2 1 ½ ) l=2 ml = 2 ms = +- 1/2 (3 2 2 ½ ) (3 2 1 -½ ) (3 2 2 -½ ) TIME OF COMPLETION_______________ NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 3650, Exam 2 Version 1 Total Weight: 100 points Section 1 October 31, 2008 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on six (6) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and three (3) calculation problems. Work five (5) multiple choice and three (3) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: PROBLEM 10:00 a.m. 10:50 a.m POINTS CREDIT 1-6 25 7 25 8 25 9 25 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Hydrogen atoms can emit lines with visible colors from red to violet. These four visible lines emitted by hydrogen atom are produced by electrons a. That start in the n = 2 level. b. That end up the n = 2 level. (5) c. That end up in the n = 3 level. d. That start in the ground level. 2. According to the Pauli’s exclusion principle, how many electrons in an atom may have a particular set of quantum numbers? a. 1. b. 2. (5) c. 6. d. 10. 3. The orbital angular momentum quantum number can take which of the following values for any given value of the principal quantum number, n? a. l = 0, 1, 2,….. b. l = 0, 1, 2, …., n. (5) c. l = 0, 1, 2, …., n – 1. d. l = 0, 1, 2, …., n + 1. 4. Which of the following values are associated with the electron spin quantum number, ms? a. ± 1/2. b. 0. (5) c. ± 1. d. ± 2. 5. Which ones of the atomic transition on the sodium energy level diagram below are NOT allowed by the selection rules? a. a and c. b. b and f. (5) c. c and d. d. d and g. 6. Given the energy diagram above, estimate the wavelength of a photon emitted by a sodium atom as the electron goes through the transition d. a. 800 nm. (5) b. 620 nm. c. 450 nm. d. 75 nm. 7. Titanium (Z = 22) is in its ground state. a. What is its electronic configuration? 1s22s22p63s23p64s23d2 b. List all the possible combinations of quantum numbers (n, l, ml, and ms) an electron can have while in the subshell which is only partially filled in the ground state. 3d subshell: n = 3, l = 2 l=2 ml = -2 ms = +- 1/2 (3 2 -2 ½ ) l=2 ml = -1 ms = +- 1/2 (3 2 -1 ½ ) l=2 ml = 0 ms = +- 1/2 (3 2 0 ½ ) l=2 ml = 1 ms = +- 1/2 (3 2 1 ½ ) l=2 ml = 2 ms = +- 1/2 (3 2 2 ½ ) (3 2 -2 -½ ) (3 2 -1 -½ ) (3 2 0 -½ ) (3 2 1 -½ ) (3 2 2 -½ ) 8. Using Bohr’s model, estimate the wavelength of the K line for a. calcium (Z = 20), and EK = - (Z-1)2 E0/12 = - (19)2 (13.6 eV)/12 = - 4,910 eV EL = - (Z-1)2 E0/12 = - (19)2 (13.6 eV)/22 = - 1.227 eV E = EL – EK = 3,683 eV = hc/E = 0.337 nm b. cadmium (Z = 48) EK = - (Z-1)2 E0/12 = - (47)2 (13.6 eV)/12 = - 30,042 eV EL = - (Z-1)2 E0/12 = - (47)2 (13.6 eV)/22 = - 7,510 eV E = EL – EK = 22,532 eV = hc/E = 0.0550 nm 10. Find the energy of the electron in the ground state of singly ionized helium (Z = 2). E1 = - (Z)2 E0/12 = - (2)2 (13.6 eV)/12 = - 54.4 eV What is the shortest possible wavelength which is emitted as a result of electron returning to the ground state? Shortest wavelength corresponds to the transition from n = ∞: E = E∞ – E1 = 54.4 eV = hc/E = 22.8 nm What is the longest possible wavelength which is emitted as a result of electron returning to the ground state? Lonest wavelength corresponds to the transision from n = 2: E2 = - (Z)2 E0/22 = - (2)2 (13.6 eV)/22 = - 13.6 eV E = E2 – E1 = 40.8 eV = hc/E = 30.4 nm TIME OF COMPLETION_______________ NAME__SOLUTION___________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 3650, Exam 2 Version 1 Total Weight: 100 points Section 1 November 13, 2009 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on six (6) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and three (3) calculation problems. Work five (5) multiple choice and three (3) calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 10:00 a.m. 10:50 a.m PROBLEM POINTS 1-6 25 7 25 8 25 CREDIT 9 25 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The Pauli exclusion principle: a. Implies that in an atom no two electrons can have the same set of quantum numbers. b. Says that no two electrons in an atom can have the same orbit. (5) c. Excludes electrons from atomic nuclei. d. Excludes protons from atomic orbits. 2. Which one of these statements is true? a. The principal quantum number of the electron in a hydrogen atom does not affect its energy. b. The principal quantum number of an electron in its ground state is zero. (5) c. The orbital quantum number of an electron state is always less than the principal quantum number of that state. d. The electron spin quantum number can take on any one of the four different values. 3. The orbital angular momentum quantum number can take which of the following values for any given value of the principal quantum number, n? a. l = 0, 1, 2,….. b. l = 0, 1, 2, …., n. (5) c. l = 0, 1, 2, …., n – 1. d. l = 0, 1, 2, …., n + 1. 4. How many of oxygen’s eight electrons are found in the p state? a. 0. b. 2. (5) c. 4. d. 6. 5. The total number of states of hydrogen with principal quantum number n = 4 is: a. 4. b. 16. (5) c. 32. d. 36. 6. Given the sodium energy diagram above, estimate the wavelength of a photon emitted by a PHYS 3650 Exam 2, Version 1 Fall 2005 19 sodium atom as the electron goes through the transition d. a. 800 nm. (5) b. 620 nm. c. 450 nm. d. 75 nm. 9. Nickel (Z = 28) is in its ground state. a. What is its electronic configuration? 1s22s22p63s23p64s23d8 b. List all the possible combinations of quantum numbers (n, l, ml, and ms) an electron can have while in the subshell which is only partially filled in the ground state. n = 3; l = 2; ml = -2, -1, 0, 1, 2; ms = ±1/2 (3, 2, -2, ±1/2) (3, 2, -1, ±1/2) (3, 2, 0, ±1/2) (3, 2, 1, ±1/2) (3, 2, 2, ±1/2) 10. Using Bohr’s model, estimate the wavelength of the K line for a nickel target (Z = 28). EK = - (Z-1)2 E0/12 = - (28-1)2 (13.6 eV)/12 = - 9,914 eV EL = - (Z-1)2 E0/22 = - (28-1)2 (13.6 eV)/22 = - 2.479 eV E = EL – EK = 7,435 eV = hc/E = 0.167 nm What is the wavelength of the K line? PHYS 3650 Exam 2, Version 1 Fall 2005 20 EK = - (Z-1)2 E0/12 = - (28-1)2 (13.6 eV)/12 = - 9,914 eV EM = - (Z-9)2 E0/32 = - (28-9)2 (13.6 eV)/32 = - 546 eV E = EM – EK = 9,368 eV = hc/E = 0.132 nm 11. Find the energy of the electron in the ground state of doubly ionized lithium (Z = 3). E1 = - (Z)2 E0/12 = - (3)2 (13.6 eV)/12 = - 122 eV What is the shortest possible wavelength which is emitted as a result of electron returning to the ground state? E = 0 – E1 = 122 eV = hc/E = 10.1 nm What is the longest possible wavelength which is emitted as a result of electron returning to the ground state? E2 = - (Z)2 E0/22 = - (3)2 (13.6 eV)/22 = - 30.6 eV E = E2 – E1 = 91.8 eV = hc/E = 13.5 nm TIME OF COMPLETION_______________ NAME____SOLUTION_________________________ DEPARTMENT OF NATURAL SCIENCES PHYS 3650, Exam 2 PHYS 3650 Exam 2, Version 1 Fall 2005 Section 1 21 Version 1 Total Weight: 100 points March 28, 2011 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on six (6) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and three (3) calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: Stop: 3:00 p.m. 3:50 p.m. PROBLEM POINTS 1-6 25 7 25 8 25 9 25 TOTAL 100 CREDIT PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. PHYS 3650 Exam 2, Version 1 Fall 2005 22 1. In a Compton scattering experiment, a photon of energy E is scattered from an electron at rest. After the scattering event occurs, which of the following statements is true? (A) The frequency of the photon is greater than E/h. (B) The energy of the photon is less than E. (C) The wavelength of the photon is less than hc/E. (D) The momentum of the photon increases. (E) None of those statements is true. (5) 2. Which of the following phenomena most clearly demonstrates the wave nature of electrons? (A) The photoelectric effect. (B) The blackbody radiation. (5) (C) The Compton effect. (D) Diffraction of electrons by crystals. (E) None of these answers. 3. A proton, an electron, and a helium nucleus all move at speed v. Rank their de Broglie wavelengths from largest to smallest. (A) electron, proton, helium nucleus. (B) proton, helium nucleus, electron. (C) proton, electron, helium nucleus. (D) helium nucleus, electron, proton. (E) helium nucleus, proton, electron (5) PHYS 3650 Exam 2, Version 1 Fall 2005 23 4.What is the longest wavelength in the Lyman Series? (A) 45.60 nm. (B) 91.20 nm. (C) 121.5 nm. (D) 240.1 nm. (E) 356.2 nm. 5. To which of the following values of n does the longest wavelength in the Balmer series correspond? (A) 1. (B) 3. (5) (C) 5. (D) 7. (E) ∞ . 6. Which one of the following is the correct expression for the energy of a photon? (A) E = h/f. (B) E = h/c. (5) (C) E = h. (D) E = hc/. (F) None of these. 7. The work function for aluminum is 4.08 eV. (a) Find the cutoff wavelength for aluminum. PHYS 3650 Exam 2, Version 1 Fall 2005 24 0 hc 1240 eV nm 304 nm 4.08 eV (b) What is the lowest frequency of light incident on aluminum that releases photoelectrons from its surface? f 0 c 0 3.00 108 m / s 9.87 1014 Hz 304 nm (c) If photons of energy 5.81 eV are incident on aluminum, what is the maximum kinetic energy of the ejected photoelectrons? KE hf 5.81 eV 4.08 eV 1.73 eV 8. In the Compton effect, a 0.100-nm photon strikes a free electron in a head-on collision and knocks it into the forward direction. The rebounding photon recoils directly backward. Find a. The wavelength of the scattered photon. h (1 cos ) (0.00243 nm)(1 cos(180o )) 0.00486 nm me c 0.100 nm 0.00486 nm 0.10486 nm 0 b. The energy of the scattered photon. E hc 1240 eV nm 11,825 eV 0.10486 nm c. The kinetic energy of the recoiling electron. (Hint: subtract the final energy of the photon from its initial energy to find the kinetic energy of the electron.) E0 hc 0 1240 eV nm 12,400 eV 0.100 nm PHYS 3650 Exam 2, Version 1 Fall 2005 25 KE E0 E 0.600 keV 9. (a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the Paschen series (n’= 3). 1 R H ( R 1 1 2) H 2 9 3 818 nm E hc 1240 eV nm 1.52 eV 818 nm (b) 1 R H ( Calculate the wavelengths for the three longest wavelengths in this series. 1 1 ) 32 4 2 1870 nm 1 R H ( 1 1 ) 32 5 2 1278 nm 1 R H ( 1 1 ) 32 6 2 1091 nm PHYS 3650 Exam 2, Version 1 Fall 2005 26 PHYS 3650 Exam 2, Version 1 Fall 2005 27 PHYS 3650 Exam 2, Version 1 Fall 2005 28