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Modern Physics: Problem Set #5 Physics 320 Fall (12) 2017 Modern Physics: Problem Set #5 The Bohr Model ("This is all nonsense", von Laue on Bohr's model; "It is one of the greatest discoveries", Einstein on the same). 2 4 mek e 1 ⎛ 1 ⎞ hc L = mvr = n! ; E n = – 2 2 ≈ –13.6eV⎜ 2 ⎟ ; λn →m = 2! n ⎝ n ⎠ | E m – E n | Notes: Exam 1 is next Friday (9/29). This exam will cover material in chapters 2 through 4. The exam€ is closed book, closed notes but you may bring in one-half of an 8.5x11" sheet of paper with handwritten notes€ and equations. € Due: Friday Sept. 29 by 6 pm Reading assignment: for Monday, 5.1-5.4 (Wave function for matter & 1D-Schrodinger equation) for Wednesday, 5.5, 5.8 (Particle in a box & Expectation values) Problem assignment: Chapter 4 Problems: 54. Bohr model for hydrogen spectrum (identify the region of the spectrum for each λ) 57. Electron speed in the Bohr model for hydrogen [Result: ke 2 /n!] A1. Rotational Spectra: The figure shows a diatomic molecule with bond- length d rotating about its center of mass. According to quantum theory, the d m ! ! ! € angular momentum ( ) of this system is restricted to a discrete set L = r × p m of values (including zero). Using the Bohr quantization condition (L=n!), determine the following: a) The allowed€ rotational energies of the molecule. [Result: n 2!2 /md 2 ] b) The wavelength of the photon needed to excite the molecule from the n to the n+1 rotational state. [Result: λ=2πcmd 2/(2n+1)!] c) The wavelength of the photon need to excite €a nitrogen molecule (N2) from its ground state (n=0) to the n=1 rotational state (for N2, d=0.11 nm). [You can check your result by noting that rotational spectra for small molecules are generally in the microwave region: 0.1mm→10cm]. A2. Spiral death time of an atom (in five easy steps): In Rutherford's planetary model of the atom electrons orbit around a very small massive nucleus. Classically, such an atom will have a finite lifetime due to radiative energy loss of the electrons, causing them to spiral in towards the nucleus. Here we consider the hydrogen atom and assume the electron initially follows a circular orbit of radius r ≈ 0.05 nm. o a) Use the Coulomb force law to show that the electron acceleration is given by ke2 a = . mr2 - continued on back - € b) Use the fact that a = v2/r for a circular orbit to show that the total electron energy can be written as ke2 E = K + V = − . 2r c) An accelerating charge will lose energy due to radiation. Show that this will lead to a decrease in the electron orbital radius described by dr 2r2 dE € = . dt ke2 dt d) The power radiated by an accelerating charge is given by the Larmor formula as dE 2kq2a2 P = = − 3 € dt 3c (negative here since the electron loses energy). Insert this into the part (c) result to obtain the following expression: 2 2 € 2 4 ⎛ ke ⎞ r dr = − ⎜ ⎟ cdt . 3⎝ mc 2 ⎠ e) Define t = 0 to be when r = ro. Integrate the above expression from r = ro to r = 0 (left side) and t = 0 to t = T (right side) to show that the "spiral death time" of the Rutherford hydrogen atom is given by€ 2 r3 ⎛ mc 2 ⎞ T = o ⎜ ⎟ . 4c ⎝ ke2 ⎠ f) Compute a numerical result for T and comment on why most physicists were skeptical of the Rutherford picture of the atom. € .
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