Chemistry, The Central Science , 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 7 (Zumdahl) Electronic Structure of Atoms. John D. Bookstaver St. Charles Community College Cottleville, MO. Waves.
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Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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s1, = waves/s = Hz (hertz)
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3.00 108 m/s.
c =
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E = h
where h is Planck’s constant, 6.626 10−34 Js.
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c =
E = h
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Solution: label your given(s) & your unknown.
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Another mystery in the early 20th century involved the emission spectra observed from energy emitted by atoms and molecules.
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E = nh
Where n=main energy level
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[
]
Where Z= nuclear charge (atomic #) and
n = main energy level and
RH = the Rydberg constant, 2.18 10−18 J
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nf2
( )

E = −RH
12
ni2
The Nature of EnergyThe energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation:
ni and nf are the initial and final energy levels of the electron.
And Z = 1 (atomic # of hydrogen!)
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Substitute & Solve:
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h
h
E
4. Substitute values (c, h, E) to solve for .
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mv
=
The Wave Nature of MatterThis formula is on the AP exam formula sheet
Please note: the base units for Joules are kg●m2/s2 . This will be important for calculations
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SOLUTION:
Givens: melectron = 9.11 x 1031 kg
velectron = 5.97 x 106 m/s
Unknown =
CONNECTION between givens & unknown:
= h
mv
Substitute given & constant values, solve for .
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Substitute given & constant values, solve for .
= h
mv
(9.11 x 1031 kg)(5.97 x 106 m/s)
= 1.22 x 1010
Where did the exponent go?
J= kg●m2/s2
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6.41) Determine the wavelength of the following objects
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4
(x) (mv)
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a) electron with a v = 0.100 m/s
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−l ≤ ml≤ l.
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Observing a graph of probabilities of finding an electron versus distance from the nucleus, we see that s orbitals possess n−1 nodes, or regions where there is 0 probability of finding an electron.
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“For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”
Another translation: “Don’t pair electrons in degenerate orbitals until there is 1 electron in each orbital.”
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Some irregularities occur when there are enough electrons to halffill s and d orbitals on a given row.
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For instance, the electron configuration for copper is
[Ar] 4s1 3d5
rather than the expected
[Ar] 4s2 3d4.
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