Solutions for Homework Problems in Quantum Physics, Assignments of Quantum Mechanics

Download Solutions for Homework Problems in Quantum Physics and more Assignments Quantum Mechanics in PDF only on Docsity! SOLUTIONS for Homework #1 1. The photon energy for a given wave length λ equals Eγ = h̄ω = 2πh̄c λ ≈ 60 keV, (1) where it is useful to memorize a simple numerical relation h̄c = 197 MeV · fm ≈ 2× 10−5 eV · cm. (2) Standard conservation laws for the Compton effect determine the energy transferred from the photon to the electron originally at rest, ∆E = Eγ 1 + 2(Eγ/mc2) sin2(ϑ/2) ≈ 51 keV, (3) where m is the electron mass, mc2 =511 keV, and ϑ is the photon scat- tering angle. The binding (ionization) energy of the lowest, the so-called K-, shell in the Mo atom, Z = 42, can be roughly estimated as Eb = 1Ry × Z2 ≈ 24 keV. (4) Then eq. (3) determines the value of final kinetic energy of the electron, Ekin = ∆E − Eb ≈ 27 keV. (5) /A precise result is 31 keV since the empirical value of the wave length λK for the K-edge of the absorption of X-rays for Mo is λK = 0.619 Å that corresponds to the ionization energy of 20 keV, rather than 24 keV of our estimate in eq. (4)./ 2. From the de Broglie wave length λ = h/p we obtain v = 2πh̄ mλ ; (6) this gives 2.5×10−12c = 7.5×10−2 cm/sec for the electron and 4×10−5 cm/sec for the neutron. Modern experimental techniques allows one to perform experiments with atomic waves. But for macroscopic bodies the quan- tum wave lengths are extremely small: for a human being of weight 50 kg moving with the speed of 1 cm/sec we obtain λ ∼ 10−31 cm, i.e. quantum effects in the translational motion are invisible. 3. Average energy of an atom at this temperature, according to the classical Maxwell distribution, is (3/2)T where temperature is expressed in ener- getic units (1 eV = 11600K). From here we obtain the average velocity, v/c ≈ 2× 10−11, and the de Broglie wave length λ ≈ 6× 10−5 cm. If this 1 wave length is comparable to the size r0 of a small cube accomodating just one atom, then the density of the gas is n ≈ r−30 ≈ 0.5×10−12 cm−3. This is a typical atomic density in the traps used for studies of the Bose-Einstein condensate. /This estimate gives only the conditions for the transition from the Maxwell distribution to the Bose-Einstein distribution. At very small temperature, all atoms occupy the ground state available in the trap so that our initial estimate for atomic energy becomes invalid, and typical atomic velocities are determined by the size of the trap in accordance with the uncertainty relation./ 4. The potential energy of an electron in the field of the screened center is U(r) = −Ze 2 r e−κr. (7) Consider a circular electron orbit of radius r and speed v. The equilibrium condition for this orbit reads mv2 r = Ze2κ r ( 1 + 1 κr ) e−κr, (8) or, applying the quantization rule L2 = (mvr)2 = n2h̄2, (9) we find Ze2mκ ( 1 + 1 κr ) r2e−κr = n2h̄2. (10) The left hand side of eq. (10) exponentially falls off for large distances, κr  1. Therefore there is no solutions for large values of n, and the number of bound states should be finite. The maximum allowed radius can be found from the maximum of the left hand side, which is given by the positive root of the equation r2 − r κ − 1 κ2 = 0 ; r = 1 + √ 5 2κ . (11) Of course, one could guess with no calculations that the maximum radius of the orbit should be of the order rD = 1/κ. The maximum quantum number, corresponding to the number of levels supported by the screened potential, is now determined from n2max = (3 + √ 5)(1 + √ 5) 4 e−(1+ √ 5)/2 mZe 2 κh̄2 ≈ 0.84 rD aB Z, (12) where aB/Z = h̄2/(me2Z) is the Bohr radius of the lowest bound orbit in the pure Coulomb potential of the charge Z. Although the semiclassical 2