Final Exam Practice Solutions

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1 Physics 390 Final Exam Practice Solutions These are a few problems comparable to those you will see on the exam. They were picked from previous exams. I will provide a sheet with useful constants and equations for the exam. The last two pages are the Formula sheets for the Final Exam.. 8 pts The 209 Bi nucleus has a spin of 9/2. a 4 pts What is the angular momentum of this nucleus in units of h? b 4 tps Suppose that it were possible to make a beam of completely ionized 209 Bi nuclei and pass it through a Stern-Gerlach apparatus. How many distinct blobs would you observe on the detector screen? a J = jj + h = 9 h = 99 h = 4.97 h b The magnetic field in the apparatus defines a z-axis, and there will be one blob for each possible projection of the total momentum vector along this axis. So there will be 2j + = 0 spots, corresponding to J z = 9/2, 7/2, 5/2, 3/2, /2, -/2, -3/2-5/2, -7/2, -9/ pts In a scattering experiment it was found that 2 C has a nuclear radius of 2.7 fm. The experiment is then repeated with another, unknown element and it is found the the nuclear radius is twice as big. What is the mass number of this unknown element? The nuclear radius of the first nucleus is given by R = r 0 A /3 and So R 2 = r 0 A / R2 A 2 = A = = 96. R

2 3. 8 pts Consider the following energetically possible transitions of an electron in an atom: 4p 3p 2 3d 2s 3 4s 2p 4 4d 3p Which of these transitions is/are allowed? Explain your reasoning. 4p 3p: is not allowed, because it is a l = 0 transition. 3d 2s: is not allowed, because it is a l = 2 transition. 4s 2p: is allowed, because it is a l = transition. 4d 3p: is allowed, because it is a l = transition pts The proton has a magnetic moment of µ p = ev/t. a 5 pts What is the energy difference between the spin-up and spin-down orientations when a proton is placed in a magnetic field of.5 T? b 7 tps In magnetic resonance imaging, a person with a body temperature of 30 K is placed in a.5 T magnetic field. Compute the ratio of N /N, where N is the number of protons in the spin-up state parallel to the field state, and N is the number in the spin-down state. State clearly any assumptions you make. a The energy of a magnetic dipole in a magnetic field is E = µ B. If we use the B field to define the z-axis, then the energy of the proton is E = µ p z B = g p m s µ N B = µ p B. The energy difference between the spin-up and spin-down orientations is E = E E = 2µ p B = ev/t.5 T = ev, with the spin-up state parallel to the field being the lower-energy state. b If we assume that the magnetic dipoles do not interact with each other we can treat them as free dipoles. Furthermore, the protons are sufficiently far apart that we can regard them as distinguishable by virtue of their position and use Maxwell-Boltzmann statistics, f MB = A e E/kT. There is no density of states factor to worry about, since the spin-up and spin-down states are non-degenerate. Then So N = N A e E /kt, N = N A e E /kt. N = e E /kt N e = E /kt e E/kT + E kt = ev ev/k30 K = So the excess of protons in the spin-up state is only a few parts per million. But because there are so many protons in the body, this small excess is enough to be useful for MRI. 2

3 5. 8 pts If the energy levels in an atom are filled in the sequence s, 2s, 2p, 3s, 3p, 4s, 3d. a 5 pts What is the atomic number of the element that is in its ground state, has all these levels filled, and all higher levels empty? b 3 tps Suppose that an electron is knocked out of the s level. From which of these levels could an electron drop down to fill the resulting unoccupied state? State your reasoning. a Each shell holds 22l + electrons. So the s shells hold 2 electrons, the p shells hold 6, and the d shell has 0. This atom has = 30 electrons. Assuming it is a neutral atom, this is the element Zn. b The transitions obey the selection rule l = ±. Since the s states have l = 0, the electron must come from an l = state. So transitions from any of the p levels are allowed pts Sketch the n =, n = 2, and n = 3 energy levels for a hydrogen atom in a magnetic field. Indicate three possible transitions with solid lines. Indicate three forbidden or at least highly suppressed transitions with dotted lines, and state why they are forbidden. Ignore fine and hyperfine structure. The energy levels and the allowed transitions are shown below. All other transitions are forbidden because they violate the selection rules l = ± or m l = ±, 0. n=3 l=0 l= l=2 m l 0 m l 0 m l n=2 n= 3

4 7. 6 pts When the sun runs out of fusion fuel, it will no longer be able to resist the compression of gravity and will collapse to become a white dwarf star. In this state it will have about the current mass of the Sun kg in a sphere with a radius a bit bigger than that of the Earth 0 7 m say. a 6 pts Estimate the Fermi energy of electrons in this remnant of the Sun. b 5 tps Get an order of magnitude estimate for the total energy of these electrons. Hint: how does the average energy of electrons compare to the Fermi Energy? Is it roughly the same, or very different? c 5 tps Given that the pressure can be calculated from P = de total /dv, estimate the pressure in this object a The Fermi energy E F of an electron is E F = h2 3 2m e 8π N 2/3 = hc2 3 N 2/3. V 2m e c 8π2/3 2 V For N, we can assume that it is all Hydrogen m H = kg and obtain N = M sun / kg = Or if we assume it is all Helium, N He =.5 N H. Putting this all together we get E F = MeV m MeV 3 2/ /3 = 0.6 MeV, 8π 4π/ or 0.0 MeV for Helium which is nearly its rest mass. b Since E m = 3/5 E F, E total is roughly c The pressure is E total = 3 5 E F N electrons MeV = MeV P = de total /dv = h 2 2m e 3 2/3 N 5/3 V 5/3 = N/m 2. 8π 8. 6 pts Protons and neutrons can interact by exchanging pions which are mesons with a rest mass of 40 MeV/c 2. What is the range of this interaction? The range of the interaction is R = hc 97.3 MeV fm = =.4 fm. mc2 40 MeV 4

5 Useful constants and equations e = C 4πǫ 0 = N m 2 /C 2 e2 4πǫ 0 =.44 ev nm c = m/s h = J s = ev s h = h 2π hc = 240 ev nm hc = 97.3 MeV fm ev = J R = m m e = kg = 5 kev/c 2 = u neutral 2 6 m p = kg = MeV/c 2 =.0073 u C atom mass = u u = 93.5 MeV m n = kg = MeV/c 2 =.0087 u a 0 = nm E 0 = 3.6 ev α = /37 σ = W/m 2 K 4 Heisenberg: p x x h E t h Atomic Physics: Hydrogen Atom: E = Generalized Balmer Formula: 3.6 ev n 2 with n =, 2, 3, l < n l m +l λ = R n 2 f n 2 i Angular Momentum: L 2 = ll + h 2 l = 0,, 2, 3,.. orbital Magnetic Moment: orbital: µ = q 2 M L L z = L cosθ = m h l m l l integer steps for electron: µ z = 2 m s for proton: µ z = g p m s e h 2 m e = µ B = ev/t e h 2 m p = µ N = ev/t Energy of magnetic dipole in B field: E = µ B Energy of particle in 3-dim square well: E nxn yn z = n2 x + n 2 y + n 2 zh 2 8mL 2 Molecular excitations: vibrational: E = n + L2 hω rotational: E = 2 2 I = ll + 2 I h 2 5

6 Statistical Physics: Maxwell Boltzmannc distribution: f MB = A e E/kT Bose-Einstein distribution: f BE = Fermi - Dirac distribution: f FD = B e E/kT e E E F /kt + Boltzmann constant k = J/K = ev/k For gas of free fermions: ge = 8 2πm 3/2 E E F = h2 2m h 3 3N 2/3 E m = 3 8πV 5 E F Nuclear Physics: nuclear radius: R = m A /3 average binding energy / nucleon 8 MeV range of interaction: R = h mc = hc mc 2 Decay law: N = N 0 e t τ with < t >= τ = λ and T /2 = ln 2 λ Binding energy BZ,A = [Z m p + N m n M atom Z,A] c 2 Particle Physics: Baryon: Q Q Q Meson: Q Q Quarks: u d c s t b q = q = 3 Leptons: e ν e µ ν µ τ ν τ q = q = 0 Cosmology: Luminosity of star: L = 4πr 2 f with f = apparent brightness of star difference in apparent magnitude: m m 2 = 2.5 logf /f 2 6

Final Exam Tuesday, May 8, 2012 Starting at 8:30 a.m., Hoyt Hall Duration: 2h 30m

Final Exam Tuesday, May 8, 2012 Starting at 8:30 a.m., Hoyt Hall. ------------------- Duration: 2h 30m Chapter 39 Quantum Mechanics of Atoms Units of Chapter 39 39-1 Quantum-Mechanical View of Atoms 39-2