Download The Bohr Model of the Hydrogen Atom

The Bohr Model of the Hydrogen Atom
“A physicist is just an atom’s way of looking at itself.” Niels Bohr.
The classical picture of the atom as a tiny solar system with the nucleus like the sun and
electrons like planets is known to be unstable. In 1913 Niels Bohr proposed a semiclassical
model of the hydrogen atom that was stable and predicted observational effects that were
in agreement with experimental observations. Although Bohr’s model was superseded by
the Schrödinger equation a decade later, it still provides a useful picture of the atom that
captures many of the essential properties (such as the emission and absorption of photons)
without adding extra complexities such as orbital patterns. In this set of notes we examine
the Bohr model of the atom, including its motivation, derivation, and ultimately its failure.
The correct quantum mechanical picture of the atom from Schrödinger’s equation is discussed
in a later set of notes.
1
Observations of the Hydrogen Atom.
By the end of the nineteenth century, scientists seemed to have a pretty good idea of how
nature worked. It was during this century that Maxwell unified electricity and magnetism
with light in his theory of electromagnetism, and the theory of heat embodied in statistical
mechanics and thermodynamics was worked out by luminaries such as Boltzmann (and
Maxwell). However, there were still some nagging problems that physicists of the day were
unable to solve, such as the Blackbody spectrum and theory of specific heats.
A major research area of the time was spectroscopy, which studied how light was emitted
or absorbed by materials. In the 1860s Gustav Kirchhoff (working with Robert Bunsen)
formulated his three laws of spectroscopy,
1. A hot solid, liquid, or dense gas (like a blackbody) will emit a full range of radiation;
one obtains a continuous spectrum.
2. A hot diffuse gas emits a discrete series of wavelengths, giving a emission spectrum.
3. A cool gas in front of a continuous source will absorb specific, discrete wavelengths,
yielding an absorption spectrum.
A continuous spectrum would basically be a rainbow - all the different wavelengths of light
(although some could be brighter than others). An image of an emission spectrum for the
simplest element, hydrogen, is seen in Fig. 1 [1], while the spectrum from the Sun, showing
absorption lines, is seen in Fig. 2 [2]. We see that, in the emission case, we see that there is
basically no light at all, except for some very specific colors, and each element (or molecule)
has its own unique emission spectrum. This uniqueness means that a spectrum can serve as
a fingerprint, for example identifying elements in stars, based the emitted or absorbed light.
This is, in fact, what we see in Fig. 2. In this case, the continuous spectrum of light emitted
from the core of the Sun works it way out through the outer atmosphere of the Sun where the
atmospheric gas (mostly hydrogen and helium) absorb specific wavelengths of light, leaving
dark absorption lines in the spectrum. By comparing these specific wavelengths of light to
experimentally-determined values, we can identify the components of a star’s atmosphere
(in fact, helium was discovered in 1868 in just this way).
1
Figure 1: The hydrogen spectrum [1]. Hydrogen, when heated or electrified, emits a very
particular spectrum of visible light with discrete and specific wavelengths, which we see
as specific colors. There are a number of other spectral lines that are in the infrared and
ultraviolet region, that we don’t see here.
1.1
The Hydrogen Spectrum.
By the beginning of the 20th century quite a variety of spectra had been observed, but
the origin of these spectra constituted a major theoretical problem. The question of why
elements only emit a discrete series of lines, and more specifically, what values those lines
should take for a given element remained unanswered. For example, the wavelengths in Fig.
1 were measured and are, from left to right, 410 nm, 434 nm, 486 nm, and 656 nm. Any
good model of the hydrogen atom should be able to calculate these wavelengths1 .
The emission lines in Fig. 1 are only those in the visible range of the spectrum. By
making measurements beyond the visible spectrum (for example, in the infrared or ultraviolet
ranges), it has been found that there are many more emission lines. A modern table of some
observed wavelengths is given in Table 1, where the meaning of ni and nf will be explained
below2 .
Lyman
(nf = 1)
ni = 1
ni = 2
ni = 3
ni = 4
ni = 5
ni = 6
ni = ∞
121.567 nm
102.572 nm
97.2537 nm
94.9743 nm
93.7803 nm
91.2∗ nm
Balmer
(nf = 2)
121.567 nm
656.464 nm
486.270 nm
434.169 nm
410.290 nm
365.5∗ nm
Paschen
Brackett
Pfund
Humphreys
(nf = 3)
(nf = 4)
(nf = 5)
(nf = 6)
102.572 nm 97.2537 nm 94.9743 nm 93.7803 nm
656.464 nm 486.270 nm 434.169 nm 410.290 nm
1874.56 nm 1281.44 nm 1093.50 nm
1874.56 nm
4049.98 nm 2624.38 nm
1281.44 nm 4049.98 nm
7462∗ nm
1093.50 nm 2624.38 nm 7462∗ nm
820∗ nm
1458∗ nm
2278∗ nm
3281∗ nm
Table 1: The observed wavelengths of light emitted from hydrogen [3]. Only the wavelengths
in the second column between 400 . λ . 700 nm correspond to the visible part of the
spectrum.
Each column (called a series) is named after the scientist that first discovered the series;
for example, the set of ultraviolet wavelengths in the first column is called the Lyman series,
after Theodore Lyman. The Balmer series is particularly important since four of these
1
Of course, at this time it was not conclusively known that atoms even existed!
The entries marked with a * were not included in the data table provided by [3], and so their values here
were calculated.
2
2
Figure 2: The solar absorption spectrum [2]. Being a dense hot glowing ball of gas, the sun
emits a continuous spectrum. However, gases in the outer layers of the sun absorb discrete
wavelengths, leading to missing lines in the spectrum. These missing lines can then be used
to identify those gases responsible for the absorption.
wavelengths are in the visible spectrum, as seen in Fig. 1, and were the first of the hydrogen
spectra to be found (as would be expected).
Table 1 comprises a wide range of specific wavelengths, mostly outside the visible spectrum, and there are actually many more not listed here3 , and it seems like quite a complicated
mess to accurately predict all of these numbers. Despite their best efforts, scientists were
unable to find a good theory to describe these wavelengths, but in 1888 a Swedish physicist
named Johannes Rydberg discovered an empirical formula that gave the right numbers. Rydberg found that the wavelength, λ, corresponding to a particular line obeyed a very simple
result,
!
1
1
1
= RH
− 2 ,
(1)
λ
n2f
ni
where RH was an empirical constant that Rydberg used to get the correct answer. It’s
experimental value is now known to be
RH = 1.09737315685 × 107 m−1 .
(2)
Rydberg found that if he took nf = 2 and ni = 3, 4, 5, or 6, then he could correctly predict
the Balmer series wavelengths (which were the only ones known at the time). It was later
3
The listed wavelengths are not exactly correct because of tiny effects having to do with magnetic interactions between the proton and electron, as well as relativistic effects.
3
understood that taking different combinations of integers would correctly give the other
series; however, the theoretical model underlying Eq. (1) would have to wait until a deeper
understanding of the atom was found.
2
Rutherford’s Experiment.
The concept of the atom is often said to have arisen with the Greeks; in the 5th century
BC the Greek philosophers Democritus and Leucippus conceived of the “indivisible” bits of
matter, but these vaguely scientific ideas have little to do with the modern picture of an
atom. In the early 1800s John Dalton revived the atomic idea to help understand chemical properties, before J.J. Thompson (who also discovered the electron) invented his plum
pudding model in 1904.
Thompson’s idea was that the atom was like a “plum pudding,” or a slightly more familiar
analog would be a cookie dough, in which the positive charge was distributed throughout
the atom (like the raw dough), while the negative electrons were embedded in the positive
charge (like the chocolate chips), leaving an overall neutral atom. It then fell to physicists
to test this model.
In a series of experiments from 1908-1910 Ernest Rutherford led Hans Geiger and Ernest
Marsden in performing what have come to be known as the Geiger-Marsden experiments (or
also the “Gold-Foil” experiments). Rutherford had already won the Nobel Prize in Chemistry
for his discovery of the concept of radioactive half-life, as well as his investigations into alpha
and beta radiation. He and his assistants set up the experimental apparatus diagrammed in
Fig. 3.
Figure 3: Rutherford’s experiment [4]. A beam of α particles (helium nuclei) are emitted
from a radon source and allowed to strike a gold foil. The angle of scattering of the alpha
particles from the foil is measured by a fluorescent screen that flashed whenever a particle
would impact it.
The Geiger-Marsden experiments were a series of measurements of the scattering angles of
alpha particles from different sorts of atoms (in particular, gold). Essentially, the experiments
involved the firing of alpha particles (helium nuclei) at a thin gold foil. The alpha particles
4
were emitted from a radon source which was known to be an alpha emitter, and sent down
a long tube to keep them coming out straight. The particles then impacted a thin gold foil;
most of the alpha particles went though the foil, only deflected through a very small angle
(which is what one would expect from Thompson’s model), but a some of the particles were
scattered at much larger angles - a few (about 1/8000) even rebounded back! Rutherford
was amazed at this discovery, saying “It was quite the most incredible event that has ever
happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a
piece of tissue paper and it came back and hit you.”
Rutherford’s amazement was because this large-angle scattering is completely unexpected
for Thompson’s model, but exactly what one would expect if the the atom was mostly empty
space, but the positive charge was concentrated in a tiny “nugget” at the center of the atom.
We can see the difference is scattering in Fig. 4. In the top illustration we see a beam
of alpha particles firing on a gold atom in the cookie-dough model; the orange part of the
gold atom represents the positive charge distribution, while the negative charges are the
purple bits. The positively-charged alpha particles are deflected through small angles while
they are passing through the atom, much like a BB would hardly deflect when fired through
cookie dough. Thompson’s model predicts a small angular distribution for the alpha particle
scattering.
Figure 4: The comparison of scattering of alpha particles by gold atoms for Thompson’s
model (top image), versus the results from experiment (bottom image). Thompson’s model
predicts the alpha particles are scattered through only small angles, while observations give
some large-angle results.
5
The situation is different in the second illustration,
however. In this case the positive charge of the gold
atom is concentrated in the center in a “nucleus,”
while the negative charge is somewhere around the
positive charge. The fact that most of the alpha particles pass straight through the atom tells us that it
is mostly empty space, suggesting that the electrons
are far away from the nucleus, and Rutherford suggested that they orbited the nucleus giving rise to the
familiar “solar system” model of the atom seen in the
figure to the right.
In the solar system model of the atom the positively-charged nucleus sits at the center
of the atom (like the sun in the solar system), while the negatively-charged electrons orbit
(like planets). The atom seen in the figure would correspond to lithium, with three each
protons, neutrons, and electrons, although the neutron wasn’t discovered until more than
twenty years after these experiments.
3
Instability of the “Solar System” Model.
Rutherford’s solar system model of the atom is nice - it’s pretty, and there is an aesthetic
pleasure in imagining that the very large and very small worlds reflect each other. There is
one small problem, however: this atom is completely unstable! The instability arises from
the fact that, as the electron goes around in its orbit, it’s accelerating, and accelerating
charges radiate energy (this is the basis for radio transmitters, for example). Since the
electron radiates energy (in the form of electromagnetic radiation - light, in other words) it
loses the same amount of energy, which can only come from its kinetic energy as it revolves
around the nucleus.
Since the electron loses kinetic energy, it
slows down, meaning that it’s not moving fast
enough to keep from falling in to the nucleus
- thus, the electron should spiral down into
the nucleus, giving off light, as seen in the figure to the right. This means that every atom
should have its electrons sitting down in the
nucleus, and that chemistry (which depends
on the interaction of electrons between atoms)
could not exist - in other words, no chemistry,
no biology, no Universe, as we know it!
If the lifetime of the atom was long, like the a hundred times the age of the Universe, then we
wouldn’t care about this instability. However it turns out that the lifetime is much shorter
than this, as we can check, now.
We will work out the lifetime for hydrogen, with the general case being straightforward.
According to classical electrodynamics (see Appendix A), the Larmor formula gives the
6
power, P , radiated by the electron when it has an acceleration a,
P =−
2 qe2
a2 ,
3 4π0 c3
(3)
where qe is the charge on the electron, 0 is the permittivity of free space, and c is the speed
of light. What we are looking for is how long the electron takes to spiral down from some
initial radius, r0 , to the center of the atom, say r = 0, and so we really need the radius as a
function of time4 . With this in mind we can write
P =
dE dr
dE
=
× .
dt
dr
dt
Now, we can determine both the energy and acceleration of the electron. The electric force
acts on the electron giving an acceleration a = FE /m, or
a=
qe2
,
4π0 mr2
and since the centripetal acceleration is a = mv 2 /r, we have a velocity
s
qe2
.
v=
4π0 mr
The energy is the kinetic energy plus the potential energy,
1
q2
q2
E = mv 2 − e = − e ,
2
4π0 r
8π0 r
where we have plugged in for the velocity in terms of the radius from above and combined the
kinetic and potential energies. Note that the total energy is negative, which means that the
electron and proton are bound together. We can now finally go back to the power expression,
first finding the derivative of the energy with respect to radius
q2
dE
= + e 2,
dr
8π0 r
and hence
2
qe2 dr
2 qe2
qe2
.
=−
8π0 r2 dt
3 4π0 c3 4π0 mr2
Solving for the derivative gives the rate of in-spiral of the electron,
dr
1
qe4
K
=−
≡ − 2,
2 2 3
2
dt
12π0 m c r
r
where the combination of constants
qe4
K≡
= 3.1744 × 10−21 m3 /s = 3.1744 × 109 Å3 /s,
12π20 m2 c3
4
Throughout this set of notes we ignore the motion of the much heavier nucleus. Including this effect
very slightly changes some numbers around the third decimal place, or so.
7
where we have expressed our final answer in terms of angstroms, since that’s the typical size
of atoms. So, all we have to do is separate our differential equation and integrate, from a
radius r0 at time t = 0 to radius r = 0 at time τ , which we take to be the lifetime of the
atom,
Z τ
Z
1 0 2
r3
dt = −
r dr ⇒ τ = 0 .
K r0
3K
0
It turns out that the orbital radius of the electron in its ground state (the closest orbital) is
r0 ≡ aB = 0.529 Å,
where aB is called the Bohr radius, and is derived below. So, plugging in this radius gives
τ = 1.556 × 10−11 seconds!
So, all hydrogen atoms should have collapsed in about a trillionth of a second after they were
formed! It turns out that starting with a different atom, or putting the electrons further
away (in a higher orbital) doesn’t make much difference, and so every atom should have
collapsed billions of years ago! Thus, atoms are a classical impossibility!
4
Bohr’s Model.
Clearly, Rutherford’s model of the atom needs some work. Fortunately, Niels Bohr formulated a new theory, only a few years later in 1913. Bohr understood that the solar system
model had problems with stability, and he also knew of both Planck’s work regarding the
discrete quanta of the electromagnetic field and Einstein’s interpretation of quanta as photons. Bohr was able to alter Rutherford’s model slightly, while incorporating the idea of
photons, in a way that ensured the stability of the atom. Bohr’s was a semiclassical theory,
that propelled forward the “‘quantum revolution,” where the phrase “semiclassical” will be
explained below.
The main idea of Bohr’s theory is that the electrons can only orbit at certain discrete
distances (and hence, with a very definite energy), and not anywhere that they want. This
means that the electrons can’t spiral down into the nucleus, since that would mean that
they would not follow those orbits. This doesn’t mean that the electrons are forever trapped
at one radius, however; electrons can make a “jump” from one orbit to another essentially
instantaneously. The electrons are forced to jump to a higher orbit if it absorbs just the
right amount of energy (in the form of light) to raise its energy up to equal that necessary to
orbit in the next level. In the same way, an electron can jump down to a lower orbit (if it’s
not in the ground state, which is the lowest-energy state) by emitting just the right amount
of energy.
Thus, Bohr postulates the following assumptions [5]:
1. That the dynamical equilibrium of the systems in the stationary states can be discussed
by help of the ordinary mechanics, while the passing of the systems between different
stationary states cannot be treated on that basis.
8
2. That the latter is followed by the emission of a homogeneous radiation, for which the
relation between the frequency and the amount of energy emitted is the one given by
Planck’s theory.
The first assumption says that classical mechanics can let us understand the stable orbits of
the electrons, but not how they jump from one orbit to another. The second assumption says
that the discrete quanta idea of Planck (later understood in terms of photons by Einstein)
is needed to understand the emission spectra produced by those orbital jumps. This is what
we mean by Bohr’s model being semiclassical.
We now want to work out the consequences of Bohr’s theory. The starting point is the
assumption that the electrons can only follow specific orbits; Bohr realized that this would
result if electrons could have only certain (quantized) values of angular momenta5 . To make
circumstances easy, we imagine that the electron of mass m orbits the nucleus in a circular
orbit of radius r at velocity v (elliptical orbits could be taken, but they don’t add anything
to the understanding). Neglecting any possible angular momentum that the electron might
have due to its own rotation (it’s spin angular momentum), the magnitude of the orbital
angular momentum is L = mvr. According to Bohr, this value is quantized, meaning that
can only come in certain discrete units. Bohr found that it was quantized in terms of Planck’s
constant, h, divided by 2π. In other words,
L = mvr = n
h
,
2π
where n = 1, 2, 3, · · · is an integer called the principal quantum number. The quantity h/2π
shows up all over the place in physics, and so it has been given it’s own symbol, ~ (called
“h-bar”),
h
~≡
(4)
= 1.054571726 × 10−34 J s,
2π
and is called the reduced Planck’s constant 6 . Thus, Bohr’s quantization condition, from
which all else follows, is that
L = mvr = n~.
(5)
Now that we have Eq. (5) we can find the allowed orbital radii, orbital velocities, and
energies. From our work above, we know that the velocity of the electron is
s
qe2
v=
,
4π0 r
and combining this expression with Eq. (5) gives the allowed orbital radii, rn , which depend
on n,
4π0 ~2 2
rn =
n ≡ aB n2 ,
(6)
mqe2
5
Bohr wrote several papers on the atomic spectra, and he didn’t actually assume quantization of angular
momentum in his first paper - he used Planck and Einstein’s ideas and derived it as a consequence (in his
second paper it was taken as a postulate, however).
6
Usually, Eq. (4) is just called Planck’s constant; h doesn’t show up, by itself, all that often!
9
where the Bohr radius, aB is given by
4π0 ~2
aB ≡
= 0.52917721092 × 10−10 m,
2
mqe
(7)
or aB ≈ 0.529 Å. So, the first thing we see is that the atom has the diameter of about an
angstrom - so far, so good! The next thing that catches our eye about Eq. (6) is that the
orbits depend on n, meaning that the smallest orbit (corresponding to the ground state),
when n = 1 is at the Bohr radius, while the next orbit (when n = 2) is four times bigger,
and so on. A single proton can never have an electron orbiting at 1.5aB ! Since the orbits
are quantized, this ensures their stability.
We can figure out how fast the electron is orbiting by taking Eq. (6) and plugging it
back into our expression for the velocity. Doing so yields
qe2 1
,
vn =
4π0 ~ n
which, again, depends on n. Thus, the velocities are also quantized; the electron in the
ground state (n = 1) is twice as fast as one in the n = 2 state (called the first excited state).
The velocity is often given as a fraction of the speed of light, so we can write
vn
qe2 1
α
=
≡ ,
c
4π0 ~c n
n
(8)
where α is called the fine structure constant, or electromagnetic coupling constant, and has
tremendous importance for physics, particularly quantum electrodynamics. The dimensionless constant has a value of α = 7.2973525698 × 10−3 , but no one ever remembers this value;
instead physicists remember it in the much simpler form
α=
1
1
≈
.
137.035999074
137
(9)
This constant will be useful in later work for us, but for right now we simply note that
the speed of the electron in the ground state is v1 = αc ≈ c/137, which is fast, but not so
fast that we have to worry about relativistic effects! This means that we’re okay in using
Newtonian physics to work out our analysis.
Finally, we can now work out the energy of the electron in its orbit, which we have already
worked out to be
q2
E=− e .
8π0 r
Plugging in for the radius from Eq. (6) gives us the allowed, quantized, energies,
En = −
mqe4 1
.
32π 2 20 ~2 n2
10
(10)
The mess of constants in the energy is often grouped together and called one Rydberg of
energy (after Rydberg, whom we met above), and has the value 2.197872 × 10−18 Joules, or
13.60569 electron volts (eV),
Ry =
mqe4
≈ 13.6 eV.
32π 2 20 ~2
(11)
With this we can write the energy in the very simple form
En = −
13.6
eV.
n2
(12)
Once again, we find that the total energy of the hydrogen atom is negative, which is
correct for a bound state, and En is a measure of how hard the proton is holding on to the
electron. We also find that the energy depends on the integer n, such that E1 = −13.6
eV, E2 = −3.4 eV, etc. The decrease of energy with increasing n is easy to understand increasing n means that the electron is in a higher orbital (further away from the proton), and
the hydrogen atom isn’t as tightly bound as it would be if the electron were in a lower orbit.
In the limit that n → ∞, then the energy goes to zero, representing an unbound, or ionized,
hydrogen atom - the electron has been completely removed. Thus, we can understand the
Rydberg of energy in another way; it’s the amount of energy needed to ionize the hydrogen
atom when it’s in its ground state.
The interesting thing about Eq. (12) is that these values represent the only energies that
the atom can have! Because the energies depend on the quantized radii, the energies are
also quantized, meaning that we can never find a hydrogen atom with -5 eV of energy. This
is the key that lets us understand the hydrogen spectra!
4.1
Solving the Hydrogen Spectra
We are now finally in a position to completely understand the origin of the discrete spectrum
that started this whole process! The central idea is Bohr’s second postulate about the
emission of radiation, and the quantized energies in Eq. (10). Suppose that we have an
electron that starts off in the n = 2 state, with energy E3 = −1.51 eV, and it makes a
transition down to the n = 2 state, with energy E2 = −3.4 eV. The electron loses 1.89 eV
of energy when it makes that transition in the form of a photon streaming away. Einstein
told us that the energy of a photon is related to its frequency, ν, by
E = hν,
(13)
where h is Planck’s constant. We can solve Eq. (13) for the wavelength of the light using
λν = c
hc
~c
λ=
= 2π ,
E
E
where we have rewritten the wavelength in terms of ~ to show a useful trick for calculations.
A handy number to remember is that
~c = 197.327 eV nm ≈ 200 eV nm.
11
(14)
So,
197
≈ 655 nm,
λ3→2 ≈ 2π
1.89
which is almost exactly the wavelength that we see in Table 1! As we’ll see, this idea will
work for all the different transitions!
We can now work out the wavelength for a general transition from an initial state ni to
a final state nf . The difference in energy between the two states is given by
!
1
1
∆E = −13.6
− 2 eV.
n2f
ni
If we start off in a higher orbital and then transition down to a lower orbital, then nf < ni ,
giving a net negative energy change. This means that energy left the atom, carried away by
an emitted photon. On the other hand, a photon could come in with just the right energy
and get absorbed by the electron. The electron would then use that extra energy to jump
up to a higher orbital, where nf > ni , leading to a net positive energy change. In one case
the photon of a definite energy (and so, therefore, wavelength) is emitted, while in the other
a definite-wavelength photon is absorbed. This is the qualitative origin of the emission and
absorption spectra! It also explains why the values in Table 1 match along the diagonal.
All we have to do now is calculate the photon wavelengths and hope that we get the right
numbers. Once again, ∆E = 2π~c/λ, and so
~c
.
∆E
Inverting this relation to find λ−1 gives (ignoring the minus sign in front of the 13.6 eV)
!
1
∆E
13.6
1
1
=
=
− 2 ,
λ
2π~c
2π~c n2f
ni
λ = 2π
which looks an awfully lot like Eq. (1), we just have to plug in the numbers to check. Doing
so gives
13.6
≈ 0.010969 nm−1 = 1.0969 × 10−7 m−1 ,
2π~c
which is essentially Eq. (2)! So, based on Bohr’s theory, we can pretty much say we
understand the hydrogen atom! But do we, really?
4.2
Extension to Other Atoms
Bohr’s model can be extended to other one-electron atoms (say singly-ionized helium, or
doubly-ionized lithium, etc.), obtaining extremely good agreement with observed spectra.
The extension is extremely straightforward, and is simply due to the fact that the nucleus
of the new atom has a different charge. If we call the charge on the nucleus Zqe , where
the atomic number Z tells us how many protons are present, then every factor of qe2 in our
formulae simply changes to Zqe2 . Thus, the allowed radii in Eq. (6) become
rn =
4π0 ~2 2 aB 2
n =
n,
mZqe2
Z
12
(15)
while the velocities are
vn
Zqe2 1
Zα
=
=
,
c
4π0 ~c n
n
(16)
with energy
En = −
mZ 2 qe4 1
13.6Z 2
=
−
eV.
32π 2 20 ~2 n2
n2
(17)
The effect of the increased charge in the nucleus changes each of these values in ways that
are easy to understand. The radius shrinks because the electron is more strongly attracted
to the nucleus, but its velocity has to increase to keep from falling in. Finally, since the
energy of the electron depends on the square of the velocity (which depends linearly on Z),
the energy depends on Z 2 .
As an example, suppose we consider singly-ionized helium (often called He II), for which
Z = 2. We would expect to find the electron four times closer than in the case of hydrogen,
at r1 = 0.25aB , and also moving four times faster. Crucially, we can calculate some numbers
that we can compare to experiment. The ionization energy of He II should be 13.60569×22 =
54.4228 eV, which we compare with an experimental value [6] of 54.41776311 eV, showing
excellent agreement. Finally, there is a visible-light transition from 4 → 3 in He II that
gives a calculated wavelength of λ = 468.848 nm, which we compare with experiment [7]7
λtheo = 486.5 nm, which is (again) an excellent agreement. Similar checks can be made for
doubly-ionized lithium (Li III), and so on.
5
The Failure of the Bohr Model.
The Bohr model of the atom seems to be extremely good at describing the atom. It has so
far successfully explained all of the different observed wavelengths of hydrogen (at least, at
the level of precision to which we are working), and even other ions, such as He II. However,
we were careful to say that this model could be extended to other single-electron atoms,
which is not the natural state for atoms. Suppose we consider neutral helium (He I), with
two electrons. Now the energy is made up of five terms: the kinetic energies of each electron,
plus the attraction of each electron for the nucleus, plus the electrical energy between the
two electrons!
1
1 2qe2
1 2qe2
1
qe2
1
−
+
.
EHe = mv12 + mv22 −
2
2
4π0 r1
4π0 r2
4π0 |~r1 − ~r2 |
It is this last term that really complicates the analysis, and makes it basically impossible to
do exactly. We can still figure out a few things, though. Experimentally, the ground state
energy of helium is found to be −78.975 eV, while the ionization energy (the amount of
energy needed to remove one electron) is 24.587 eV. Clearly, once we remove one electron
7
Note that there are five different wavelengths listed for the transition 4 → 3. This has to do with the fact
that the correct theory of atoms (which is actually the Dirac equation, and not the Schrödinger equation)
tells us that the atomic energies actually depend on other factors, such as the spin of the electron, that we
have neglected. We take for our experimental value just one decimal place (for which all the wavelengths
agree), but a more complete theory should include these complications.
13
then the Bohr model works again and we expect that it would take another 54.42 eV to
remove the last electron (as we calculated above). We can see that 78.975 − 24.587 = 54.39
eV, which is extremely close. So, we can understand some things about helium, but where
did the first two numbers come from?
It turns out that the Bohr model is incapable of correctly describing any atom with more
than one electron, and, in fact, fails to describe even hydrogen when pushed any harder. The
problem with Bohr’s model is that there is too much classical physics in it, and nature isn’t
classical. Despite the numerical successes of the Bohr model, there are a number of issues
with it. For example, it is found that if the hydrogen atom is placed in a magnetic field,
then additional spectral lines are seen. Or, if very careful measurements are made spectra
that look like single lines could actually be two very slightly different lines. Furthermore,
there’s no understanding of why hydrogen tends to be diatomic. Bohr’s model can’t explain
these observations, and the more scientists learned about the atom, the less correct Bohr’s
model looked8 . We could go on forever explaining what else this model doesn’t work for, but
it’s enough simply to say that the correct (so far as we know) answer, quantum mechanics,
required a completely new way of thinking, incredibly different than classical physics, and
was more than a decade off from Bohr’s work. We will discuss quantum mechanics soon.
Appendix
A
Derivation of Power from a Radiating Charge.
In this appendix we want to work out an expression for the power radiated by an accelerating
point charge, which is the Larmor formula given in Eq. (3). The correct derivation from
electrodynamics for a charge moving at any velocity (even relativistically) is rather involved
and more complicated than we need. Instead, we will limit ourselves to a non-relativistic
particle and use a trick due to J.J. Thompson.
Suppose we have a point charge, q, sitting at rest and imagine the electric field lines
radiating outward from it. For a stationary point charge the field lines are radial outward
and static, radiating no power (there is no magnetic field for a stationary particle and so the
Poynting vector would be zero). In this case, the electric field is given by Coulomb’s law,
~ =
E
1 q
r̂.
4π0 r2
Now, suppose that we instantaneously (or at least very very quickly) move the particle to
the right for a time dt, as seen in the figure below. The field lines must be continuous, but
they can get kinks 9 .
8
Despite this, he Nobel Prize in Physics 1922 was awarded to Niels Bohr ”for his services in the investigation of the structure of atoms and of the radiation emanating from them”. Even though his model wasn’t
completely correct, he got the ball rolling!
9
There is an extremely good simulation of a radiating charge on the PhET website, http://phet.
colorado.edu/en/simulation/radiating-charge, where you can get a very good idea about what the
field lines look like as the charge moves.
14
In this figure, we imagine that the the charge
has field lines (seen in the dashed green lines)
which spread out for some long time. Then,
the charge is accelerated (with a constant acceleration a) for a time dt, picking up a change
in velocity dv = adt, and is shown in the red
arrow. The field lines spread out from the new
position, but since the light travels at a finite
speed, c, the new field lines only reach a certain radial distance before they have to match
back up to the old field lines originating from
the old position of the charge. Before and after the acceleration, the field lines are radial
and unchanging, but during the acceleration
the field lines kink and so are changing - this
changing electric field induces a changing magnetic field leading to a radiated power.
t
θ
E
cd
θ
cT
We can find this changing electric field as follows. We zoom in on the changing fields, seen
to the right. The charge accelerates for a time
dt, picking up a speed dv, and then moves at
a constant speed for a time T . The field radiates our at some angle, θ, from the original
position of the charge for a time T , reaching
a distance of r = cT . During that time the
charge moves a distance dvT , given by the red
arrow. Because of the acceleration, the field
kinks through a distance cdt, which gives a
tangential component of the field Eθ , in addition to the radial component, Er .
E
dvTsinθ
Er
dvTsinθ
θ
dvT
θ
From the geometry we can see that the ratio of the components of the field, Eθ /Er is the
same as the ratio dvT sin θ/cdt, or, since a = dv/dt,
T
ar
Eθ
= a sin θ = 2 sin θ,
Er
c
c
where we have noted that T = r/c.
15
Since the radial field is just the Coulomb field
we find that the tangential field is
Eθ =
q a
sin θ.
4π0 c2 r
(A-1)
Notice that the field falls away inversely as the
first power of the distance, not as the square of
the distance; this is what tells us this field corresponds to radiation with an emitted power.
The angular dependence of the field is seen in
the plot to the right and has a sort of donut
shape coming from the sine of the angle, as
well as an azimuthal symmetry, since nothing
depends on φ.
For electromagnetic radiation flowing in the direction k̂ = ~k/k, where ~k is the wave
vector, then the magnetic field is at right angles to both the electric field, and the direction
of flow of the radiation,
~
~ = k̂ × E,
B
c
giving a Poynting vector
!
1
1
k̂
~= E
~ ×B
~ =
~ = 1 E 2 k̂ = 0 cE 2 k̂,
S
×
×E
µ0
µ0
c
µ0 c
where we have recalled that the speed of light c−2 = µ0 0 . The intensity, which is the amount
of radiated energy passing through a unit area per second (that is, the power per area), is
given by the magnitude of the Poynting vector. So, for the field in Eq. (A-1) we find
2
q a
q2
a2
dP
= 0 c
sin
θ
=
sin2 θ.
I=
dA
4π0 c2 r
16π 2 0 c3 r2
This is the power radiated out through a small area, dA, in other words, how much power
passes through a small set of angles dθ and dφ. To determine the total power radiated
through all angles we just need to integrate the power, recalling that dA = r2 sin θdθdφ,
Z 2π Z π
Z π
Z
q 2 a2
q 2 a2
3
P = IdA =
dφ
sin θ dθ =
sin3 θ dθ.
3
16π 2 0 c3 0
8π
c
0
0
0
The integral over θ has a value of 4/3, which is easily seen by first writing sin3 θ = (1 − cos2 θ) sin θ,
expanding in two integrals, and then making a u-substitution of u = cos θ in the second integral. So, we finally find the Larmor formula, Eq. (3) (we don’t have a negative sign since
we only are interested in the magnitude of the power radiated),
P =
2 q2
a2 .
3 4π0 c3
16
(A-2)
References
[1] The hydrogen spectrum picture may be found on the Wikipedia website, http://en.
wikipedia.org/wiki/Hydrogen_spectral_series
[2] The solar absorption spectrum picture may be found on the NASA picture of the day
website, http://apod.nasa.gov/apod/ap000815.html
[3] W. L. Wiese and J. R. Fuhr, “Accurate Atomic Transition Probabilities for Hydrogen,
Helium, and Lithium,” J. Phys. Chem. Ref. Data 38, 565 (2009), http://www.nist.
gov/srd/upload/jpcrd382009565p.pdf.
[4] Giancoli, D. C. (2008). Physics for scientists and engineers with modern physics. Upper
Saddle River, N.J., Pearson Education.
[5] N. Bohr, “On the Constitution of Atoms and Molecules,” Phil. Mag. 26, 1 (1913).
[6] The helium ionization energy may be found on the NIST website, http://physics.
nist.gov/cgi-bin/ASD/ie.pl
[7] The helium transition wavelengths may be found on the NIST website, http://
physics.nist.gov/PhysRefData/Handbook/Tables/heliumtable4.htm
17