Chapter 11

The Bohr Atom

Topics

The experiments of Thomson and Rutherford. The radiation problem and Bohr’s
postulates. De Broglie waves and Bohr’s quantisation condition, experiments of
Davisson and Germer and G.P. Thomson, particles as waves and the wave-particle
duality. The Bohr model of the atom and the Balmer formula. Successes and failures.

11.1 The Experiments of Thomson and

Rutherford

The next step is the application of the ideas of

quanta to the understanding of atomic structure.
This was the great achievement of Niels Bohr, but
it relied heavily upon the discoveries of the elec-
tron by J.J. Thomson and of the nuclear structure
of atoms by Ernest Rutherford.

11.1.1 J.J. Thomson and the Electron

The electron was discovered by J.J. Thomson in

1897 and the experiment by which he established
that its mass-to-charge ratio is only about one two
thousandth of that of hydrogen ions was carried
out in the course Fields, Oscillations and Waves.
The electron was the first sub-atomic particle to
be discovered. As Thomson stated in his discovery
paper,

1
The Bohr Atom 2

‘. . . [the cathode rays constituted] a new

state, in which the subdivision of mat-
ter is carried very much further than in
the ordinary gaseous state.’

In summary, Thomson’s great achievements were:

• The demonstration that the cathode rays, which

became known as electrons, were the particles
which carried an electric current, were iden-
tical to the particles emitted in β decay and
those emitted in the photoelectric effect.

• The demonstration, with Barkla, from X-ray

scattering experiments that the number of
electrons in an atom is about half its atomic
weight, except for hydrogen. The number of
electrons and, consequently, the amount of
positive charge in the atom increases by units
of the electronic charge.

• The invention of the ‘plum-pudding’ model

of the atom, designed to minimise the loss of
energy by electrons in orbit within the atom
(see Section 11.2.1).

11.1.2 Rutherford and the Nuclear Atom

Rutherford had been very impressed by the fact

that fast α-particles, which are emitted in radioac-
tive decays and which are helium nuclei with mass
almost four times that of the hydrogen atom, could
pass through thin films rather easily, suggesting
that much of the volume of the atoms is empty
space. Furthermore, he found that some of the α-
particles were deflected through very large angles
indeed and a few of them were sent back more or
less in the direction of incidence of the α-particles.
Rutherford realised that it required a very consid-
erable force to send the α-particle back along its
track – the α-particles were travelling at 10,000 km
s−1 . As he stated,

‘It was quite the most remarkable event

that has ever happened to me in my life.
It was almost as if you fire a fifteen-inch
The Bohr Atom 3

shell at a piece of tissue paper and it

came back and hit you.’

His guess was that all the positive charge was con-
tained in a compact nucleus and that it was the
electrostatic repulsion of the positively charged α-
particle by the positively charged nucleus which
was the origin of the repulsive force. The α-particle
scattering experiments, which he carried out with
Marsden and Geiger in 1910-11, showed that the
predicted scattering law, that the probability of
scattering through an angle φ per unit solid angle
is
1 φ
N (φ) ∝ 4 cosec4 (11.1)
v0 2
was precisely followed even for very large deflec-
tions. This is the famous cosec4 (φ/2) law which
Rutherford derived theoretically – this form of scat-
tering is known as Rutherford scattering.
They had, however, achieved much more. The fact
that the scattering law was so precisely obeyed,
even for large angles of scattering, meant that the
inverse square law held good to very small distances
indeed. An upper limit to the size of the nucleus
could be found from the maximum angle for which
the cosec4 (φ/2) law holds good. They found that
the nucleus had to have size less than about 10−14
m. This is very much less than the sizes of atoms,
which are typically about 10−10 m.

11.2 The Radiation Problem and Bohr’s

Postulates
Figure 11.1. The geometry of Rutherford
scattering of an α-particle by a positively
The first successful application of the concept of
charged nucleus. The α-particle follows a
quantisation to atoms was made by the Danish
hyperbolic path with the nucleus in the outer
physicist Niels Bohr in 1913. Once Rutherford had
focus.
demonstrated that atoms consist of a compact nu-
cleus with a positive charge surrounded by a cloud
of electrons, each with negative charge e, it was
natural to assume that the negatively charged elec-
trons moved in bound orbits about the nucleus.
There was, however, one very profound problem
with this picture, according to classical physics.
The Bohr Atom 4

An accelerated charged particle radiates electro- The Radiation Problem

magnetic waves. It is straightforward to show that, According to classical electrodynamics, an
in the case of the hydrogen atom, as the electron accelerated electron loses energy at a rate
loses energy, it moves into an orbit of smaller ra- µ ¶
dius, loses energy more rapidly and spirals into the dE e2 a2
− = = 5.7 × 10−54 a2 W,
nucleus within about 10−10 seconds. This cannot dt 6π²0 c3
be correct since atoms certainly exist.
where a is the acceleration of the elec-
This problem was solved by Bohr in an remarkable tron. For an electron orbiting a hydro-
leap of the imagination. He realised that to solve gen nucleus, for example, at a radius r =
the problem he had to adopt the concept of quan- 0.5 × 1010 m, you may wish to show that
tisation as expounded by Planck and Einstein. He the centipetal acceleration is a ≈ 1023 m
noted the key point of their work, that only a finite s−2 . The time it takes the electron to lose
number of energy states are allowed and not the in- its energy is therefore roughly
finite number allowed according to classical physics.
E
He therefore introduced the concept of stationary T ≈ ¯¯ ¯ ≈ 4 × 10−11 second.
dE ¯
states within the atom, which corresponded to the ¯ ¯
¯ dt ¯
quantised energy states which were introduced by
Planck and Einstein to account for the spectrum
of black-body radiation. This postulate abolishes
the problem of the radiation loss of the electrons
because they are not allowed to assume a conti-
nous set of energy states – they can only have the
energies associated with the stationary states. He
postulated that the electron could only lose energy
by radiation by making transitions between the sta-
tionary states - the concept of quantum jumps. This
was the process responsible for the origin of spec-
tral lines.
The postulates which Bohr made in order to apply
the concept of quanta to atoms were as follows:

1. Each electron in an atom can only exist in

certain stable circular orbits of definite energy
which are known as stationary states. These
energy levels have energies E1 , E2 , E3 , . . . The quantisation of angular momen-
tum according to the old quantum
2. The angular momentum J of an electron in
theory
a stationary state is quantised such that the
only allowed values are those for which h
J = me vn rn = n = nh̄
h 2π
J = me vn rn = n = nh̄ (11.2) where n can only take integral values, n =
2π
1, 2, 3, . . . .
where n can only take integral values, n =
1, 2, 3, . . . . We have introduced the quantity
h̄ = h/2π, which is called ‘h-bar’. This quan-
The Bohr Atom 5

tity turns up ubiquitously in quantum me-

chanics and you should feel equally at home
with either h or h̄.
3. Transitions are possible between different sta-
tionary states and, when the electron makes
a transition between those with energies Ei
and Ef , a photon is emitted with energy
The energy of a photon emitted in a
E = hν = Ei − Ef (11.3) transition between stationary states

Notice that the key step is the introduction of the E = hν = Ei − Ef

quantisation of angular momentum, postulate 2.
This is not quite as arbitrary as it might seem since
the dimensions of Planck’s constant h are
[h] = J s = [M ][L]2 [T ]−2 × [T ]
= [M ][L][T ]−1 [L] = [mvr],
that is, the dimensions of h are those of angular
momentum.

11.3 De Broglie Waves and Bohr’s Quan-

tisation Condition

Bohr’s argument for quantising angular momentum

originated from the need to quantise the energy lev-
els in the atom. In fact, the rule for quantising an-
gular momentum can be derived from the inspired
guess made by Louis-Victor de Broglie in his doc-
toral dissertation of 1924. He suggested that, just
as light waves have particle properties, so parti-
cles have wave properties. There was no experi-
mental evidence for this hypothesis, but de Broglie
showed how Bohr’s quantisation rules could be de-
rived from this hypothesis.
Einstein had shown that the light-quantum, or pho-
ton, has energy E = hν and momentum hν/c. De
Broglie made the hypothesis that, exactly as in the
case of photons, the electron has wave properties
such that its wavelength is
h
λ= (11.4)
p
where p is the momentum of the electron. Vectori-
ally, p = (h/λ)i, where i is the unit vector in the
The Bohr Atom 6

direction of propagation of the electron. The wave-

length λ is known as the de Broglie wavelength of
the electron.
In the theory of black-body radiation, we required
the wavelengths of permissible modes of oscillation
within a box to be such that only a finite number
of half wavelengths fit across any of its dimensions.
De Broglie applied a similar condition to the waves
associated with the electrons in their orbits about
the nucleus. He argued that the stationary states
of the electrons in the atom should be such that
there is an integral number of wavelengths around
the orbit. His reasoning was that, unless there is
a finite number of complete wavelengths round the
orbit, the waves would interfere destructively. His
picture is illustrated schematically in Figure 11.2
which show the quantisation of the de Broglie waves
for the n = 3 and n = 6 stationary states.
Adopting de Broglie’s hypothesis,
nh nh
nλ = 2πrn = 2πrn = 2πrn . (11.5)
p me vn
Therefore,
Figure 11.2. Illustrating De Broglie’s
h
Ln = mn vn rn = n = nh̄. (11.6) construction for the quantisation of angular
2π
momentum states in hydrogen-like atoms.
This is precisely Bohr’s condition for the quantisa-
tion of angular momentum.
A copy of de Broglie’s dissertation reached Ein-
stein, who immediately appreciated its deep signif-
icance. Through Einstein, the Austrian theoretical
physicist Erwin Schrödinger learned of the idea of
de Broglie waves and, from it, developed the the- The de Broglie wavelength of a
ory of wave mechanics, one of the first successful particle of momentum p
realisations of quantum mechanics.
h
λ=
Evidence for the wave nature of the electron was p
only discovered in 1927 in two classic experiments.
The Davisson-Germer Experiment Clinton Davis-
son studied the elastic scattering of electrons by
metal surfaces. The term elastic means that the
electrons are reflected from the surface of the ma-
terial without loss of energy. He found that the
intensity of reflected electrons varied in a peculiar
The Bohr Atom 7

fashion with the angle of reflection. The key ex-

periment was carried out using a carefully cleaned
nickel crystal.
If the incident beam of electrons has wavelength λ,
we can use the standard procedures of wave reflec-
tion by a crystal lattice to work out the intensity
of scattered electrons in different directions. Fig-
ure 11.3 shows the construction for Bragg reflec-
tion from a crystal surface. We adopt the simplest
picture of a regular cubic lattice and consider the
interference of the reflected beam from the surface
layer of the crystal. Constructive interference of
the reflected waves is obtained when the path dif-
ference between the waves from neighbouring sites
is a whole number of wavelengths

d sin θ = nλ n = 1, 2, 3, . . . (11.7)

In the experiment carried out by Davisson and Lester Figure 11.3. Illustrating Bragg reflection
Germer, a pronounced maximum was observed at from the surface of a regular lattice.
an angle of 50◦ for electrons of kinetic energy 54
eV. The momentum of a 54 eV electron is related
to its kinetic energy by E = p2 /2me and so

p = (2me E)1/2 = 3.97 × 10−24 kg m s−1 . (11.8)

This can be compared with the momentum inferred

from de Broglie’s hypothesis. The separation be-
tween the atoms in the crystal structure was known
to be 0.215 nm and so, taking n = 1, the de Broglie
wavelength of the electron must be λ = d sin θ =
0.165 nm. Therefore, its momentum was inferred
to be

p = h/λ = 4.02 × 10−24 kg m s−1 , (11.9)

in other words, essentially perfect agreement.

Figure 11.4. Illustrating the results of

The G.P. Thomson Experiment George Thom- Davisson and Germer’s electron scattering
son was the son of J.J. Thomson. At almost exactly experiments from cleaned crystal surfaces.
the same time that Davisson and Germer completed
their experiments, Thomson observed diffraction
rings in the scattering of 15 keV electrons through
a thin metal foil. The pattern of the rings was of
exactly the same form as would have been expected
The Bohr Atom 8

if the electrons had wavelength λ = h/p. Examples

of the diffraction patterns due to the scattering of
X-rays and electrons of the same momenta by a
target of powdered aluminium are shown in Figure
11.5.
It has been remarked that

Thomson, the father, was awarded the

Nobel prize for having shown that the
electron is a particle, and Thomson, the Figure 11.5. The diffraction patterns
son, for having shown that the electron produced by X-rays (left) and electrons
is a wave. (right) of the same momenta on passing
through a target of powdered aluminium.
These experiments were conclusive evidence that
electrons have wave properties. Similar experiments
are now carried out routinely using α particles, neu-
trons and so on.
The fact that particles have wave properties is the
second aspect of the wave-particle duality. A new
formulation of the fundamental laws of physics was
needed which could encompass both aspects of par-
ticles and waves and we will show how this can be
done in the next chapter.

11.4 The Bohr Model of the Atom

Bohr adopted his three postulates in order to de-

termine the stationary states of the hydrogen atom.
The quantisation condition for angular momentum
of the stationary state with quantum number n is

me vn rn = nh̄ (11.10)

Equating the centripetal acceleration to the accel-

eration due to the electrostatic force between the
nucleus and the electron for circular orbits,

me vn2 e2
= (11.11)
rn 4π²0 rn2
e2
me vn2 = (11.12)
4π²0 rn
From the relations (11.10) and (11.12), we find ex-
pressions for the radii and speeds of the electrons
The Bohr Atom 9

in their stationary states

µ 2¶ µ ¶
2 4π²0 h̄ 2 ²0 h 2
rn = n =n (11.13)
m e e2 πme e2
µ 2 ¶ µ 2 ¶
1 e 1 e
vn = = (11.14)
n 4π²0 h̄ n 2²0 h

We can therefore find the radius a0 of the orbit of

the ground state of the hydrogen atom, n = 1. a0
is called the Bohr radius.
4π²0 h̄2
a0 = = 0.529 × 10−10 m (11.15)
me e2
This is a very convenient unit for atomic sizes. The
speed of the electron in its ground state, n = 1, is
e2
v1 = = 2.2 × 106 m s−1 , (11.16)
2²0 h
only about one hundredth of the speed of light. We
do not need to bother about relativistic corrections
in the first approximation.
We can now find the energies of the stationary
states of the electrons in the hydrogen atom. We
recall that, for circular orbits, the kinetic energy of
the electron is 12 me v 2 and the electrostatic poten-
tial energy is −e2 /4π²0 r. Therefore, remembering
that the orbit is in centripetal balance,
me v 2 e2
= , (11.17)
r 4π²0 r2
the total energy of the orbit is
1 e2 e2
E = me v 2 − =− (11.18)
2 4π²0 r 8π²0 r
Thus, for the nth stationary state, since rn = n2 (²0 h2 /πme e2 ),

e2 1 me e4
En = − =− 2 2 2 (11.19)
8π²0 rn n 8²0 h
When the electron makes a transition from the sta-
tionary state m to n with m > n, the energy of the
emitted photon is
µ ¶
me e4 1 1
E = hν = Em − En = 2 2 − .
8²0 h n2 m2
(11.20)
The Bohr Atom 10

Note that the final state has the more negative en-
ergy. Therefore, the frequencies of the lines in the
hydrogen spectrum are expected to be
µ ¶
m e e4 1 1
ν= 2 3 − (11.21)
8²0 h n 2 m2 The Balmer series of hydrogen
µ ¶
In the case of the Balmer series, the lines originate me e4 1 1
ν= 2 3 −
from transitions from energy levels with m > 2 into 8²0 h 22 m2
the n = 2 energy level and then
µ ¶
m e e4 1 1
ν= 2 3 − . (11.22)
8²0 h 22 m2

Substituting for the values of the constants, we find

µ ¶
15 1 1
ν = 3.29 × 10 − . (11.23)
22 m2

Miraculously, this is precisely the expression for the

Balmer series of hydrogen. Historical note. One pleasant discovery
included in Bohr’s paper of 1913 was the
The expression for the Balmer series is often writ-
explanation of the Pickering series which
ten in terms of wavelengths rather than frequencies
had been discovered in astronomical spec-
and then it becomes
tra by the astronomer Edward Pickering.
µ ¶
1 1 1 The Pickering series is a series of singly
= R∞ − (11.24)
λ 22 m2 ionised helium observed in the spectra of
hot stars. Since the nuclear charge is
where R∞ is known as the Rydberg constant and twice that of hydrogen, the spectrum is
has value R∞ = 1.097 × 107 m−1 . The subscript shifted to four times shorter wavelengths
∞ means that it is assumed that the mass of the as compared with that of hydrogen. The
nucleus is infinite, as has been assumed in our cal- Pickering series corresponds to transitions
culation. into the n = 4 energy level, similar to
If we had applied the rules of quantisation of angu- the Bracket series of hydrogen but at four
lar momentum to the system consisting of both the times shorter wavelengths.
electron and the proton, we would have written

h̄n = me ve re + mN vN rN (11.25)

with the electron and nucleus orbiting their com-

mon centre of mass. It is a useful exercise to show
that for this case,
mN R me R
re = and rN = (11.26)
me + mN me + mN
where R = re + rN is the distance between the
electron and the nucleus. If the angular velocity
of the electron and the nucleus about their centre
The Bohr Atom 11

of mass is ω, the condition for the quantisation of

angular momentum becomes Historical note The necessity of using
the reduced mass was important histori-
h̄n = µωR2 (11.27) cally. Bohr argued that singly-ionised he-
where µ = me mN /(me + mN ) is known as the re- lium atoms would have exactly the same
duced mass of the system. This is left as a revi- spectrum as hydrogen, but the wave-
sion exercise in mechanics. The stationary states lengths of the corresponding lines would
are found from the corresponding energy equation be four times shorter, as observed in the
with the kinetic energy being given by 21 Iω 2 = −En Pickering series. Fowler objected, how-
where I = µR2 . ever, that the ratio of the Rydberg con-
stants for singly-ionised helium and hydro-
The energy level diagram for the hydrogen atom gen was not 4, but 4.00163. Bohr realised
is shown in Figure 11.6. The various energy lev- that the problem arose from neglecting the
els are shown with their appropriate names. The contribution of the mass of the nucleus to
transitions associated with the series are known as the computation of the moments of iner-
follows: tia of the hydrogen atom and the helium
ion. When the reduced masses of hydro-
• The Lyman series: n = 2, 3, 4, . . . to m = 1 gen and ionised helium are used, the ratio
• The Paschen series: n = 4, 5, 6, . . . to m = 3 of the Rydberg constants is 4.00160.
In his biography of Bohr, Pais tells the
• The Brackett series: n = 5, 6, 7, . . . to m = 4 story of Hevesy’s encounter with Einstein
in September 1913. When Einstein heard
• The Pfund series: n = 6, 7, 8, . . . to m = 5
of Bohr’s analysis of the Balmer series
of hydrogen, Einstein remarked cautiously
The energy levels can be written in the form
that Bohr’s work was very interesting, and
1 m e e4 13.6 important if right. When Hevesy told
En = − 2 2 2
= − 2 eV (11.28) him about the helium results, Einstein re-
n 8²0 h n
sponded,
where n = 1, 2, 3, . . . is known as the principal
quantum number. This expression indicates how ‘This is an enormous achieve-
much energy is necessary to remove an electron ment. The theory of Bohr
from any stationary state of the atom. In par- must then be right.’
ticular, to ionise an atom, that is, to remove an
electron from the ground state, n = 1 to E = 0, an
energy E1 = 13.6 eV is required, corresponding to
the transition from n = 1 to n = ∞.
The Bohr theory of the hydrogen atom was a quite
remarkable achievement and it indicated clearly that
it is essential to incorporate quantum concepts into
the theory of atomic structure. It was apparent,
however, that it was an incomplete theory. This
theory, often called the old quantum theory, was
an uneasy mixture of classical and quantum con-
cepts without any proper theoretical underpinning.
There were also a number of basic problems which
remained unsolved.
The Bohr Atom 12

1. It was not at all clear why the electrons in the

atoms did not radiate. Bohr had to suppose
that the classical laws of electromagnetic ra-
diation did not apply on the atomic scale.

2. The theory does not allow us to calculate

transition rates from one state to another,
nor does it tell us whether or not all possible
transitions between energy levels are allowed.

3. The theory worked extremely well for hydro-

gen and other one electron ions, such as ionised
helium, but could not account for the spectra
of other atoms.

4. The theory could give no account of the bond-

ing between molecules.

5. Many of the spectral lines are actually dou-

blets, that is, two very closely spaced lines,
for example, the D-lines of sodium.

The period from 1913 to 1925 was one in which the Figure 11.6. The term diagram for hydrogen
fundamentals of classical physics had been thor- showing the first four series.
oughly undermined by discoveries which necessi-
tated the introduction of the concept of quanta
and yet no satisfactory quantum theory was avail-
able. The Bohr model was patched up in vari-
ous ways but these were ad hoc adjustments to a
fundamentally flawed theory. It was the discov-
ery of wave and quantum mechanics in 1925-6 by
Schrödinger, Heisenberg and their colleagues which
finally rewrote the fundamental laws of physics.