Lecture 17: Bohr Model of the Atom
• Reading: Zumdahl 12.3, 12.4
• Outline
– Emission spectrum of atomic hydrogen.
– The Bohr model of the hydrogen atom.
– Extension to higher atomic number: Atoms
• Problems (5th Ed, Chapter 12)
– 2, 30, 31, 32, 33, 34
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Light is emitted from H (Atoms)
http://Jersey.uoregon.edu/vlab/elements/Elements.html
IF “Quantized” spectrum
ΔE
ΔE
“Continuous” spectrum
(Sunlight)
Any ΔE is
possible
Only certain
ΔE are
allowed
Atoms relax to lower energy states by giving off energy as light.2
Emission spectrum of H
Light Bulb (Sunlight)
Hydrogen Lamp
(video)
3
Quantized, not continuous
Emission
Photon Emission
• An atom gives up energy to
create a photon. The energy is
conserved.
• Electrons relax from one energy
level to another by emitting a
photon.
• The energy of the photon
matches the energy difference
of the electron
• With
−ΔE Atom = E photon = hν = hc / λ
λ = 440nm ( Blue Light )
6.6 ⋅10−34 ⋅ 3 ⋅108
−19
=
4.5
⋅
10
E photon = hν = hc / λ =
J
−9
440 ⋅10
4
−19
ΔE Atom = −4.5 ⋅10 J
Emission spectrum of H (cont.)
We can use the emission spectrum to determine the
energy levels for the hydrogen atom.
Warning: Light oscillations not drawn to scale. 5
Balmer Formula for Frequencies
• Joseph Balmer (1885) first noticed that the
frequency of visible lines in the H atom spectrum
could be reproduced by:
1 1
ν∝ 2− 2
2 n
n = 3, 4,5"
• The above equation predicts that as n increases,
the frequencies become more closely spaced.
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Rydberg Formula
• Johann Rydberg extends the Balmer Formula (because
there were more lines discovered at other frequencies as
photographic paper to detect UV and IR were invented)
• For Emission
⎛ 1 1 ⎞
ν = Ry ⎜ 2 − 2 ⎟
⎝ n2 n1 ⎠
n2 = 1, 2, 3, & "
n1 = n2 + 1, n2 + 2" ∞
n1 > n2
R y = 3.29 ⋅ 1015 sec-1
• This suggests that the energy levels of the H atom
are proportional to 1/n2
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H atom energy levels
• Use Rydberg formula and assume energy
conservation (as with photoelectric effect)
ΔEHAtom = E2 − E1 = − hν photon
ν photon
⎛ 1 1 ⎞
= Ry ⎜ 2 − 2 ⎟
⎝ n2 n1 ⎠
⎛ 1 1 ⎞
E2 − E1 = ( − hRy ) ⎜ 2 − 2 ⎟
⎝ n2 n1 ⎠
Then
Atom emits energy
(exothermic), energy
comes out as photon.
n1 > n2
⎛ 1 ⎞
⎛ 1 ⎞
⎛ 1 ⎞
E2 = ( − hRy ) ⎜ 2 ⎟ and E1 = ( − hRy ) ⎜ 2 ⎟ or En = − ( hRy ) ⎜ 2 ⎟ < 0
⎝n ⎠
⎝ n2 ⎠
⎝ n1 ⎠
N.B.: There could be an offset constant but it is zero; show later.8
The Bohr Model
• Niels Bohr uses the emission spectrum of
hydrogen to develop a quantum model for H.
• Central idea: electron orbit the “nucleus” and only
certain orbits (orbitals) are allowed.
• Atoms emit light by electrons moving from one
state (or orbit) to another.
• Bohr used Coulomb’s law of charge attraction
between e- and nucleus.
• However, classical physics is unable to explain
why an H atom doesn’t simply collapse.
P. E .
Vcoulombic
2
′
e Z
=−
<0
r
Potential energy is always negative
(i.e. attractive), V=0 at infinite
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separation
Circular Orbits and Planetary Motion
• For planets (gravity substitutes for charge attraction) the
total energy, for a plant going around the sun at a fixed
distance, r, (or an electron around nucleus) is:
E = KE + PE = T + V
• KE and PE ( or T and V) balance, giving stable circular
orbits for planets. The Virial Relation. T = − 1 V
2
• If the distance were a multiple of a fundamental distance
then we would have a formula for all energy levels:
2
′
Ze
E = KE + V = 12 V = − 12
r
ao n 2
If : r = rn =
Z
⎛ e′2 ⎞ Z 2
E = En = − ⎜
⎟ 2
2
a
⎝ o⎠n
Consistent with Rydberg formula, even
have minus sign, because electrons are
bound and stable (like planets)
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More fundamental Connection
• Total Energy (E) is connected to2 PE (V) and KE (T) as:
E = T +V =
1 mv 2
2
Ze′
−
r
T = − E = − 12 V
2
V
V
−
E = 12 V ⋅
=−
2T
4T
• All of these connections are true in Quantum Mechanics
(for real) if we average over all possible places where the
electron can be. So no wasted motion here.
m ( Ze′ )
V
E=−
=−
2
4T
2 ( mvr )
2 2
2
mvr = n=
E = En
= − 12
m ( Ze′
(=)
)
2 2
2
1
⋅ 2
n
Now Bohr makes his famous assumption that
the angular momentum comes in units of
Planck’s constant and obtains a statement
about his effective radius (now called the
Bohr radius) in terms of known constants.
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The Bohr Model
Bohr model for the H atom is capable of reproducing the energy levels
given by the empirical formulas of Balmer and Rydberg.
hRy =
m ( e′
)
2 2
2 ( =)
2
= 12
e2
4πε o ao
= 2.178 ⋅10−18 J
ao = 53 picometers ( pm) 12 Ang
ε o = 8.8 ⋅10−12 ( S .I .)
F
= 1.6 ⋅10−19 C
e=
NA
Z = atomic number (1 for H)
n = integer (1, 2, ….)
⎛ Z2 ⎞
En = − hRy ⎜ 2 ⎟
⎝n ⎠
Bohr could account for the Rydberg constant at a very fundamental level, in
terms of the mass and charge of the electron, and no other parameters. He even
accounted for the shift of the Rydberg constant (0.1%) when deuterium was
substituted for hydrogen.
1
1
1
m
=
me
+
M
12
“Classical” Motion of an Electron
Compare with Refrigerator Magnets.
Prob Z12.33
• E.g.: An electron is in n=3 orbit (orbital). The radius of that
orbit is ~2 times that of the n=2 orbit.
•Why is the energy negative? (What is the reference state?)
•If the electron “hops” from n=3 to the n=2 orbit what happens?
(i.e. How is energy conserved)? (This is the “red” transition.)
•Is the electron held more tightly now?
•Is its total energy lower?
•Is its potential energy lower?
•Is its kinetic energy lower?
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•Is it considered “more stable”?
The Bohr Model
2
⎛
⎞
Z
−18
En = −2.178 ⋅10 J ⎜ 2 ⎟
⎝n ⎠
• E<0 means electrons are bound to
Nucleus (lower E is more stable).
•Energy levels get closer together
as n increases
• at n = infinity, E = 0
Meaning the electron is not bound
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Energy Changes
We can use the Bohr model to predict what ΔE is between
any two energy levels. Same as we discussed with the
Rydberg formula (because it is the same formula).
ΔE = E final − Einitial = E2 − E1
⎛ 1 ⎞
⎛ 1 ⎞
−18
−18
ΔE = −2.178x10 J⎜⎜ 2 ⎟⎟ − (−2.178x10 J)⎜ 2 ⎟
⎝ n initial ⎠
⎝ n final ⎠
ΔE = −2.178 x10
−18
⎛ 1
1
J⎜ 2 − 2
⎜n
⎝ final ninitial
⎞
⎟⎟ n final < ninitial
⎠
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Energy Conservation and Photons
• Example: At what wavelength will emission from
n = 4 to n = 1 for the H atom be observed?
ΔE = −2.178 ⋅10
−18
⎛ 1
1
J⎜ 2 − 2
⎜n
⎝ final ninitial
1
ΔE = −2.178 ⋅10
EPhoton = 2.04 x10
−18
−18
J=
⎞
⎟⎟ = − hν photon
⎠
4
1⎞
⎛
J ⎜1 − ⎟ = −2.04 x10−18 J
⎝ 16 ⎠
hc
λ
λ = 9.74 x10−8 m = 97.4nm
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This photon is in the UV and can’t be seen by the human eye.
An Example of Energy Transition and Photon
• Example: What is the longest wavelength (i.e. lowest
frequency) of light needed to removal the e- from H in its
lowest energy state? (Prob Z12.32, Z12.34)
•N.B: This is absorption not emission.
⎛ 1
⎞
1
ΔE = −2.178x10−18 J⎜⎜ 2 − 2 ⎟⎟
⎝ n final n initial ⎠
∞
1
ΔE = −2.178x10−18 J (0 −1) = 2.178x10−18 J
ΔE = 2.178 x10
−18
J = hν =
hc
λ
λ = 9.13x10−8 m = 91.3nm
This is high UV, the largest possible energy transition (in a
single jump). Beyond this there is the continuum and extra
kinetic energy is imparted to the electron.
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Include Higher Z (ions)
• The Bohr model can be extended to any single
electron system….must keep track of Z
(atomic number).
2
⎛
Z ⎞
−18
E = −2.178 ⋅10 J ⎜ 2 ⎟
⎝n ⎠
Z = atomic number
n = integer (1, 2, ….)
• Examples: He+ (Z = 2), Li+2 (Z = 3), etc.
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Higher Z
•Electrons are held more tightly by nuclei with greater charge.
(Can only allow one electron, though)
•Example: At what wavelength will emission from
n = 4 to n = 1 for the He+ atom be observed?
⎛ 1
⎞
1
ΔE = −2.178x10−18 J (Z 2 )⎜⎜ 2 − 2 ⎟⎟
⎝ n final n initial ⎠
2
ΔE = −2.178x10
−18
ΔE = 8.16x10
J=
−18
hc
λ
1
4
⎛
1⎞
J (4)⎜1− ⎟ = −8.16x10−18 J
⎝ 16 ⎠
λ = 2.43x10−8 m = 24.3nm
λ H > λ He
+
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Where is this model right and wrong? (Z12.2)
• Bohr’s assumption that the energy of the photon
exactly matches the energy difference inside the
atom (molecule) is exactly correct.
• The description of the columbic potential energy
was correct.
• Obtaining the empirical Rydberg constant in terms
of fundamental quantities was correct.
• The Bohr model’s successes are limited:
• Doesn’t work for multi-electron atoms.
• The “electron racetrack” picture is incorrect.
• That said, the Bohr model was a pioneering,
“quantized” picture of atomic energy levels.
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