atomic structure lecture part 1/3
TRANSCRIPT
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ATOMIC STRUCTURE
Lesson by Dr.Chris University of Phayao, August 2016
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WHAT WE WILL LEARN … PART 1:
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ATOM SYMBOLS
Ne 20
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Atomic number Z
= no. of protons = no. of electrons
Mass number A
= no. of protons + neutrons
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EXAMPLES
How many protons, electrons and neutrons are in:
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SOLUTION
Cl can have 18 or 20 neutrons
35.45 is a mix of 2/3 35Cl and 1/3 37Cl
Element number
= no. of protons = no. of electrons
Mass number, not integer !
=> mix of ISOTOPES with different no. of neutrons !
Z
A
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ISOTOPES
Nearly all elements have isotopes,
that means the same elements
(no. of protons = Z) has different no. of
neutrons, and therefore different mass
Example:
Copper exists to 69.2% of 63Cu and the rest
of 65Cu with masses 62.93 and 64.93
what is the atomic mass of the mixture ?
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REVIEW BOHR ATOM MODEL
How do we know the structure of an atom ? 7
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All waves have: frequency and wavelength
symbol: n (Greek letter “nu”) l (Greek “lambda”)
units: “cycles per sec” = Hertz “distance” (nm)
• All radiation: l • n = c
where c = velocity of light = 300’000 km/sec
ELECTROMAGNETIC RADIATION
Note: Long wavelength
small frequency
Short wavelength
high frequency increasing
wavelength
increasing
frequency
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Example: Red light has l = 700 nm.
Calculate the frequency n .
= 3.00 x 10
8 m/s
7.00 x 10 -7
m = 4.29 x 10
14 Hz n =
c
l
• Wave nature of light is shown by classical
wave properties such as
• interference
• diffraction
ELECTROMAGNETIC RADIATION (2)
What is the energy in cm-1 ?
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ENERGY UNITS
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WHERE ARE THE ELECTRONS IN AN ATOM ? WE CAN KNOW FROM LIGHT SPECTRA !!
3 kinds of spectra:
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BRIGHT COLORS FROM
HYDROGEN MATCH THE MISSING
COLORS IN SUNLIGHT Hydrogen spectrum
Solar spectrum
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Li
Na
He
K
Cd
H
A
B
C
D
Identify the elements in the mixture A – D !
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BOHR’S IDEA
“Allowed” orbits
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WHERE DO THE LINES COME FROM ?
Bohr (1913)
emission spectra of hydrogen gas
Lines correspond to energies that are emitted by electrons:
emitted 15
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ELECTRONS ARE “FIXED” ON ORBITS !
Electrons can move between distinct energy levels,
they cannot exist just anywhere in the atom = quantum
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How many lines in the emission spectrum and at which energies (in cm-1) ?
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Solution: 3 levels 3 lines
Transition A:
∆E = E3 – E2 =
-20’000 + 50’000 cm-1 =
30’000 cm-1 =
λ = 1/30’000cm-1 * 107 nm/1 cm = 333 nm
We can express energy as wavenumber, because h and c are constant:
= const * 1/λ = const * ν 19
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THE ENERGIES OF ORBITALS
There are 2 forces acting on a circling
electron:
• Electrostatic attraction
• Centripetal force
They must be equal to keep the
electron stable:
The kinetic energy is then: 20
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BOHR’S POSTULATE
A circling electric charge would emit energy, what is not
the case of real electrons.
Bohr postulated that electrons can live in certain energy
levels without loss of energy:
He assumed that the orbital angular momentum L = r *
p must be a multiple of h:
L = me * v * r = n * h/2 = n * h
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From this we can calculate the orbital radius r:
with a0 as Bohr Radius
= 52.4 pm
-> the energy levels of the electron in a H atom are:
E1 = 13.6 eV
( 1eV = 8065.6 cm-1)
Calculate:
• an energy diagram for the H-atom electron
• the energy difference between the ground
and first excited state
• the wavelength of light emitted
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1 eV => v = 8065.6 cm-1
l = 1240 nm
10.2 eV : v = 82.270 cm-1
l = 121.6 nm
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RYDBERG EQUATION
From which energy level does an electron
come to n=2 when visible light of 410 nm
is emitted ?
What is the Ionization energy of hydrogen from this formula ?
-1
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n = 2, RH = 1.0974 10-2 nm-1
1/λ = 1.0974 10-2 * ( 1/4 – 1/n2) = 1/410
1/n2 =1/4 - 1/(410 * 0.011)
n2 = 36
=> n =6
The electron goes from n=6 to n=2 and releases an energy that equals 410 nm
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“HYDROGEN LIKE” ELEMENTS
The Bohr Theory can applied to all atoms
with one valence electron, like He(+)
Compare the energy of a transition
n=6 n=2 for H and He(+) in cm-1
http://www.fxsolver.com/browse/formulas/ Rydberg+formula+for+any+hydrogen-like+element
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DETAILS
Summary
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LESSON 2
QUANTUM THEORY: THE ELECTRON AS A WAVE
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WHY A NEW THEORY ?
The Bohr Theory could explain the nature of
the Hydrogen atom quite precisely
BUT: elements with more than one electron
could not be handled by this theory
AND: the classical theory cannot explain
why there are only certain orbits that are
“allowed” for electrons. 29
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ELECTRON BEAM INTERFERENCE ELECTRONS SHOW WAVE-BEHAVIOUR !
For example, if two slits
are separated by 0.5mm
(d), and are illuminated
with a 0.6μm
wavelength laser (λ),
then at a distance of 1m
(z), the spacing of the
fringes will be 1.2mm.
n λ / d
d
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ELECTRONS AS WAVES
DeBroglie: (“ de Broy “) 1924
The energy of a photon from Einstein’s
theory:
E = c * p = h *c / l
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RELATION PARTICLE - WAVE
Means every moving particle can be
considered as wave.
In the macroscopic world, the wavelength is
so small that it is not measurable.
Example:
a gun projectile with m=10 g moves with
800 m/sec => what is its wavelength ?
Compare to an electron with m = 9.1 10-31 kg
(h = 6.626 x 10-34 J*s / 1J = 1 kg m2/s2) 32
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CALCULATION
λ = (6.626 x 10^-34 J*s)/(0.01 kg)(800 m/s)
1 J = 1 kg*m^2/s^2
λ = 0.83 x 10^-34 m = 8.3 10-24 nm
(compare: x-ray around 1 nm)
But for an electron it is:
λ = (6.626 x 10-34 J*s)/(9 10-31kg)(800 m/s)
= 9.2 10-7 m = 920 nm similar red light
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LIGHT WAVES = PHOTONS
When light is emitted (like from the sun or
a heated metal), the energy of the photons
is discrete - comes in “packets” of
E = h ν = h c/λ = m c2
Energy is not continuous, but quantisized !
(one quantum of energy is h ν )
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EXPLAINS WHY ELECTRONS CAN ONLY EXIST ON CERTAIN ORBITS
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CONSEQUENCES: UNCERTAINTY !
When we treat electrons as waves, then we
cannot determine the position of the electron
exactly. The position (as particle) depends on
its speed – we describe this relation by the
Heisenberg Uncertainty Relation:
∆x m ∆v >= h/2π ( h )
(“Energy multiplied by time is constant” Murphy’s Law)
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ENERGY OF “WAVE-ELECTRONS”
The model based on electrons circling around a nucleus is not satisfying, even it can explain the hydrogen atom.
A new model based on standing electron waves was developed:
Model the behaviour of a particle in a restricted space (“particle in a box”)
Quantization comes from the fact that a wave has to “fit” into the boundaries
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Only wavelengths are allowed which “fit” into the box length L:
Use in DeBroglie: (h = Planck constant)
Therefore the kinetic energy is:
n is the main quantum number indicating the energy level 38
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EXAMPLE
Calculate the first 2 energy levels for an
electron in a box with L = 300 pm (ca.
circumference of H-atom).
me = 9.1 * 10-31 kg h = 6.6 * 10-34 m2 kg/s
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APPLICATION
The simple box model can help us to estimate the color of linear chromophores
We can assume that 2 electrons
can move along this molecule
chain (“resonance”)
If C-C and N-C is 1.40 A, which
light will be absorbed by this molecule ? 40
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ANSWER
The molecule “length” is about 7 * 1.4 A
We have 8 electrons, the transition should occur between level 5 and 4:
=> 352 nm
http://www.umich.edu/~chem461/Ex3.pdf
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CALCULATIONS
In the box model, an electron goes from n=2
level to the ground state and emits red light
of λ = 694 nm . What is the length of the box ?
(h = 6.63 E-34 Js, me = 9.11 E-31 kg, c= 3 E8 m/s 1 J = 1 kg·m2/s2.)
λ = 694 nm
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ANSWER
http://www.physics.umd.edu/courses/Phys270/Jenkins/Hwksolns13_Publishers.pdf
(compare: r(H) = 0.053nm => L is about the double of the circumference )
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SUMMARY Summary
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