Download ONE-ELECTRON ATOMS: SPECTRAL PATTERNS Late 19th

Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
ONE-ELECTRON ATOMS: SPECTRAL PATTERNS
Late 19th century: Balmer, Rydberg, Lyman, Paschen, and others determine the emission spectra
of hydrogen atoms. They find a wide range of spectral lines, but they notice a pattern in the
frequencies of the lines: they can all be written in the form
ν = 3.29 x 1015 s-1 (n-2 – m-2)
or in terms of the energy of the emitted photons:
ε = hν = 13.60 eV (n-2 – m-2) = 2.180 x 10-18 J (n-2 – m-2)
where n and m are positive integers, and m > n.
In fact, once they notice the pattern, they notice that for each possible pair of integers, they find
an emission, and for no other frequencies.
Simplest analysis: since for emission, ε = Einitial – Efinal, the experimental data suggest that
the energy levels of a Hydrogen atom are given by
En = –(2.180 x 10-18 J) n-2 for n = 1, 2, 3, ... (the ground state has n = 1)
---- SHOW ENERGY LEVEL TRANSPARENCY (OXTOBY, Fig. 15.14, p. 540) ---If m is the initial state and n the final state (so that Em > En), then
Em – En = –(2.180 x 10-18 J) m-2 – [–(2.180 x 10-18 J) n-2] = 2.180 x 10-18 J (n-2 – m-2)
as observed. Note that the energy levels are very unevenly spaced. They rapidly move closer
together as n increases. As a result, there is a wide range of energy differences, and thus the range
of spectral lines is very wide. The resulting emission spectrum of the Hydrogen atom (including
all transitions of the atom) looks like this:
---- SHOW PORILE TRANSPARENCY ---The spacings between energy levels becomes very closely packed, so the light absorbed or emitted
in transitions between levels of large ni and nf is of very low frequency (long wavelengths.)
The only ones in the visible range (400 nm to 700 nm or so) are:
initial state
3
4
5
6
final state
2
2
2
2
∆E
(1/4 - 1/9) 2.180 x 10-18 J = 3.03 x 10-19 J
(1/4 - 1/16) 2.180 x 10-18 J = 4.06 x 10-19 J
(1/4 - 1/25) 2.180 x 10-18 J = 4.58 x 10-19 J
(1/4 - 1/36) 2.180 x 10-18 J = 4.84 x 10-19 J
λ
656.3 nm (red)
486.2 nm (blue)
434.2 nm (violet)
410.2 nm (far-violet)
Obviously, there is an infinite number of spectral lines, since there is an infinite number of pairs of
allowed electron states. It is convenient (arbitrarily) to group the lines by the final (lower energy)
state, called a “series”. For example, the visible lines in an emission spectrum are part of the
Balmer series, for which nf = 2; the Lyman series has nf = 1, the Paschen has nf = 3, and so on.
The range of photon energies is quite different in the different series. For the Balmer series,
for example, the photon of lowest energy (smallest ν, largest λ) comes from the transition ni = 3,
nf = 2; while that of largest energy (largest ν, smallest λ) is the transition in the limit ni → ∞, i.e.,
an unbound state, so it is the opposite of ionization. The limiting cases for each named series are
given below. I’ve given all the values in units of 10-19 J, to make comparison easier:
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
Lyman series:
largest ν: initial state = ∞ final state = 1
∆E = E∞ – E1 = –(1/∞ – 1/1) 2.180 x 10-18 J = (1) 2.180 x 10-18 J = 21.80 x 10-19 J
smallest ν: initial state = 2 final state = 1
∆E = E2 – E1 = –(1/4 - 1/1) 2.180 x 10-18 J = (3/4) 2.180 x 10-18 J = 16.35 x 10-19 J
Balmer series:
largest ν: initial state = ∞ final state = 2
∆E = E∞ – E2 = –(1/∞ – 1/4) 2.180 x 10-18 J = (1/4) 2.180 x 10-18 J = 5.45 x 10-19 J
smallest ν: initial state = 3 final state = 2
∆E = E3 – E2 = (1/9 - 1/4) 2.180 x 10-18 J = (5/36) 2.180 x 10-18 J = 3.03 x 10-19 J
Paschen series:
largest ν: initial state = ∞ final state = 3
∆E = E∞ – E3 = –(1/∞ – 1/9) 2.180 x 10-18 J = (1/9) 2.180 x 10-18 J = 2.42 x 10-19 J
smallest ν: initial state = 4 final state = 3
∆E = E4 – E3 = (1/16 - 1/9) 2.180 x 10-18 J = (7/144) 2.180 x 10-18 J = 1.06 x 10-19 J
Brackett series:
largest ν: initial state = ∞ final state = 4
∆E = E∞ – E4 = –(1/∞ – 1/16) 2.180 x 10-18 J = (1/16) 2.180 x 10-18 J = 1.36 x 10-19 J
smallest ν: initial state = 5 final state = 4
∆E = E5 – E4 = (1/25 - 1/16) 2.180 x 10-18 J = (9/400) 2.180 x 10-18 J = 0.491 x 10-19 J
Pfund series:
largest ν: initial state = ∞ final state = 5
∆E = E∞ – E5 = –(1/∞ – 1/25) 2.180 x 10-18 J = (1/16) 2.180 x 10-18 J = 0.872 x 10-19 J
smallest ν: initial state = 6 final state = 5
∆E = E6 – E5 = (1/36 - 1/25) 2.180 x 10-18 J = (11/900) 2.180 x 10-18 J = 0.266 x 10-19 J
Note that the lines in the Lyman series (in the ultraviolet) are widely separated from all the others,
and there is a smaller separation between the lines in the Balmer series (mostly in the visible) and
the Paschen series. But the Paschen, Brackett, and Pfund series overlap, as do all the remaining
(even lower frequency) series with higher final energies. This kind of overlap makes it difficult to
make sense of the emission spectrum!
25
Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
Absorption is the opposite of emission. Consider, for example, an electron initially in the state n =
2, with energy E2. It can only absorb photons of energy ε = E3 – E2, or ε = E4 – E2, or ε = E5 –
E2, ...; i.e., for any value of nf > ni. As nf → ∞, the energies of the photons become very
similar. If the photon has energy ≥ E∞ – E2 = –E2, then the electron leaves the atom, and the extra
energy from the photon beyond that needed to escape, [ε – (–E2)] in this case, is retained as the
electron’s kinetic energy, as in the photoelectric effect. Since there is no quantization of the
electron’s kinetic energy (assuming the electron is in a very large container), the absorption
spectrum becomes a continuous spectrum for ε > –E2.
Other one-electron atoms: cations with Z > 1
More generally, it is possible to obtain emission spectra from one-electron ions, such as He+,
Li2+, Be3+, etc., with nuclear charges +Ze (Z = 2, 3, ...). The frequencies of the lines in the
spectra are analogous to those for hydrogen, but with the frequencies multiplied by Z2. This leads
to the more general formula for the energies levels in a one-electron atom:
En = –2.180 x 10-18 J (Z2/n2)
for which hydrogen (Z = 1) is a special case.
Exercise: Show that the minimum energy needed to ionize a hydrogen atom with its electron
initially in the n = 2 state is exactly equal to the minimum energy that can be absorbed by a He+ ion
in its ground state (n = 1).
Now that we understand the relation between energy levels and emission and
absorption spectra, we need to address the question:
Why should the energies of one-electron atoms be limited to this discrete set of values?
In particular,
• Why do their “allowed” energies vary as the inverse square of a positive integer?
• Why do their “allowed” energies vary as the square of the nuclear charge?
• Why this particular numerical prefactor (2.18 x 10-18 J)?
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
E(r) for a one-electron atom: Rutherford
Let’s reexamine Rutherford’s model for a one electron atom. The electron feels only two
forces: the electrostatic attraction to the nucleus (of charge +Ze), and the centrifugal force in the
opposite direction. For a stable orbit, these two forces must balance, i.e., they must sum to zero.
Coulomb force: Recall that we derived the force from the Coulomb potential energy:
and thus
ECoul = Q1Q2/αr = (+Ze)(–e)/αr = – Ze2/αr
FCoul = – dECoul/dr.
FCoul = – Ze2/αr2
(1)
(2)
(3)
Centifugal force: From classical mechanics, the centrifugal force depends on the velocity (v)
and the mass (me) of the electron, and on the radius of its orbit (r):
Fcent = mev2/r
(4)
Setting these sum of these two forces (Eq. 3 and 4) to zero allows us to relate the velocity (really
the speed) of the electron to the radius of the orbit:
and thus
– Ze2/αr2 + mev2 = 0
(5)
r = Ze2/αmev2
(6)
v2 = Ze2/αmer
(7)
It will be useful to rewrite the kinetic energy in terms of r, rather than v; using Eq. (7) for v2:
or:
Ekin = mev2/2 = me(Ze2/αmer)/2 = Ze2/2αr
(8)
The total energy of the electron in the atom is the sum of its kinetic and potential contributions,
which, using Eqs. (1) and (8), can be written entirely as a function of the radius of the orbit, r:
Etotal = Ekin + Epot = Ze2/2αr – Ze2/αr = –Ze2/2αr
Note that
(9)
Epot = –2Ekin, so Etotal = –Ekin = (1/2)Epot
As the radius of the orbit decreases, the energy becomes more and more negative. As the
radius increases, the energy increases as well, with the limit of infinite radius (the electron escaping
from the nucleus = ionization) corresponding to the zero of energy. Note that as the radius
increases, the velocity decreases, since from Eq. (6), r is proportional to 1/v2.
Following classical physics, orbits of all radii should be allowed, and from Eq. (9), all
energies should thus be allowed. (We’ll ignore, for the moment, the problem of a charged particle
going in a circular path, which should radiate and lose energy, as discussed earlier.) A continuum
of possible energies would result in a continuous emission spectrum, as for a white light source,
but only discrete lines are observed from atoms, and as discussed above, for one-electron atoms,
these lines are consistent with electron energies restricted to En = –(2.180 x 10-18 J) Z2/n2, n =
1, 2, 3, ... So: we need some justification for restricting the “stable orbits” of the Rutherford
atom to only a discrete set with radii r such that –Ze2/2αr = Eelectron = –(2.180 x 10-18 J) Z2/n2.
27
Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
Niels Bohr (1913) Still uses the “orbit” idea, i.e., thinking about the electron as a particle:
- electrons only exist in circular orbits around a massive nucleus, with a well-defined radius,
the energy (or energy “level”) of the electron related the radius as in the above formula. This is
called a “state” of the electron.
- label states in order of increasing energy; electron can change from one state to another of
higher energy by absorbing a photon of energy exactly equal to the energy difference between the
initial and final states of the electron. Thus an electron in the ground state (the lowest energy state)
cannot emit light.
KEY ASSUMPTION: only those orbits are allowed in which the product of the linear
momentum (mv) and the circumference (2πr) is a multiple of Planck’s constant:
(mev)(2πr) = nh n = 1, 2, 3, ...
(10)
Put more elegantly, the angular momentum of the electron, mvr, is a multiple of h/2π
mevr = n(h/2π) = nh n = 1, 2, 3, ...
(11)
h/2π comes up so frequently, that it is given its own symbol: “h-bar” h
In other words, to account for the experimental emission spectra of one-electron atoms, electrons
must, for some as yet unknown reason, obey “quantization of angular momentum”.
Does it work?
Starting with mevr = nh/2π, so vr = nh/2πme, or
v2r2 = n2h2/4π2me2
(12)
From the relation between r and v for stable orbits: r = Ze2/αmev2 and thus
v2r = Ze2/αme
(13)
Dividing Eq. (12) by Eq. (13) gives
r = (n2h2/4π2me2) / (Ze2/αme)
r = n2h2α / 4π2meZe2 = (h2α / 4π2mee2) n2/Z
(14)
And now we can calculate the energy by substituting Eq. (14) into Eq. (9):
E = –Ze2/2αr = –(Ze2/2α) r-1
= –(Ze2/2α) [(4π2mee2 / h2α) (Z/n2)]
= –(2π2mee4/h2α2) Z2/n2
(15)
which, when I plug in all the constants, gives
E = –(2.180 x 10-18 J) Z2 /n 2
in precise agreement with the energy levels deduced from experimental emission spectra of oneelectron atoms.
What are the allowed values of r?
From Eq. (14), r = ro n2/Z, where
h2α / 4π2mee2 = ro = 0.529Å is the “Bohr radius”. So, for a hydrogen atom (Z = 1):
r1 = ro, r2 = 4ro, r3 = 9ro, etc. The radius increases as the square of the quantum number n.
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
But: there are still problems with the Bohr atom...
- Only works for one-electron atoms
- Why these assumptions? (why quantize angular momentum)
- and, there’s still the question of the stability of a charged particle moving in a circular orbit.
NEW IDEA: deBroglie (1924)
He thinks about light, which has zero mass, which can exhibit either wave-like (classical) or
particle-like (nonclassical; photons) behavior. So, he wonders if perhaps particles (which have
mass) might have the same duality of behavior: although our experience is usually with their
classical particle-like behavior, maybe they can act like waves, too; particularly for the very low
mass particles, like electrons. In other words, both light and matter can exhibit particle-like and
wave-like behavior. If so, he argues, electrons (and perhaps other, very light particles) ought to
have wave-like properties: they won’t be localized, but exist over some spatial extent; they will
have a wavelength, etc. They should, in principle, exhibit interference in a diffraction experiment!!
Electrons have momentum p = mv. Is it possible to determine, for example, their
“wavelength” from the momentum? deBroglie guesses the relation between p and λ for particles
with mass (like electrons) by analogy with light. The energy of a photon is ε = hc/λ. For light: ε
= pc. (This is the extreme relativistic limit; in the nonrelativistic limit, for particles with mass, E =
mv2/2 = pv/2.) So, pc = hc/λ, or
λ = h/p
If this is analogously valid for particles, then we can calculate the wavelength of a particle, like an
electron, as
λ = h/p = h/(mev)
What about electron “orbits” in atoms? He interprets the circular orbit as a circle on the entirety of
which the wave is defined, rather than as the trajectory of a point particle, as Bohr had envisioned.
Then, unlike the straight-line waves we have talked about, the wave interferes with itself. If I
start at a point on the circle, and go around the circle exactly once (i.e., a distance given by the
circumference = 2πr), I come back to exactly the same spot. If the circumference is an integral
multiple of the wavelength, then the wave function will have the same value at a given point in
space.
----- SHOW OXTOBY TRANSPARENCY ----If not, the wave function will have many different values at the same point in space, which would
make no sense. So, the condition for waves on a circular ring is
2πr = nλ
But, using deBroglie’s conjecture for the relation between the electron’s wavelength and its
momentum λ = h/mev, we get
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
2πr = n(h/mev)
which we can rewrite as
mvr = n(h/2π)
which is identical to Bohr’s assumption of quantization of angular momentum (Eq. 11). In other
words, Bohr’s assumption arises naturally if we assume that electrons have wave-like properties.
Consequences of wave behavior of particles:
Typical wavelengths:
Remember, Ekin = p2/2m, so p = (2mEkin)1/2. Thus λ = h/mv = h/p = h/(2mEkin)1/2.
Wavelength increases with decreasing mass, kinetic energy and velocity. Macroscopic objects
have such large masses, that they have extremely small wavelengths, which are much smaller than
their own size. However, objects of very low mass such as electrons can have very long
wavelengths, particularly at not too large velocities. Even neutrons, which have mass about 2000
times that of an electron, can have fairly large wavelengths (of order the size of an atom or larger)
if they have low Ekin. Since low Ekin corresponds to low temperatures, they are sometimes called
“cold neutrons” or “slow neutrons”.
(1) e- in a hydrogen atom with n = 1:
Remember, Ekin = p2/2m, so p = (2mEkin)1/2. For an electron in a hydrogen atom, we
showed (see line after Eq. 9) that Ekin = -2Epot = –Etotal, so Ekin = 2.18 x 10-18 J, and thus
λ = h/(2meEkin)1/2 ≈ 3.3Å = 0.33 nm
which is of the same order of magnitude as the size of the atom. (To do the unit conversion in the
above equation, remember that 1 J = 1 kg m2 s-2, so J1/2 = kg1/2 m s-1)
Remember (from Chem 3/5) how the average kinetic energy of molecules depends on T:
Eavg (per particle) ≈ (3/2)kBT, so T = (2/3)E/kB, where kB = R/No. Thus, typical energies
per particle at room temperature (300K) correspond to energies E ~ 0.040 eV; an energy of 1 eV
corresponds to typical particle energies at T = 7700K. Since Ekin = mv2/2, v is proportional to
Ekin1/2; since λ = h/(2meEkin)1/2, the particle wavelength is proportional to Ekin-1/2.
particle
energy (eV)
temp (K)
mass (kg)
velocity (m/s)
wavelength (nm)
electron
electron
electron
100
1.0
0.040
7.7 x 105
7700
300
9.11 x 10-31
9.11 x 10-31
9.11 x 10-31
5.9 x 106
5.9 x 105
1.2 x 105
0.12
1.2
6.0
neutron
neutron
neutron
100
1.0
0.040
7.7 x 105
7700
300
1.67 x 10-27
1.67 x 10-27
1.67 x 10-27
1.4 x 105
1.4 x 104
2.8 x 103
0.0029
0.029
0.14
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
Evidence of wave-like properties for particles with mass: DIFFRACTION
Davisson and Germer (1927): generate a well-collimated beam of particles (such as electrons)
which are filtered to have a very narrow range of velocities. If the relation between velocity and
wavelength is valid, then they should be “monochromatic”, i.e., all of the same wavelength,
should pass through a diffraction grating and create a diffraction pattern. The electrons should
obey the same constraint for constructive interference as for light, so electrons should be observed
only at angles: θ = sin-1(mλ/d), where λ = h/mev = h/(2meEkin)1/2
CAUTION!! There are two uses of the symbol “m” here: as the order of diffraction and the
particle mass. If you get them confused, it should be obvious since the units won’t make sense.
To get a reasonable separation between the predicted diffraction spots, the wavelength should
be of the same order of magnitude as the spacing d. Typical electrons emitted from radioactive
decay, which was the source of the electron beams, often have high energies (>> 1 eV),
corresponding to λ of order 0.1 nm or less. So, you need a diffraction grating with “slits” that are
only a few Å apart. Is it possible to construct such a grating?
Easy - use a crystal, in which the atoms are spaced in regular 3-dimensional arrays, with rows
only a few Å (a few tenths of a nm) apart. Davisson and Germer generated a collimated beam of
electrons of precisely known Ekin, which impinged on a piece of nickel metal (of accurately known
atomic spacing) and found that the electrons only arrived on the screen at angles predicted precisely
by the diffraction formula θ = sin-1(mλ/d).
In fact, diffraction of particles (neutron and electron), along with diffraction of light (usually xrays, i.e., at very high frequency) have become among the most powerful techniques used to
determine the structure of molecular and ionic crystalline solids.
Exercises:
1) At what angles would you expect to see diffraction spots, using a grating with d = 0.485 nm,
and a collimated beam containing electrons with two different kinetic energies: Ekin = 50 eV
and 100 eV?
2) Suppose you perform a diffraction experiment using a beam of electrons in which the smallest
angles at which you see electrons are 15° and 19°. Is the beam “monochromatic”?
3) Suppose you have a mixture of electrons and muons (a subatomic particle with the same charge
as an electron), all travelling at the same velocity in a collimated beam in a diffraction
experiment. The first order spot is observed at 42° for the electrons and at 0.185° for the
muons. Calculate the mass of the muon. (You don’t need to know the values of h, v, or d.)
Answers: (1) 14.6°, 21.0°, 30.4°, 45.7°, 49.3°
(2) no
31
(3) mµ = 207me = 1.89 x 10-28 kg
Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
Heisenberg Indeterminacy Principle
If a particle such as an electron behaves like a wave, then we expect to have difficulty saying
precisely “where” in space it is localized. Just how well can we localize a particle? While it may
be difficult to describe the location of an electron within a few Å in any direction, it is certainly
associated with one atom, and not with another a few meters away.
Using a variety of arguments, Heisenberg was able to demonstrate that if one measures
carefully the position of a particle (say, by bombarding it with other particles and looking at the
effect of collisions), one loses information about the momentum of the particle, and vice versa. In
fact, it was possible to show that the product of the indeterminacy (lack of knowledge) of the
position in, say, the x- direction (∆x) and the indeterminacy of the component of the momentum in
that direction (∆px) is always at least h/4π:
∆x ∆px ≥ h/4π
There are various ways of explaining this; I’ll use one that differs from that in your text (which is
the standard way.) It doesn’t give the value of the constant on the right hand side, but it shows
that as ∆x increases, ∆px decreases.
Suppose you have a light wave that extends very far in space along a line (although
eventually, since it must be in a box of some size, its amplitude must be attenuated at the ends.)
Then it is nearly perfectly sinusoidal and its wavelength can be determined to high accuracy by
finding the precise distance it must be shifted in the direction of propagation until the wave
superposes identically on itself. But, suppose you create a wave “packet” by attenuating the wave
after only a few wavelengths, as shown below, and you try to line up the wave with itself by
shifting to the right or left by one wavelength. It gets a bit imprecise, since you can’t line them up
perfectly, but it’s close.
Now, if I attenuate the wave much more severely, I get something which dies out after a distance
comparable to the wavelength, so superposing becomes very imprecise:
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
So, as you localize the wave in space, i.e., the position becomes more determined, the wavelength
becomes less well defined; and it is possible to show that it does so in inverse proportion: ∆x is
proportional to 1/∆λ. But the deBroglie relation says that λ determines the momentum px. So as
the indeterminacy ∆λ increases, the indeterminacy ∆px increases in proportion [from calculus, if p
= h/λ, then ∆p = –(h/λ2) ∆λ.] Thus, ∆x is inversely proportional to ∆px, or
∆x ∆px = constant.
Back to atoms:
In the Bohr atom, the electron was delocalized along the circular ring, but its radius was precisely
defined. Suppose we have an electron in the n=1 state of a hydrogen atom, with r = ro = 0.529Å,
i.e., we specify its radial position to 0.001 Å = 10-13 m. By the Heisenberg Principle:
∆p ≥ h/(4π∆x)
But ∆p = ∆v/m, so the indeterminacy of the velocity of the electron is
∆v = hme/(4π∆x) = 5.8 x 108 m/s
which is greater than the speed of light!! This clearly cannot be; it results from trying to localize
the particle in an orbit of precise radius.
So, it seems we have to throw out the idea of “orbits” (trajectories) completely;
the electron cannot correctly be viewed as localized precisely in any of the spatial dimensions.
Wave-particle duality. What else can we learn from our ideas about waves, in terms of the
interpretation for particles? For waves, the wave function (electric field) was not directly
measurable, but the square of the wave function was the intensity, which was measurable. For
particles, we need to define a “wave function” that can be superposed with other wave functions in
space, the square of which has a physical interpretation: the probability density of finding the
particle at that point in space.
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
Let’s let the greek letter psi (ψ) represent the “wave function” of the particle. Like the electric field
of a light wave it will have a numerical value (which, in analogy with light, can be positive or
negative) at each point in three-dimensional space: ψ(x,y,z). The sign will matter for
superposition of multiple wave functions; the part we can observe will be related to ψ2. In 1927,
Erwin Schrödinger postulated an equation (developed further by Heisenberg and Dirac) the
solutions to which describe all that can be known about the behavior of the particle:
–(h2/8π2m) [∂2ψ/dx2 + ∂2ψ/dy2 + ∂2ψ/dz2] + V(x,y,z) ψ = E ψ
where, remember, ψ(x,y,z) is a function, not a number. This equation is of the form Ekin + Epot
= Etotal: the first term on the left hand side represents Ekinetic, the second Epotential, and the right
hand side is the total energy E. Apologies for switching from Epot to V, but that’s the notation.
The term ∂2ψ/dx2 represents the second derivative of the function ψ with respect to x, i.e.,
you take the derivative of y to get a new function ∂ψ/∂x, and take the x-derivative of this new
function ∂ψ/∂x to get ∂2ψ/dx2. (For example, if ψ = x3, then dψ/dx = 3x2, and d2ψ/dx2 = 6x.)
V(r) is the potential energy experienced by the particle at a position in space (x, y, z), and E is the
energy (as yet undetermined).
Basically, this equation says the following: find a function ψ(x,y,z) such that when you take
its second derivative to get a new function, multiply by –(h2/8π2m) and then add the result the
product of V(x,y,z) and ψ(x,y,z), you get the original function multiplied by a number, the value
of which is the energy of the particle. So a solution to the Schrödinger Equation (S.E.) gives you
a specific function ψ(x,y,z) and an energy E of the particle in a state specified by that function.
For many potentials, such as the Coulomb interaction between an electron (charge -e) and the
nucleus (charge +Ze) in a one-electron atom, all that matters is the distance r between the two
particles, so we can write V(x,y,z) = V(r) = –Ze2/αr. We will be interested in the solutions to this
equation for one-electron atoms, then multi-electron atoms, and ultimately, molecules. All these
are 3-D problems, and thus involve some pretty involved mathematical techniques.
To illustrate how one solves the S.E. and interprets the solutions, let’s use the simplest
example - the motion of a particle in a one-dimensional “box”. (Oxtoby does this well: see pp.
547-550.) By this, we mean the particle is constrained to reside on a line of length L (on the xaxis, from x = 0 to x = L), where there are no forces on the particle (F(x) = 0) and thus V(x) = 0
for 0 < x < L. Outside of this line, the particle is not allowed to exist - the potential energy is
infinitely large: V(x) = ∞ for x < 0 or x > L. The S.E. now has a much simpler form:
–(h2/8π2m) ∂2ψ/dx2 = E ψ
This equation can be “solved” mathematically to give a set of wave functions and associated
energies. I’ll walk you through it:
What function, when you take its derivative twice, gives you back the original function, but
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Lecture Notes: Quantum Mechanics and Atomic Structure (2)
Chem 6 Spring '00
with the opposite sign? Only sines and cosines. So we expect that the wave functions might be
ψ(x) = a sin(2πx/λ + c)
where λ represents the wavelength (as yet undetermined, as are the coefficients a and c).
Boundary Conditions: if ψ2 is to be interpreted as a probability, then ψ(x) must be zero outside of
the box. In fact, ψ(x) must be a continuous function of x, so at the edges (walls) of the box it
must also be zero: ψ(0) = ψ(L) = 0. These are called “boundary conditions”, because they
specify the value of the wave function at the boundaries of the box. The only way to guarantee this
condition is if the sine wave goes to zero at these points. To make ψ(0) = 0, use pure sine waves
(not cosine), or equivalently, set c = 0 in the above equation: ψ(x) = a sin(2πx/λ)
Now to satisfy the boundary condition ψ(L) = 0, make sure that the length of the box
corresponds either to half a wavelength, a full wavelength, 3/2 wavelengths, and so on, i.e.,
L = n(λ/2), where n = 1, 2, 3, ... is a positive integer (NOT ZERO!!).
Thus 1/λ = n/2L, and the wave function becomes
ψn(x) = a sin(2πx/λ) = a sin(nπx/L)
Note that the finite size of the box creates boundary conditions that are the source of quantization:
only a discrete (enumerable) set of wavelengths is possible; I’ve labeled the function with the
subscript n to indicate that. The energy corresponding to this wave function will also be labeled
En. The only thing we haven’t figured out yet is “a”, the amplitude of the wave function.
---- SHOW OXTOBY TRANSPARENCY ---Plug this function back into the S.E.:
–(h2/8π2m) ∂2ψn/dx2 = En ψn
Take the second derivative: ∂2ψn/dx2 = –(nπ/L)2 a sin(nπx/L) = –(n2π2/L2) ψn(x)
and thus the SE becomes: (h2/8π2m) (n2π2/L2) ψn(x) = En ψn(x)
The functions are the same on both sides (as they must be), so the coefficients must be equal. Thus
E n = h2 n 2 /8mL 2 , n = 1, 2, 3, ...
are the allowed energy levels for the motion of the particle in a one-dimensional box. The energies
(and wave functions) are quantized in the sense that there is a discrete set. The integer n is called a
quantum number. Note that for this problem in which there is one spatial variable (x), only one
quantum number is needed to specify the solution to the SE. The solution is a stable state of the
particle, described entirely by a wave function ψn(x) and its energy “in that state”, En.
Incidentally: deBroglie’s relation p = h/λ can be predicted from the S.E. for this case:
If E = h2n2/8mL2, and E = Ekin (since V(x) = 0 in this case) = p2/2m, then p2/2m = h2n2/8mL2
and thus p2 = h2n2/4L2, and thus p = hn/2L. But the boundary condition for the particle in the
box is 2L = nλ, so
p = hn/2L = hn/nλ, or p = h/λ .
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