Transcript Lecture 17: Bohr Model of the Atom

Plan for Fri, 31 Oct 08
• Lecture
– Emission spectrum of atomic hydrogen (7.3)
– The Bohr model of hydrogen (7.4)
– Quantum mechanical model (7.5)
• Quiz 5
Emission
Photon Emission
• Relaxation from one
energy level to another
by emitting a photon.
• With DE = hc/l
• If l = 440 nm,
DE = 4.5 x 10-19 J
Atomic Emission
When we heat a sample of an element, the atoms become excited. When
the atom relaxes it emits visible light. The color of the light depends on
the element.
Li
When the light emitted
from excited atoms is
passed through a
prism, we see discrete
bands of color at
specific wavelengths.
Na
K
Ca
Sr
H
Li
Ba
Emission spectrum of H (cont.)
Light Bulb
Hydrogen Lamp
Quantized, not continuous
Emission spectrum of H (cont.)
We can use the emission spectrum to determine the
energy levels for the hydrogen atom.
The Bohr Model
• Niels Bohr uses the emission spectrum of
hydrogen to develop a quantum model for H.
• Central idea: electron circles the “nucleus” in
only certain allowed circular orbitals.
• Bohr postulates that there is Coulombic attraction
between e- and nucleus. However, classical
physics is unable to explain why an H atom
doesn’t simply collapse.
The Bohr Model of
the atom
Principle Quantum number: n
An “index” of the energy levels
available to the electron.
The Bohr Model (cont.)
2

Z
18
E  2.178x10 J 2 
n 

• Energy levels get closer together
as n increases
• at n = infinity, E = 0
Identify the energy level diagram that best
represents hydrogen:
A
B
C
D
The Bohr Model (cont.)
• We can use the Bohr model to predict what DE is
for any two energy levels
DE  E final  E initial
18
DE  2.178x10

 1 
 1 
18
J 2   (2.178x10 J) 2 
 ninitial 
 n final 
18
DE  2.178x10
 1
1 
J 2  2 
 n final n initial 
The Bohr Model (cont.)
• Example: At what wavelength will emission from
n = 4 to n = 1 for the H atom be observed?
18
DE  2.178x10
 1
1 
J 2  2 
 n final n initial 
1
4

1
DE  2.178x10 J1   2.04x1018 J
 16 
18

18

DE  2.04 x10 J 
hc
l
l  9.74 x108 m  97.4nm
The Bohr Model (cont.)
• Example: What is the longest wavelength of light
that will result in removal of the e- from H?
 1

1
DE  2.178x1018 J 2  2 
 n final n initial 
1



DE  2.178x1018 J0 1  2.178x1018 J
18
DE  2.178x10 J 
hc
l
l  9.13x108 m  91.3nm
The n = 4 to n = 1 transition in hydrogen corresponds
to a wavelength of 97.4 nm. At what wavelength is
the n = 4 to n = 2 transition expected?
A. l41 > l42
B. l41 < l42
C. l41  l42
D. l42  0
Extension to Higher Z
• The Bohr model can be extended to any single
electron system….must keep track of Z
(atomic number).
2

Z
18
E  2.178x10 J 2 
n 
Z = atomic number
n = integer (1, 2, ….)
• Examples: He+ (Z = 2), Li+2 (Z = 3), etc.
Extension to Higher Z (cont.)
• Example: At what wavelength will emission from
n = 4 to n = 1 for the He+ atom be observed?
 1

1
DE  2.178x1018 J Z 2  2  2 
 n final ninitial 
2
1
4
 1
DE  2.178x10 J41   8.16x1018 J
 16 
hc
18
l  2.43x108 m  24.3nm
DE  8.16x10 J 
l
l H > l He
18



So what happened to the Bohr Model?
Although it successfully described the line spectrum of hydrogen
and other one-electron systems, it failed to accurately describe
the spectra of multi-electron atoms.
The Bohr model was soon scrapped in favor of the Quantum
Mechanical model, although the vocabulary of the Bohr model
persists.
However, Bohr pioneered the idea of quantized electronic energy
levels in atoms, so we owe him big.
Thanks Niels Bohr!
Quantum Concepts
• The Bohr model was capable of describing the
discrete or “quantized” emission spectrum of H.
• But the failure of the model for multielectron
systems combined with other issues (the
ultraviolet catastrophe, workfunctions of metals,
etc.) suggested that a new description of atomic
matter was needed.
Quantum Concepts (cont.)
• This new description was known as wave
mechanics or quantum mechanics.
• Recall, photons and electrons readily demonstrate
wave-particle duality.
• The idea behind wave mechanics was that the
existence of the electron in fixed energy levels
could be though of as a “standing wave”.
Quantum Concepts (cont.)
• What is a standing wave?
• A standing wave is a motion in
which translation of the wave does
not occur.
• In the guitar string analogy
(illustrated), note that standing
waves involve nodes in which no
motion of the string occurs.
• Note also that integer and halfinteger values of the wavelength
correspond to standing waves.
Quantum Concepts (cont.)
• Louis de Broglie suggests that for the e- orbits envisioned
by Bohr, only certain orbits are allowed since they satisfy
the standing wave condition.
h
h

p mv
l  wavelength
l
m  mass
v  velocity
h  Planck's constant
not allowed
Quantum Concepts (cont.)
• Erwin Schrodinger developed a mathematical
formalism that incorporates the wave nature of
matter:
Hˆ   E
• H, the “Hamiltonian,” is a special kind of function
that gives the energy of a quantum state, which is
 by the wavefunction, Y.
described
Quantum Concepts (cont.)
• What is a wavefunction?

= a probability amplitude
• Probability of finding a particle
in space:
Probability =

*
With the wavefunction, we can
describe spatial distributions.

The Probability Distribution for the Hydrogen 1s Orbital in ThreeDimensional Space (b) The Probability of Find the Electron at
Points Along a Line Drawn From the Nucleus Outward in Any
Direction for the Hydrogen 1s Orbital
Hydrogen’s Electron
Cross Section of the Hydrogen 1s
Orbital Probability Distribution
Divided into Successive Thin
Spherical Shells (b) The Radial
Two Representations of the Hydrogen 1s, 2s,
Probability Distribution
and 3s Orbitals (a) The Electron Probability
Distribution (b) The Surface Contains 90% of
the Total Electron Probability (the Size of the
Oribital, by Definition)
Quantum Concepts (cont.)
• Another limitation of the Bohr model was that it
assumed we could know both the position and
momentum of an electron exactly.
• Werner Heisenberg development of quantum
mechanics leads him to the observation that there is a
fundamental limit to how well one can know both the
position and momentum of a particle.
h
Dx  Dp 
4
Uncertainty
in position

Uncertainty in
momentum
where…
Dp  D(mv)  mDv
Quantum Concepts (cont.)
• Example:
What is the uncertainty in velocity for an electron in a
1 Å radius orbital in which the positional uncertainty
is 1% of the radius. (1 Å = 10-10 m)
Dx = (1 Å)(0.01) = 1 x 10-12 m
34
6.626x10
J.s

h
23
Dp 


5.27x10
kg.m /s
12
4 Dx
4 1x10 m
Dp 5.27x1023 kg.m /s
7m
Dv 


5.7x10
s
m
9.11x1031 kg
huge
Quantum Concepts (cont.)
• Example (you’re quantum as well):
What is the uncertainty in position for a 80 kg student
walking across campus at 1.3 m/s with an uncertainty
in velocity of 1%.
Dp = m Dv = (80kg)(0.013 m/s) = 1.04 kg.m/s
34
6.626x10
J.s

h
Dx 

 5.07x1035 m
4Dp 4 1.04kg.m /s
Very small……we know where you are.
