Introduction to Quantum Mechanics - Test 3, Fall 2004 | PHY 4604, Exams of Physics

Download Introduction to Quantum Mechanics - Test 3, Fall 2004 | PHY 4604 and more Exams Physics in PDF only on Docsity! PHY4604–Introduction to Quantum Mechanics Fall 2004 Practice Test 3 November 22, 2004 These problems are similar but not identical to the actual test. One or two parts will actually show up. 1. Short answer. (a) Recall the Bohr energy levels of the Hydrogen atom are En = − m2h̄2n2 ( e2 4π²0 )2 (1) Compare the wavelengths of the 2p → 1s transitions in i. hydrogen The energy which must be carried away by a photon in the transition is E2 − E1 = hc/λ, so λH = hc/(E2 − E1) = 8h̄ 3 3m ( 4π²0 e2 )2 ii. deuterium (mass of nucleus 2× that of H). The mass which appears in the Bohr formula is the electron mass m, which is replaced by the reduced mass µ = mmp/(m+mp) if one allows for the proton motion. Changing mp → 2mp doesn’t change much, so to a good approximation the photon wavelength doesn’t change, λD ' λH . iii. positronium. In this case the positron is not heavier than the electron, but in fact has the same mass. The reduced mass of 2 identical particles of mass m is m/2, so we obtain the correct answer from the Hydrogen expression by substituting m → m/2, or λpos = 2λH . (b) An electron is in the ground state of tritium, a form of heavy hydrogen where the nucleus has two neutrons in addition to a proton. A nuclear reaction now changes the nucleus instantaneously to 3He, i.e. two protons 1 and a neutron. Calculate the probability that the electron remains in the ground state of 3He. You may need ψ100 = (2/ √ 4πa30 ) exp−r/a0, with a0 = h̄ 2/(mZe2), where Z is the nuclear charge. Z, the nuclear charge or number of protons, is 1 for tritium and 2 for 3He. Thus the Bohr radius for 3He is twice as small as for tritium or hydrogen, a 3He 0 ' 12at0. The electron in the 3He ion is more tightly bound to the nucleus. We’re told the system starts out in the tritium ground state, ψt100 = 2√ 4π(aH0 ) 3 e−r/a t 0 , which can be expressed as a linear combination of any complete set of states in Hilbert space, for example the eigenstates of the 3He Hamiltonian, ψt100 = aψ 3He 100 + bψ 3He 200 + cψ 3He 210 + ... meaning the electron in the tritium ground state will be found after a measurement in the 3He ground state with probability amplitude a = 〈ψt100|ψ 3He 100 〉 = 4 4π 1√ (at0) 3(a 3He 0 ) 3 ∫ d3r e −r ( 1 at 0 + 1 a 3He 0 ) = 4 (at0) 3 √ 8 ∫ ∞ 0 dr r2 e−3r/a t 0 = 4 √ 8 ∫ dy y2e−3y = 16 √ 2 27 = 0.838 and probability = |a|2 = 0.702. (c) An electron moving in the Coulomb field of a proton is in a state described by the wave function Ψ = 1 6 [4ψ100 + 3ψ211 − ψ210 + √ 10ψ21−1] (2) i. What is the expectation value of the energy? 〈Ψ|H|Ψ〉 = 1 36 (16〈100|H|100〉+ 9〈211|H|211〉+ 〈210|H|210〉 + 10〈21 − 1|H|21 − 1〉) = 1 36 (16E1 + 9E2 + 1E2 + 10E2) = 1 36 (16E1 + 20E2). Note this is an example of general rule 〈ψ|O|ψ〉 = ∑n on|cn|2, where the on are the eigenvalues of the Hermitian operator O, and the cn the expansion coefficients of ψ in the basis of O eigenstates. 2