Download Problem Set 05

Physics 320
Fall (12) 2016
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).
L = mvr = n ; E n = –
⎛ 1 ⎞
hc
me k 2e 4 1
≈ –13.6eV⎜ 2 ⎟ ; λn →m =
2
2
⎝ n ⎠
| Em – En |
2 n
Notes: Exam 1 is next Friday (9/30). This exam will cover material in chapters 2 through 4. The
€ is closed book, closed notes but you may bring in one-half of an 8.5x11" sheet of paper
exam
€
€
with handwritten notes
and equations.
Due: Friday Sept. 30 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 bondlength d rotating about its center of mass. According to quantum
theory, the
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 
angular momentum ( L = r × p ) of this system is restricted to a discrete set
of values (including zero). Using the Bohr quantization condition (L=n),
determine the following:
d
m
m
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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 ro ≈ 0.05 nm.
a) Use the Coulomb force law to show that the electron acceleration is given by
ke 2
a=
.
mr 2
- continued on back -
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b) Use the fact that a = v2/r for a circular orbit to show that the total electron energy can be
written as
ke 2
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 2r 2 dE
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=
.
dt ke 2 dt
d) The power radiated by an accelerating charge is given by the Larmor formula as
dE
2kq 2 a 2
P=
=−
dt
3c 3
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(negative here since the electron loses energy). Insert this into the part (c) result to obtain the
following expression:
2
4 ⎛ ke 2 ⎞
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2
r dr = − ⎜ 2 ⎟ cdt .
3 ⎝ mc ⎠
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
ro3 ⎛ mc 2 ⎞
T = ⎜ 2 ⎟ .
4c ⎝ ke ⎠
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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