ONE-ELECTRON ATOMS: SPECTRAL PATTERNS Late …chem6/9/QMAtomsLectureNotes2.pdf · ONE-ELECTRON...

Lecture Notes: Quantum Mechanics and Atomic Structure (2) Chem 6 Spring '00

ONE-ELECTRON ATOMS: SPECTRAL PATTERNSLate 19th century: Balmer, Rydberg, Lyman, Paschen, and others determine the emission spectraof hydrogen atoms. They find a wide range of spectral lines, but they notice a pattern in thefrequencies 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 findan emission, and for no other frequencies.

Simplest analysis: since for emission, ε = Einitial – Efinal, the experimental data suggest thatthe 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 closertogether as n increases. As a result, there is a wide range of energy differences, and thus the rangeof spectral lines is very wide. The resulting emission spectrum of the Hydrogen atom (includingall transitions of the atom) looks like this:

---- SHOW PORILE TRANSPARENCY ----The spacings between energy levels becomes very closely packed, so the light absorbed or emittedin 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 final state ∆E λ3 2 (1/4 - 1/9) 2.180 x 10-18 J = 3.03 x 10-19 J 656.3 nm (red)

4 2 (1/4 - 1/16) 2.180 x 10-18 J = 4.06 x 10-19 J 486.2 nm (blue)

5 2 (1/4 - 1/25) 2.180 x 10-18 J = 4.58 x 10-19 J 434.2 nm (violet)

6 2 (1/4 - 1/36) 2.180 x 10-18 J = 4.84 x 10-19 J 410.2 nm (far-violet)

Obviously, there is an infinite number of spectral lines, since there is an infinite number of pairs ofallowed 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 theBalmer 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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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:


∆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


∆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:


∆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


∆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:


∆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


∆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:


∆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


∆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!

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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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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:

ECoul = Q1Q2/αr = (+Ze)(–e)/αr = – Ze2/αr (1)

FCoul = – dECoul/dr. (2)

and thus FCoul = – Ze2/αr2 (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:

– Ze2/αr2 + mev2 = 0 (5)

and thus r = Ze2/αmev2 (6)

or: 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:

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 (9)

Note that 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.

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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/2πmevr = 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/n2

in precise agreement with the energy levels deduced from experimental emission spectra of one-

electron 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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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 = nλBut, using deBroglie’s conjecture for the relation between the electron’s wavelength and its

momentum λ = h/mev, we get

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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 100 7.7 x 105 9.11 x 10-31 5.9 x 106 0.12

electron 1.0 7700 9.11 x 10-31 5.9 x 105 1.2

electron 0.040 300 9.11 x 10-31 1.2 x 105 6.0

neutron 100 7.7 x 105 1.67 x 10-27 1.4 x 105 0.0029

neutron 1.0 7700 1.67 x 10-27 1.4 x 104 0.029

neutron 0.040 300 1.67 x 10-27 2.8 x 103 0.14

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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 x-

rays, 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 (3) mµ = 207me = 1.89 x 10-28 kg

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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/4πThere 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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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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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 x-

axis, 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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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

En = h2n2/8mL2, 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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