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Rutherford Model 1911
Positive charge is concentrated in a very small nucleus. So aparticles can sometimes approach very close to the charge Ze
in the nucleus and the Coulomb force
1 ( Ze)( Ze)
F=
4πε o
r2
Can be large enough to cause large angle deflections.
Nuclear model of the atom
Detailed calculations now show that there is
a non-negligible probability of large angle
scattering
Stability problem
(I) If electrons are stationary, there is nothing to prevent
them from responding to the Coulomb force of attraction
and being sucked into the nucleus. But the atomic diameter
would then be ~ nuclear diameter.
(II) If the electrons are in orbit, they are continuously
being accelerated and therefore, classically, should emit
radiation. They should lose energy and spiral into the
nucleus!
Bohr’s postulates
Bohr 1913: circumvented stability problem by
making two postulates:
1. An electron in an atom moves in a circular orbit for
which its angular momentum is an intergral multiple of h
2. An electron in one of these orbits is stable. But if it
discontinuously changes its orbit from one where
energy is E i to one whose energy is Ef , energy is
emitted or absorbed in quanta satisfying:
Ei − E f = hν
Analysis based on Bohr postulates: for
hydrogen-like atom (1 electron)
Force of
attraction
1 Ze 2 mv 2
F=
=
2
4πε o r
r
m = mass of
electron
1 2 1 1 Ze 2
KE = mv =
2
2 4πε o r
1 Ze 2
PE = −
4πε o r
1 Ze 2
Total E = −
4πε o 2r
mvr = nh
1 Ze
mv =
4πε o r
2
and
2 2
n h
1 Ze
m 2 2=
4πε o r
m r
E =−
2
r = 4πε o
2 4
1
mZ e
( 4πε o ) 2n h
2
2 2
2
2 2
n h
mZe
2
For convienience we define the “Bohr Radius” as:
a=
4πε o h
me
2
= 0.0529nm
2
And the energy unit 1 Rydberg (1Ry) as:
me
4
( 4πε o ) 2h
2
2
2
n
r=
a
Z
and
= 13.6eV
E= −
Z
2
n
2
Ry
n=1
n=2
n=3
“excited states”
“ground state”
a=radius of ground state (n=1) orbit for hydrogen (Z=1)
1 Ry=13.6eV is the ground state (n=1) binding energy for
hydrogen (agrees with experiment)
For larger Z-values, radius of orbit shrinks for given n. Binding
energies get much bigger
mvr = nh
Electron velocity: obtained from
2
2
nh
nhZe
Ze
v=
=
=
mr 4πε o n 2h 2 4πε o nh
2
Z=1, n=1
e
6
−1
= 2.2 ×10 ms
4πε o nh
This is less than 1% of c ( 3 × 108 ms −1 ) suggesting that a
calculation involving non-relativistic mechanics is valid
Emission Spectra
Electron makes transition from initial qu. State ni to final state
nf . The frequency ν of the photon emitted satisfies:
mZ e  1
1 
hν = Ei − E f =
−
2
2  2
2
(4πε o ) 2h  n f ni 
2 4
1
Express this in terms of wavelength
1
1
mZ e  1
1 
=
−
λ (4πε o ) 2 4πh 3c  n 2f ni2 


2 4
 1

1
1
= R∞ Z 2  2 − 2 
n

λ
n
i 
 f
R∞ =
1
me
4
( 4πε o ) 4πh c
2
3
= 1.0974 ×10 m
Is the Rydberg constant
This generates various families of spectral lines.
7
−1
Note that in each case there is a series limit corresponding to
1
For the Lyman series, this is at
= R∞ = 91nm
λ
Well
within the
UV range
The Balmer series lies in the near UV and visible region,
and the others are all in the infrared.
At time of Bohr’s proposal, only Balmer-Paschen series
were known, and the series were therefore predicted in
advance of the discovery (triumph for Bohr theory).
Books
Physics of Atoms and Molecules, Bransden and Joachain,
~pp 1-50 (Longman).
Quantum Mechanics, Rae, 3rd edition (first chapter)
NOTE: THESE ARE ADVANCED TEXTS
AND I AM ONLY RECOMMENDING YOU
READ FIRST SECTIONS OF EACH BOOK.
Absorption spectrum
General formula above also applies to the case where an electron
gains the energy of a suitable photon having energy hν exactly
equal to the difference between initial and final states.
Normally, electron will start off in ground state so only the
Lyman series is observed in absorption spectrum.
Finite Nuclear Mass
In the theory above, we assumed that the nucleus is so
massive that it is effectively at rest. But the mass of the
nucleus is finite and the classical model of the atom therefore
envisages that the electron and the nucleus both revolve
around their common centre of mass.
Therefore, any theory should be modified by replacing the
mass of the electron by the “reduced mass” of the system.
mM
m
µ=
=
m + M 1+ m
M
 1

1
1
= R.Z 2  2 − 2 
n

λ
n
i 
 f
R =
µe
1
4
( 4πε o ) 4πh c
2
3
R
µ
M
1
= =
=
m
R∞ m m + M 1 + M
R∞
is the limit of R as M tends to infinity
Difference between R and R∞ is small (~ 1 part in 1800)
but shows up in precise measurements of spectral lines.
For hydrogen
R = 1.0968 ×10 m
7
Cf.
−1
R∞ = 1.0974 ×10 m
7
−1
Example
Positron “atom” =system containing 1 positron (like an e- with a
positive charge) and 1 electron.
How does emission spectrum differ from hydrogen?
Set M=m, so
R∞
R∞
R=
=
m
2
1+
M
1 R∞  1
1  1
=
−
=
×
λ
2  n 2f ni2  2


previous formula
All wavelengths are doubled
Effect on energy and radii of quantum states?
We replace m in the formulae above by
m
µ=
2
Therefore, all energies are halved in magnitude, radii
doubled.
Fine Structure
This is the splitting of the spectral lines into several
components, when measured with equipment of high
resolution. Explanation: each energy level actually consists of
several distinct states with almost the same energy.
When viewed at high resolution, transitions split.
Transitions between any two Bohr energy states involve
several spectral lines.
Theoretical explanation: modification of simple Bohr theory by
Wilson and Sommerfeld: electron orbits can be elliptical, of which
a circular orbit is a special case.
Each orbit is specified by 2 parameters instead of 1.
Geometrically by semi-major and semi-minor axes a,b, no just
radius r.
Thus, energy levels turn out to be dependent on two quantum
numbers, but only when one takes relativsitic considerations
into account.
Without relativity, we get the same formula for E as before.
Relativistic correction: electrons in very eccentric orbits have
large velocities when they are near the nucleus, so v/c is NOT
negligible.
Can show
2 2

Z
α Z
1
3 
 −  Ry
E = − 2 1 +
n  nθ 4n 
n 
2
Where n is the principal quantum number = 1, 2, 3...
nθ is the “azimuthal” quantum number = 1, 2, ….n
2
1 e
1
α=
≈
4πε o hc 137
Fine structure constant
But note that not all possible transitions occur in practice. The
“allowed” transitions are shown in the diagram: these are the
transitions for which:
nθf − nθi = ±1
This is due to the operation of a “selection rule”,
which states that transitions can only occur between
states for which nθ changes by ± 1
The only Dane ever to win the Nobel Prize in Physics
Why does the Bohr theory work?
More precisely, why does the Bohr theory generate the correct
formula for the energy levels in a hydrogen atom?
Answer: because the Bohr condition: mvr = nh inadvertently
makes use of the wave associated with an electron.
Supposing the classical orbit is circular: we assume that the
associated wave is a standing wave following the motion.
But in order for the wave to return to its initial value (i.e.
we are requiring that the wave be single valued), we require
that 2πr = circumfere nce = nλ
h
p=
λ
h
h
λ= =
p mv
2πrmv = nh
mvr = nh
The Bohr condition
Inadequacies of the Bohr Theory
1. Does well to describe hydrogen, but can be extended only to
1-electron atoms, i.e. hydrogen-like, with higher Z values. Can
treat alkali atoms with some success, but only because they have
1 electron only outside closed shells. Fails to account for spectra
of other atoms.
2. Theory does not explain rate at which transitions occur
between states, i.e. the relative intensities of spectral lines.
3. The theory is ad hoc and lacks a satisfying basis.
Superseded by Quantum Mechanics, initiated by de Broglie
(1924) and Schrodinger (1926).
In Schrodinger’s theory, the electron is described by a
wavefunction Ψ (r ) which is the solution of SchrÖdinger’s
equation:
d 2Ψ
dx 2
+
[E − V ( x)] Ψ ( x) = 0
2
2m
h
In 1D
The electron does not move on well-defined orbital paths as in
the Bohr theory, one can only speak of the probability density
2
Ψ (x ) of finding the electron in any region of space in the
neighbourhood of the nucleus.
This is the modern view of the behaviour of electrons in atoms
and in similar situations on an atomic scale. The classical view
of “particles” subject to Newtonian mechanics and
electromagnetic forces has been replaced by a wave picture in
which the behaviour of the electron is governed by the
behaviour of the associated wave.
Note: discrete wavelengths and energies are associated with
particles which are confined to a finite region.
Analogy with waves on strings. Confinement leads to
standing NOT travelling waves.
Particle confined to 1D
Suppose particle is confined to a region between x=0 and x=l.
x=0
x=l
If there is no chance of the particle lying outside this
region, the associated wave must vanish:
nλ
l=
2
n = 1, 2,3...
p2
h
E=
and p =
2m
λ
2
2
2
2 2
h n
n h
E=
=
=


2mλ2 2m  2l 
8ml 2
h
So, like the hydrogen atom, energy is quantised and the
quantum number arises from the confinement of the particle.
Energy level
diagram
n=2
Excited
states
n=1
Ground
state
E=0
Note minimum energy
E=
h2
8ml 2
= zero point energy
Even at T=0, particle must have
at least this energy
Explained by Uncertainty Principle.
E =0 ⇒
∆x ~ l
p2
= 0 ⇒ p = 0, ∆p = 0.
2m
Contradiction of Heisenberg
Uncertainty Principle
Examples of Zero Point energy
1. Nucleus can’t contain electrons. [between 1911-1933
structure of the nucleus wasn’t known, since neutron had not
been discovered. Nucleus= mixture of protons and electrons?
But then each electron would have zero point energy ~20MeV
whereas β-decay shows electron energies are all ~1MeV.
2. Liquid He doesn’t solidify at T=0, except at very high
pressure.
Wavefunctions: for particle in a 1D potential well
nπx
Ψn ( x) = A sin
l
Note: this leads to standing waves. As time proceeds,
get oscillations in the usual way.
Since Ψ 2 gives probability of particle being found, this
also varies...
Notes:
n=0 not relevant (wave doesn’t exist!)
Strong analogy with standing wave on a stretched string
But analogy not complete since in quantum case:
hν = E ⇒ ν n ∝ E n ∝ n 2
string
ν n ∝ En ∝ n
Example: calculate wavelength of photon emitted when electron in
potential well of width 1nm undergoes a transition from n=2 to n=1.
n=2
n=1
E=0
En = n
2
h
2
8ml
2
=n
(6.6 × 10
2
8 × 9.1× 10
En = n 2 .0.374eV
−34 2
−31
)
−9 2
× (10 )
E 2 = 1.496eV
E 2 − E1 = 1.122 eV
J
Wavelength
λ
given by:
hc
hν =
= 1.122eV ⇒ λ = 1.1µm
λ
Note variation with l: if we change l from 1nm to 0.5nm, due
to factor of l 2 in denominator all energies are x4, so
wavelengths reduced by a factor of 4.
Quantum dots: practical example of an electron (or several
electrons) being confined to a region 10nm thick. Such devices
are increasingly important in quantum engineering, I.e. in
devices which exploit quantum physics.
Quantum Tunneling
“Wave associated with a particle” : what equation does it satisfy?
Answer: Schrodinger’s equation. No derivation here (too
complex). Subject of equation in 1D is Ψ ( x , t ) in 1D
2
Ψ (x ) gives probability that particle will be found at
position x at a time t.
For a fixed time t, the profile of Ψ is given by:
d 2Ψ
dx
2
+
2m
h
2
[E − V ( x)] Ψ = 0
Where m is the mass, V(x) is the potential energy function,
E= total energy
Note that: E − V ( x ) = KE
E = KE + V (x )
Classically allowed region
E − V ( x) ≥ 0
Classically forbidden region
E − V ( x) ≤ 0
Nature of the solution Ψ (x ) in classically allowed region
Suppose
Ψ (x) is positive, then
d 2 Ψ ( x ) d  dΨ ( x ) 
= 

2
dx
dx
dx


d 2 Ψ( x)
dx
2
<0
dΨ ( x )
dx
slope
Also,
dΨ ( x)
dx
measures rate of change. So if
this means that the slope is decreasing
d  dΨ ( x) 

<0
dx  dx 
Ψ (x)
x
The REVERSE holds when
Ψ(x) < 0
d 2 Ψ ( x)
dx
2
>0
Therefore throughout classically allowed region Ψ (x) must
be oscillatory.
Ψ (x) in a classically forbidden region
Ψ(x) > 0
Exponential increase
d 2 Ψ ( x)
dx
2
>0
Exponential decrease:
this one makes sense
Similarly if
Ψ(x) < 0
Ψ (x)
d 2 Ψ ( x)
dx
2
<0
Near a potential barrier
Main point is that the particle has a finite probability of being in
a region which is inaccessible classically, and of “tunneling”
from one classically allowed region to another.
Known as “barrier penetration”.
W
V =0
V =W
− D / 2 < x < +D / 2
x ≤ − D / 2, x ≥ + D / 2
Wavefunction oscillatory inside, decays exponentially
outside
dψ
ψ,
dx
continuous
Applications/Illustrations
(a)
238
92 U
234
→ 90Th
+
4
2 He
α-particle
V
V0
Distance from
nucleus
Cold electron emission
Direct a strong electric field towards a metal surface
V
Potential energy of an
electron near the surface
of metal
Cold electron emission: scanning tunnelling microscope
http://eaps4.iap.tuwien.ac.at/www/surface/STM_Gallery/stm_
schematic.html
http://eaps4.iap.tuwien.ac.at/www/surface/STM_Gallery/stm_
schematic.html
Quantum puzzles and Mysteries
1. Meaning of the Uncertainty Principle.
h
∆p.∆x ≥
2
Extent to which position x is unpredictable
Sometimes expressed: impossible to determine simultaneously
the position and momentum of a particle with arbritary
accuracy (Copenhagen Interpretation).
Supported by argument involving Heisenberg microscope.
Resolving power of microscope is:
λ
∆x ≈
sin θ
Recoil momentum of particle is uncertain to an extent:
h
∆px ≈ sin θ
λ
h
λ
∆x.∆p x ≈ sin θ .
≈h
λ
sin θ
But is this a “simultaneous measurement”?
Alternative viewpoint: (statistical interpretation) ∆x is a
standard deviation for an ensemble of particles, i.e. it is
a statistical property.
2
Ψ ∝ probabilit y
Implies a statistical
population
To verify experimentally, we would have to measure x for
a large number of particles and build up a histogram.
Do the same for p x
But: no simultaneous measurements!
Check ∆x.∆p x ≥ h
2
Einstein-Podolsky-Rosen (EPR) Paradox (thought experiment)
Firstly, concept of spin: non-classical property of electrons,
protons etc.
Nearest classical analogue: spin of the Earth. But this is not a
good analogy since the electron is a point particle.
Can have: correlated system of 2 electrons,such that total spin
component in a certain direction is zero. (in the up-down
direction say). So if one electron has a “spin up” the other has
“spin-down”
Now suppose we prepare such a pair of electrons and separate
them. Measure spin of one - say “up”
other has spin “down”
Therefore, 1st measurement determines (immediately)
the result of the second, no matter how far apart the
electrons are!
But: perhaps this is the same as for a completely classical
system - e.g. 2 boxes, one with something in it, the other
empty.
So make hypothesis: “each electron in the EPR thought
experiment has a definite spin state, even before one looks”
Determined by Hidden variables
Then we find Bell’s theorem
hypothesis is correct.
predictions if this
1. Measure one particle effects other particle
2. Measure one particle only gives us knowledge of other (true
before measurement)
1. Relates to QM, 2. Is classical
Hence forced to accept that QM is
“non-local”, which is disturbing.
3. Schrodinger’s Cat
(Thought Experiment)
Logical problem. “The atom at any time has decayed or it has
not”
“The cat at any time, is either alive or dead”
Defies common sense! Can partly evade this problem by
emphasising that the wavefunction describes an ensemble
of radioactive atoms, accompanied by an ensemble of
cats. A certain fraction will be dead and the other fraction
alive.