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12.1 INTRODUCTION
By the nineteenth century, enough evidence had accumulated in favour of
atomic hypothesis of matter. In 1897, the experiments on electric discharge
through gases carried out by the English physicist J. J. Thomson (1856 –
1940) revealed that atoms of different elements contain negatively charged
constituents (electrons) that are identical for all atoms. However, atoms on a
whole are electrically neutral. Therefore, an atom must also contain some
positive charge to neutralise the negative charge of the electrons. But what
is the arrangement of the positive charge and the electrons inside the atom?
In other words, what is the structure of an atom?
The first model of atom was proposed by J. J. Thomson in 1898.
According to this model, the positive charge of the atom is uniformly
distributed throughout the volume of the atom and the negatively charged
electrons are embedded in it like seeds in a watermelon. This model was
picturesquely called plum pudding model of the atom. However
subsequent studies on atoms, as described in this chapter, showed that
the distribution of the electrons and positive charges are very different
from that proposed in this model.
We know that condensed matter (solids and liquids) and dense gases at
all temperatures emit electromagnetic radiation in which a continuous
distribution of several wavelengths is present, though with different
intensities. This radiation is considered to be due to oscillations of atoms
Chapter Twelve
ATOMS
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Atoms
and molecules, governed by the interaction of each atom or
molecule with its neighbours. In contrast, light emitted from
rarefied gases heated in a flame, or excited electrically in a
glow tube such as the familiar neon sign or mercury vapour
light has only certain discrete wavelengths. The spectrum
appears as a series of bright lines. In such gases, the
average spacing between atoms is large. Hence, the
radiation emitted can be considered due to individual atoms
rather than because of interactions between atoms or
molecules.
In the early nineteenth century it was also established
that each element is associated with a characteristic
spectrum of radiation, for example, hydrogen always gives
a set of lines with fixed relative position between the lines.
This fact suggested an intimate relationship between the
internal structure of an atom and the spectrum of
radiation emitted by it. In 1885, Johann Jakob Balmer
(1825 – 1898) obtained a simple empirical formula which
gave the wavelengths of a group of lines emitted by atomic
hydrogen. Since hydrogen is simplest of the elements
known, we shall consider its spectrum in detail in this
chapter.
Ernst Rutherford (1871–1937), a former research
student of J. J. Thomson, was engaged in experiments on
α-particles emitted by some radioactive elements. In 1906,
he proposed a classic experiment of scattering of these
α-particles by atoms to investigate the atomic structure.
This experiment was later performed around 1911 by Hans
Geiger (1882–1945) and Ernst Marsden (1889–1970, who
was 20 year-old student and had not yet earned his
bachelor’s degree). The details are discussed in Section
12.2. The explanation of the results led to the birth of
Rutherford’s planetary model of atom (also called the
nuclear model of the atom). According to this the entire
positive charge and most of the mass of the atom is
concentrated in a small volume called the nucleus with electrons revolving
around the nucleus just as planets revolve around the sun.
Rutherford’s nuclear model was a major step towards how we see
the atom today. However, it could not explain why atoms emit light of
only discrete wavelengths. How could an atom as simple as hydrogen,
consisting of a single electron and a single proton, emit a complex
spectrum of specific wavelengths? In the classical picture of an atom, the
electron revolves round the nucleus much like the way a planet revolves
round the sun. However, we shall see that there are some serious
difficulties in accepting such a model.
12.2 ALPHA-PARTICLE SCATTERING AND
RUTHERFORD’S NUCLEAR MODEL OF ATOM
At the suggestion of Ernst Rutherford, in 1911, H. Geiger and E. Marsden
performed some experiments. In one of their experiments, as shown in
Ernst Rutherford (1871 –
1937) New Zealand born,
British physicist who did
pioneering work on
radioactive radiation. He
discovered alpha-rays and
beta-rays. Along with
Federick Soddy, he created
the modern theory of
radioactivity. He studied
the ‘emanation’ of thorium
and discovered a new noble
gas, an isotope of radon,
now known as thoron. By
scattering alpha-rays from
the metal foils, he
discovered the atomic
nucleus and proposed the
plenatery model of the
atom. He also estimated the
approximate size of the
nucleus.
ERNST RUTHERFORD (1871 – 1937)
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Fig. 12.1, they directed a beam of
5.5 MeV α-particles emitted from a
214
83
Bi
radioactive source at a thin metal
foil made of gold. Figure 12.2 shows a
schematic diagram of this experiment.
Alpha-particles emitted by a
214
83
Bi
radioactive source were collimated into
a narrow beam by their passage
through lead bricks. The beam was
allowed to fall on a thin foil of gold of
thickness 2.1 × 10
–7
m. The scattered
alpha-particles were observed through
a rotatable detector consisting of zinc
sulphide screen and a microscope. The
scattered alpha-particles on striking
the screen produced brief light flashes
or scintillations. These flashes may be
viewed through a microscope and the
distribution of the number of scattered
particles may be studied as a function
of angle of scattering.
FIGURE 12.2 Schematic arrangement of the Geiger-Marsden experiment.
A typical graph of the total number of α-particles scattered at different
angles, in a given interval of time, is shown in Fig. 12.3. The dots in this
figure represent the data points and the solid curve is the theoretical
prediction based on the assumption that the target atom has a small,
dense, positively charged nucleus. Many of the α-particles pass through
the foil. It means that they do not suffer any collisions. Only about 0.14%
of the incident α-particles scatter by more than 1°; and about 1 in 8000
deflect by more than 90°. Rutherford argued that, to deflect the α-particle
backwards, it must experience a large repulsive force. This force could
FIGURE 12.1 Geiger-Marsden scattering experiment.
The entire apparatus is placed in a vacuum chamber
(not shown in this figure).
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be provided if the greater part of the
mass of the atom and its positive charge
were concentrated tightly at its centre.
Then the incoming α-particle could get
very close to the positive charge without
penetrating it, and such a close
encounter would result in a large
deflection. This agreement supported
the hypothesis of the nuclear atom. This
is why Rutherford is credited with the
discovery of the nucleus.
In Rutherford’s nuclear model of
the atom, the entire positive charge and
most of the mass of the atom are
concentrated in the nucleus with the
electrons some distance away. The
electrons would be moving in orbits
about the nucleus just as the planets
do around the sun. Rutherford’s
experiments suggested the size of
the nucleus to be about 10
–15
m to
10
–14
m. From kinetic theory, the size
of an atom was known to be 10
–10
m,
about 10,000 to 100,000 times larger
than the size of the nucleus (see Chapter 11, Section 11.6 in Class XI
Physics textbook). Thus, the electrons would seem to be at a distance
from the nucleus of about 10,000 to 100,000 times the size of the nucleus
itself. Thus, most of an atom is empty space. With the atom being largely
empty space, it is easy to see why most α-particles go right through a
thin metal foil. However, when α-particle happens to come near a nucleus,
the intense electric field there scatters it through a large angle. The atomic
electrons, being so light, do not appreciably affect the α-particles.
The scattering data shown in Fig. 12.3 can be analysed by employing
Rutherford’s nuclear model of the atom. As the gold foil is very thin, it
can be assumed that α-particles will suffer not more than one scattering
during their passage through it. Therefore, computation of the trajectory
of an alpha-particle scattered by a single nucleus is enough. Alpha-
particles are nuclei of helium atoms and, therefore, carry two units, 2e,
of positive charge and have the mass of the helium atom. The charge of
the gold nucleus is Ze, where Z is the atomic number of the atom; for
gold Z = 79. Since the nucleus of gold is about 50 times heavier than an
α-particle, it is reasonable to assume that it remains stationary
throughout the scattering process. Under these assumptions, the
trajectory of an alpha-particle can be computed employing Newton’s
second law of motion and the Coulomb’s law for electrostatic
force of repulsion between the alpha-particle and the positively
charged nucleus.
FIGURE 12.3 Experimental data points (shown by
dots) on scattering of α-particles by a thin foil at
different angles obtained by Geiger and Marsden
using the setup shown in Figs. 12.1 and
12.2. Rutherford’s nuclear model predicts the solid
curve which is seen to be in good agreement with
experiment.
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EXAMPLE
12.1
The magnitude of this force is
2
0
(2 )( )
1
4
e Ze
F
r
ε
=
π
(12.1)
where r is the distance between the α-particle and the nucleus. The force
is directed along the line joining the α-particle and the nucleus. The
magnitude and direction of the force on an α-particle continuously
changes as it approaches the nucleus and recedes away from it.
12.2.1 Alpha-particle trajectory
The trajectory traced by an α-particle depends on the impact parameter,
b of collision. The impact parameter is the perpendicular distance of the
initial velocity vector of the α-particle from the centre of the nucleus (Fig.
12.4). A given beam of α-particles has a
distribution of impact parameters b, so that
the beam is scattered in various directions
with different probabilities (Fig. 12.4). (In
a beam, all particles have nearly same
kinetic energy.) It is seen that an α-particle
close to the nucleus (small impact
parameter) suffers large scattering. In case
of head-on collision, the impact parameter
is minimum and the α-particle rebounds
back (
θ
≅ π). For a large impact parameter,
the α-particle goes nearly undeviated and
has a small deflection (
θ
≅ 0).
The fact that only a small fraction of the
number of incident particles rebound back
indicates that the number of α-particles
undergoing head on collision is small. This,
in turn, implies that the mass and positive charge of the atom is
concentrated in a small volume. Rutherford scattering therefore, is a
powerful way to determine an upper limit to the size of the nucleus.
FIGURE 12.4 Trajectory of α-particles in the
coulomb field of a target nucleus. The impact
parameter, b and scattering angle
θ
are also depicted.
Example 12.1 In the Rutherford’s nuclear model of the atom, the
nucleus (radius about 10
–15
m) is analogous to the sun about which
the electron move in orbit (radius ≈ 10
–10
m) like the earth orbits
around the sun. If the dimensions of the solar system had the same
proportions as those of the atom, would the earth be closer to or
farther away from the sun than actually it is? The radius of earth’s
orbit is about 1.5 × 10
11
m. The radius of sun is taken as 7 × 10
8
m.
Solution The ratio of the radius of electron’s orbit to the radius of
nucleus is (10
–10
m)/(10
–15
m) = 10
5
, that is, the radius of the electron’s
orbit is 10
5
times larger than the radius of nucleus. If the radius of
the earth’s orbit around the sun were 10
5
times larger than the radius
of the sun, the radius of the earth’s orbit would be 10
5
× 7 × 10
8
m =
7 × 10
13
m. This is more than 100 times greater than the actual
orbital radius of earth. Thus, the earth would be much farther away
from the sun.
It implies that an atom contains a much greater fraction of empty
space than our solar system does.
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EXAMPLE 12.2
Example 12.2 In a Geiger-Marsden experiment, what is the distance
of closest approach to the nucleus of a 7.7 MeV α-particle before it
comes momentarily to rest and reverses its direction?
Solution The key idea here is that throughout the scattering process,
the total mechanical energy of the system consisting of an α-particle
and a gold nucleus is conserved. The system’s initial mechanical
energy is E
i
, before the particle and nucleus interact, and it is equal
to its mechanical energy E
f
when the α-particle momentarily stops.
The initial energy E
i
is just the kinetic energy K
of the incoming
α- particle. The final energy E
f
is just the electric potential energy U
of the system. The potential energy U can be calculated from
Eq. (12.1).
Let d be the centre-to-centre distance between the α-particle and
the gold nucleus when the α-particle is at its stopping point. Then
we can write the conservation of energy E
i
= E
f
as
2
0 0
1 (2 )( ) 2
4 4
e Ze Ze
K
d
d
ε ε
= =
π π
Thus the distance of closest approach d is given by
2
0
2
4
Ze
d
K
ε
=
π
The maximum kinetic energy found in α-particles of natural origin is
7.7 MeV or 1.2 × 10
–12
J. Since 1/4π
ε
0
= 9.0 × 10
9
N m
2
/C
2
. Therefore
with e = 1.6 × 10
–19
C, we have,
9 2 2 –19 2
12
(2)(9.0 10 Nm / )(1.6 10 ) Z
1.2 10 J
C C
d
−
× ×
=
×
= 3.84 × 10
–16
Z m
The atomic number of foil material gold is Z = 79, so that
d (Au) = 3.0 × 10
–14
m = 30 fm. (1 fm (i.e. fermi) = 10
–15
m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10
–14
m. This
is not in very good agreement with the observed result as the actual
radius of gold nucleus is 6 fm. The cause of discrepancy is that the
distance of closest approach is considerably larger than the sum of
the radii of the gold nucleus and the α-particle. Thus, the α-particle
reverses its motion without ever actually touching the gold nucleus.
12.2.2 Electron orbits
The Rutherford nuclear model of the atom which involves classical
concepts, pictures the atom as an electrically neutral sphere consisting
of a very small, massive and positively charged nucleus at the centre
surrounded by the revolving electrons in their respective dynamically
stable orbits. The electrostatic force of attraction, F
e
between the revolving
electrons and the nucleus provides the requisite centripetal force (F
c
) to
keep them in their orbits. Thus, for a dynamically stable orbit in a
hydrogen atom
F
e
= F
c
2 2
2
0
1
4
ε
=
π
e mv
r
r
(12.2)
Simulate Rutherford scattering experiment
http://www-outreach.phy.cam.ac.uk/camphy/nucleus/nucleus6_1.htm
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Thus the relation between the orbit radius and the electron
velocity is
2
2
0
4
e
r
mv
ε
=
π
(12.3)
The kinetic energy (K) and electrostatic potential energy (U) of the electron
in hydrogen atom are
2 2
2
0 0
1
and
2 8 4
e e
K mv U
r r
ε ε
= = = −
π π
(The negative sign in U signifies that the electrostatic force is in the –r
direction.) Thus the total energy E of the electron in a hydrogen atom is
2 2
0 0
8 4
e e
E K U
r r
ε ε
= + = −
π π
2
0
8
e
r
ε
= −
π
(12.4)
The total energy of the electron is negative. This implies the fact that
the electron is bound to the nucleus. If E were positive, an electron will
not follow a closed orbit around the nucleus.
12.3 ATOMIC SPECTRA
As mentioned in Section 12.1, each element has a characteristic spectrum
of radiation, which it emits. When an atomic gas or vapour is excited at
low pressure, usually by passing an electric current through it, the emitted
radiation has a spectrum which contains certain specific wavelengths
only. A spectrum of this kind is termed as emission line spectrum and it
EXAMPLE 12.3
Example 12.3 It is found experimentally that 13.6 eV energy is
required to separate a hydrogen atom into a proton and an electron.
Compute the orbital radius and the velocity of the electron in a
hydrogen atom.
Solution Total energy of the electron in hydrogen atom is –13.6 eV =
–13.6 × 1.6 × 10
–19
J = –2.2 ×10
–18
J. Thus from Eq. (12.4), we have
2
18
0
2.2 10 J
8
ε
−
= − = − ×
π
e
E
r
This gives the orbital radius
2 9 2 2 19 2
18
0
(9 10 N m /C )(1.6 10 C)
8
(2)(–2.2 10 J)
e
r
E
ε
−
−
× ×
= − = −
π
×
= 5.3 × 10
–11
m.
The velocity of the revolving electron can be computed from Eq. (12.3)
with m = 9.1 × 10
–31
kg,
6
0
2.2 10 m/s.
4
e
v
mr
ε
= = ×
π
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consists of bright lines on a
dark background. The
spectrum emitted by atomic
hydrogen is shown in
Fig. 12.5. Study of emission
line spectra of a material can
therefore serve as a type of
“fingerprint” for identification
of the gas. When white light
passes through a gas and we
analyse the transmitted light
using a spectrometer we find
some dark lines in the
spectrum. These dark lines
correspond precisely to those wavelengths which were found in the
emission line spectrum of the gas. This is called the absorption spectrum
of the material of the gas.
12.3.1 Spectral series
We might expect that the frequencies of the light emitted by a particular
element would exhibit some regular pattern. Hydrogen is the simplest
atom and therefore, has the simplest spectrum. In the observed spectrum,
however, at first sight, there does not seem to be
any resemblance of order or regularity in spectral
lines. But the spacing between lines within certain
sets of the hydrogen spectrum decreases in a
regular way (Fig. 12.5). Each of these sets is called
a spectral series. In 1885, the first such series was
observed by a Swedish school teacher Johann Jakob
Balmer (1825–1898) in the visible region of the
hydrogen spectrum. This series is called Balmer
series (Fig. 12.6). The line with the longest
wavelength, 656.3 nm in the red is called H
α
;
the
next line with wavelength 486.1 nm in the blue-
green is called H
β
, the third line 434.1 nm in the
violet is called H
γ
;
and so on. As the wavelength
decreases, the lines appear closer together and are weaker in intensity.
Balmer found a simple empirical formula for the observed wavelengths
2 2
1 1 1
2
R
n
λ
= −
(12.5)
where
λ
is the wavelength, R is a constant called the Rydberg constant,
and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 10
7
m
–1
.
This equation is also called Balmer formula.
Taking n = 3 in Eq. (12.5), one obtains the wavelength of the H
α
line:
7 –1
2 2
1 1 1
1.097 10 m
2 3
λ
= × −
= 1.522 × 10
6
m
–1
i.e.,
λ
= 656.3 nm
FIGURE 12.5 Emission lines in the spectrum of hydrogen.
FIGURE 12.6 Balmer series in the
emission spectrum of hydrogen.
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For n = 4, one obtains the wavelength of H
β
line, etc. For n = ∞, one obtains
the limit of the series, at
λ
= 364.6 nm. This is the shortest wavelength in
the Balmer series. Beyond this limit, no further distinct lines appear,
instead only a faint continuous spectrum is seen.
Other series of spectra for hydrogen were subsequently discovered.
These are known, after their discoverers, as Lyman, Paschen, Brackett,
and Pfund series. These are represented by the formulae:
Lyman series:
1 1
1
1
2 2
λ
= −
R
n
n = 2,3,4... (12.6)
Paschen series:
1 1
3
1
2 2
λ
= −
R
n
n = 4,5,6... (12.7)
Brackett series:
1 1
4
1
2 2
λ
= −
R
n
n = 5,6,7... (12.8)
Pfund series:
1 1
5
1
2 2
λ
= −
R
n
n = 6,7,8... (12.9)
The Lyman series is in the ultraviolet, and the Paschen, Brackett,
and Pfund series are in the infrared region.
The Balmer formula Eq. (12.5) may be written in terms of frequency
of the light, recalling that
c =
νλ
or
1
c
ν
λ
=
Thus, Eq. (12.5) becomes
ν
= Rc
n
1
2
1
2 2
−
(12.10)
There are only a few elements (hydrogen, singly ionised helium, and
doubly ionised lithium) whose spectra can be represented by simple
formula like Eqs. (12.5) – (12.9).
Equations (12.5) – (12.9) are useful as they give the wavelengths that
hydrogen atoms radiate or absorb. However, these results are empirical
and do not give any reasoning why only certain frequencies are observed
in the hydrogen spectrum.
12.4 BOHR MODEL OF THE HYDROGEN ATOM
The model of the atom proposed by Rutherford assumes that the atom,
consisting of a central nucleus and revolving electron is stable much like
sun-planet system which the model imitates. However, there are some
fundamental differences between the two situations. While the planetary
system is held by gravitational force, the nucleus-electron system being
charged objects, interact by Coulomb’s Law of force. We know that an
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EXAMPLE 12.4
object which moves in a circle is being constantly
accelerated – the acceleration being centripetal in nature.
According to classical electromagnetic theory, an
accelerating charged particle emits radiation in the form
of electromagnetic waves. The energy of an accelerating
electron should therefore, continuously decrease. The
electron would spiral inward and eventually fall into the
nucleus (Fig. 12.7). Thus, such an atom can not be stable.
Further, according to the classical electromagnetic theory,
the frequency of the electromagnetic waves emitted by
the revolving electrons is equal to the frequency of
revolution. As the electrons spiral inwards, their angular
velocities and hence their frequencies would change
continuously, and so will the frequency of the light
emitted. Thus, they would emit a continuous spectrum,
in contradiction to the line spectrum actually observed.
Clearly Rutherford model tells only a part of the story
implying that the classical ideas are not sufficient to
explain the atomic structure.
FIGURE 12.7 An accelerated atomic electron must spiral into the
nucleus as it loses energy.
Example 12.4 According to the classical electromagnetic theory,
calculate the initial frequency of the light emitted by the electron
revolving around a proton in hydrogen atom.
Solution From Example 12.3 we know that velocity of electron moving
around a proton in hydrogen atom in an orbit of radius 5.3 × 10
–11
m
is 2.2 × 10
–6
m/s. Thus, the frequency of the electron moving around
the proton is
( )
6 1
11
2.2 10 m s
2
2 5.3 10 m
v
r
ν
−
−
×
= =
π
π ×
≈ 6.6 × 10
15
Hz.
According to the classical electromagnetic theory we know that the
frequency of the electromagnetic waves emitted by the revolving
electrons is equal to the frequency of its revolution around the nucleus.
Thus the initial frequency of the light emitted is 6.6 × 10
15
Hz.
Niels Henrik David Bohr
(1885 – 1962) Danish
physicist who explained the
spectrum of hydrogen atom
based on quantum ideas.
He gave a theory of nuclear
fission based on the liquid-
drop model of nucleus.
Bohr contributed to the
clarification of conceptual
problems in quantum
mechanics, in particular by
proposing the comple-
mentary principle.
NIELS HENRIK DAVID BOHR (1885 – 1962)
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It was Niels Bohr (1885 – 1962) who made certain modifications in
this model by adding the ideas of the newly developing quantum
hypothesis. Niels Bohr studied in Rutherford’s laboratory for several
months in 1912 and he was convinced about the validity of Rutherford
nuclear model. Faced with the dilemma as discussed above, Bohr, in
1913, concluded that in spite of the success of electromagnetic theory in
explaining large-scale phenomena, it could not be applied to the processes
at the atomic scale. It became clear that a fairly radical departure from
the established principles of classical mechanics and electromagnetism
would be needed to understand the structure of atoms and the relation
of atomic structure to atomic spectra. Bohr combined classical and early
quantum concepts and gave his theory in the form of three postulates.
These are :
(i) Bohr’s first postulate was that an electron in an atom could revolve
in certain stable orbits without the emission of radiant energy,
contrary to the predictions of electromagnetic theory. According to
this postulate, each atom has certain definite stable states in which it
can exist, and each possible state has definite total energy. These are
called the stationary states of the atom.
(ii) Bohr’s second postulate defines these stable orbits. This postulate
states that the electron revolves around the nucleus only in those
orbits for which the angular momentum is some integral multiple of
h/2π where h is the Planck’s constant (= 6.6 × 10
–34
J s). Thus the
angular momentum (L) of the orbiting electron is quantised. That is
L = nh/2π (12.11)
(iii) Bohr’s third postulate incorporated into atomic theory the early
quantum concepts that had been developed by Planck and Einstein.
It states that an electron might make a transition from one of its
specified non-radiating orbits to another of lower energy. When it
does so, a photon is emitted having energy equal to the energy
difference between the initial and final states. The frequency of the
emitted photon is then given by
h
ν
= E
i
– E
f
(12.12)
where E
i
and E
f
are the energies of the initial and final states and E
i
> E
f
.
For a hydrogen atom, Eq. (12.4) gives the expression to determine
the energies of different energy states. But then this equation requires
the radius r of the electron orbit. To calculate r, Bohr’s second postulate
about the angular momentum of the electron–the quantisation
condition – is used. The angular momentum L is given by
L = mvr
Bohr’s second postulate of quantisation [Eq. (12.11)] says that the
allowed values of angular momentum are integral multiples of h/2π.
L
n
=
mv
n
r
n
=
2
nh
π
(12.13)
where n is an integer, r
n
is the radius of n
th
possible orbit and v
n
is the
speed of moving electron in the n
th
orbit. The allowed orbits are numbered
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1, 2, 3 ..., according to the values of n, which is called the principal
quantum number of the orbit.
From Eq. (12.3), the relation between v
n
and r
n
is
0
4
n
n
e
v
mr
ε
=
π
Combining it with Eq. (12.13), we get the following expressions for v
n
and r
n
,
( )
2
0
1 1
4 2
n
e
v
n h
ε
=
π π
(12.14)
and
2
2
0
2
4
2
n
n h
r
m
e
ε
π
=
π
(12.15)
Eq. (12.14) depicts that the orbital speed in the n
th
orbit falls by a factor
of n. Using Eq. (12.15), the size of the innermost orbit (n = 1) can be
obtained as
2
0
1
2
h
r
me
ε
=
π
This is called the Bohr radius, represented by the symbol a
0
. Thus,
2
0
0
2
h
a
me
ε
=
π
(12.16)
Substitution of values of h, m,
ε
0
and e gives a
0
= 5.29 × 10
–11
m. From
Eq. (12.15), it can also be seen that the radii of the orbits increase as n
2
.
The total energy of the electron in the stationary states of the hydrogen
atom can be obtained by substituting the value of orbital radius in
Eq. (12.4) as
2
2 2
2
0 0
2
8 4
n
e m e
E
hn
ε ε
π
= −
π π
or
4
2 2 2
0
8
n
me
E
n h
ε
= −
(12.17)
Substituting values, Eq. (12.17) yields
18
2
2.18 10
J
n
E
n
−
×
= −
(12.18)
Atomic energies are often expressed in electron volts (eV) rather than
joules. Since 1 eV = 1.6 × 10
–19
J, Eq. (12.18) can be rewritten as
2
13.6
eV
n
E
n
= −
(12.19)
The negative sign of the total energy of an electron moving in an orbit
means that the electron is bound with the nucleus. Energy will thus be
required to remove the electron from the hydrogen atom to a distance
infinitely far away from its nucleus (or proton in hydrogen atom).
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EXAMPLE 12.5
The derivation of Eqs. (12.17) – (12.19) involves the assumption that
the electronic orbits are circular, though orbits under inverse square
force are, in general elliptical. (Planets move in elliptical orbits under the
inverse square gravitational force of the sun.) However, it was shown by
the German physicist Arnold Sommerfeld (1868 – 1951) that, when the
restriction of circular orbit is relaxed, these equations continue to hold
even for elliptic orbits.
ORBIT VS STATE (ORBITAL PICTURE) OF ELECTRON IN
ATOM
We are introduced to the Bohr Model of atom one time or the other in the course of
physics. This model has its place in the history of quantum mechanics and particularly
in explaining the structure of an atom. It has become a milestone since Bohr introduced
the revolutionary idea of definite energy orbits for the electrons, contrary to the classical
picture requiring an accelerating particle to radiate. Bohr also introduced the idea of
quantisation of angular momentum of electrons moving in definite orbits. Thus it was a
semi-classical picture of the structure of atom.
Now with the development of quantum mechanics, we have a better understanding
of the structure of atom. Solutions of the Schrödinger wave equation assign a wave-like
description to the electrons bound in an atom due to attractive forces of the protons.
An orbit of the electron in the Bohr model is the circular path of motion of an electron
around the nucleus. But according to quantum mechanics, we cannot associate a definite
path with the motion of the electrons in an atom. We can only talk about the probability
of finding an electron in a certain region of space around the nucleus. This probability
can be inferred from the one-electron wave function called the orbital. This function
depends only on the coordinates of the electron.
It is therefore essential that we understand the subtle differences that exist in the two
models:
l Bohr model is valid for only one-electron atoms/ions; an energy value, assigned to
each orbit, depends on the principal quantum number n in this model. We know
that energy associated with a stationary state of an electron depends on n only, for
one-electron atoms/ions. For a multi-electron atom/ion, this is not true.
l The solution of the Schrödinger wave equation, obtained for hydrogen-like atoms/
ions, called the wave function, gives information about the probability of finding an
electron in various regions around the nucleus. This orbital has no resemblance
whatsoever with the orbit defined for an electron in the Bohr model.
Example 12.5 A 10 kg satellite circles earth once every 2 h in an
orbit having a radius of 8000 km. Assuming that Bohr’s angular
momentum postulate applies to satellites just as it does to an electron
in the hydrogen atom, find the quantum number of the orbit of the
satellite.
Solution
From Eq. (12.13), we have
m v
n
r
n
= nh/2π
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EXAMPLE 12.5
Here m = 10 kg and r
n
= 8 × 10
6
m. We have the time period T of the
circling satellite as 2 h. That is T = 7200 s.
Thus the velocity v
n
= 2π r
n
/T.
The quantum number of the orbit of satellite
n
= (2π r
n
)
2
× m/(T × h).
Substituting the values,
n = (2π × 8 × 10
6
m)
2
× 10/(7200 s × 6.64 × 10
–34
J s)
= 5.3 × 10
45
Note that the quantum number for the satellite motion is extremely
large! In fact for such large quantum numbers the results of
quantisation conditions tend to those of classical physics.
12.4.1 Energy levels
The energy of an atom is the least (largest negative value)
when its electron is revolving in an orbit closest to the
nucleus i.e., the one for which n = 1. For n = 2, 3, ... the
absolute value of the energy E is smaller, hence the energy
is progressively larger in the outer orbits. The lowest state
of the atom, called the ground state, is that of the lowest
energy, with the electron revolving in the orbit of smallest
radius, the Bohr radius, a
0
. The energy of this state (n = 1),
E
1
is –13.6 eV. Therefore, the minimum energy required to
free the electron from the ground state of the hydrogen atom
is 13.6 eV. It is called the ionisation energy of the hydrogen
atom. This prediction of the Bohr’s model is in excellent
agreement with the experimental value of ionisation energy.
At room temperature, most of the hydrogen atoms are
in ground state. When a hydrogen atom receives energy
by processes such as electron collisions, the atom may
acquire sufficient energy to raise the electron to higher
energy states. The atom is then said to be in an excited
state. From Eq. (12.19), for n = 2; the energy E
2
is
–3.40 eV. It means that the energy required to excite an
electron in hydrogen atom to its first excited state, is an
energy equal to E
2
– E
1
= –3.40 eV – (–13.6) eV = 10.2 eV.
Similarly, E
3
= –1.51 eV and E
3
– E
1
= 12.09 eV, or to excite
the hydrogen atom from its ground state (n = 1) to second
excited state (n = 3), 12.09 eV energy is required, and so
on. From these excited states the electron can then fall back
to a state of lower energy, emitting a photon in the process.
Thus, as the excitation of hydrogen atom increases (that is
as n increases) the value of minimum energy required to
free the electron from the excited atom decreases.
The energy level diagram* for the stationary states of a
hydrogen atom, computed from Eq. (12.19), is given in
* An electron can have any total energy above E = 0 eV. In such situations the
electron is free. Thus there is a continuum of energy states above E = 0 eV, as
shown in Fig. 12.8.
FIGURE 12.8 The energy level
diagram for the hydrogen atom.
The electron in a hydrogen atom
at room temperature spends
most of its time in the ground
state. To ionise a hydrogen
atom an electron from the
ground state, 13.6 eV of energy
must be supplied. (The horizontal
lines specify the presence of
allowed energy states.)
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Fig. 12.8. The principal quantum number n labels the stationary
states in the ascending order of energy. In this diagram, the highest
energy state corresponds to n =∞ in Eq, (12.19) and has an energy
of 0 eV. This is the energy of the atom when the electron is
completely removed (r = ∞) from the nucleus and is at rest. Observe how
the energies of the excited states come closer and closer together as n
increases.
12.5 THE LINE SPECTRA OF THE HYDROGEN ATOM
According to the third postulate of Bohr’s model, when an atom makes a
transition from the higher energy state with quantum number n
i
to the
lower energy state with quantum number n
f
(n
f
< n
i
), the difference of
energy is carried away by a photon of frequency
ν
if
such that
FRANCK – HERTZ EXPERIMENT
The existence of discrete energy levels in an atom was directly verified in 1914 by James
Franck and Gustav Hertz. They studied the spectrum of mercury vapour when electrons
having different kinetic energies passed through the vapour. The electron energy was
varied by subjecting the electrons to electric fields of varying strength. The electrons
collide with the mercury atoms and can transfer energy to the mercury atoms. This can
only happen when the energy of the electron is higher than the energy difference between
an energy level of Hg occupied by an electron and a higher unoccupied level (see Figure).
For instance, the difference between an occupied energy level of Hg and a higher
unoccupied level is 4.9 eV. If an electron of having an energy of 4.9 eV or more passes
through mercury, an electron in mercury atom can absorb energy from the bombarding
electron and get excited to the higher level [Fig (a)]. The colliding electron’s kinetic energy
would reduce by this amount.
The excited electron would subsequently fall back to the ground state by emission of
radiation [Fig. (b)]. The wavelength of emitted radiation is:
34 8
19
6.625 10 3 10
4.9 1.6 10
hc
E
λ
−
−
× × ×
= =
× ×
= 253 nm
By direct measurement, Franck and Hertz found that the emission spectrum of
mercury has a line corresponding to this wavelength. For this experimental verification
of Bohr’s basic ideas of discrete energy levels in atoms and the process of photon emission,
Frank and Hertz were awarded the Nobel prize in 1925.
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Atoms
h
ν
if
= E
n
i
– E
n
f
(12.20)
Using Eq. (12.16), for E
n
f
and E
n
i
, we get
h
ν
if
=
me
h n n
f i
4
2 2 2 2
8
1 1
ε
0
−
(12.21)
or
ν
if
=
me
h n n
f i
4
2 3 2 2
8
1 1
ε
0
−
(12.22)
Equation (12.21) is the Rydberg formula, for the spectrum of the
hydrogen atom. In this relation, if we take n
f
= 2 and n
i
= 3, 4, 5..., it
reduces to a form similar to Eq. (12.10) for the Balmer series. The Rydberg
constant R is readily identified to be
R =
4
2 3
8
me
h c
0
ε
(12.23)
If we insert the values of various constants in Eq. (12.23), we get
R = 1.03 × 10
7
m
–1
This is a value very close to the value (1.097 × 10
7
m
–1
) obtained from the
empirical Balmer formula. This agreement between the theoretical and
experimental values of the Rydberg constant provided a direct and
striking confirmation of the Bohr’s model.
Since both n
f
and n
i
are integers,
this immediately shows that in
transitions between different atomic
levels, light is radiated in various
discrete frequencies. For hydrogen
spectrum, the Balmer formula
corresponds to n
f
= 2 and n
i
= 3, 4, 5,
etc. The results of the Bohr’s model
suggested the presence of other series
spectra for hydrogen atom–those
corresponding to transitions resulting
from n
f
= 1 and n
i
= 2, 3, etc.; n
f
= 3
and n
i
= 4, 5, etc., and so on. Such
series were identified in the course of
spectroscopic investigations and are
known as the Lyman, Balmer,
Paschen, Brackett, and Pfund series.
The electronic transitions
corresponding to these series are
shown in Fig. 12.9.
The various lines in the atomic
spectra are produced when electrons
jump from higher energy state to a
lower energy state and photons are
emitted. These spectral lines are called
emission lines. But when an atom
absorbs a photon that has precisely
FIGURE 12.9 Line spectra originate in
transitions between energy levels.
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EXAMPLE
12.6
the same energy needed by the electron in a lower energy state to make
transitions to a higher energy state, the process is called absorption.
Thus if photons with a continuous range of frequencies pass through a
rarefied gas and then are analysed with a spectrometer, a series of dark
spectral absorption lines appear in the continuous spectrum. The dark
lines indicate the frequencies that have been absorbed by the atoms of
the gas.
The explanation of the hydrogen atom spectrum provided by Bohr’s
model was a brilliant achievement, which greatly stimulated progress
towards the modern quantum theory. In 1922, Bohr was awarded Nobel
Prize in Physics.
Example 12.6 Using the Rydberg formula, calculate the wavelengths
of the first four spectral lines in the Lyman series of the hydrogen
spectrum.
Solution The Rydberg formula is
hc/λ
if
=
4
2 2 2 2
1 1
8
f i
me
h n n
0
ε
−
The wavelengths of the first four lines in the Lyman series correspond
to transitions from n
i
= 2,3,4,5 to n
f
= 1. We know that
4
2 2
8
me
h
0
ε
= 13.6 eV = 21.76 ×10
–19
J
Therefore,
1
19
2
m
1 1
21.76 10
1
i
i
hc
n
λ
−
=
× −
=
.
34 8 2
19 2
6 625 10 3 10
m
21.76 10 ( 1)
i
i
n
n
−
−
× × × ×
× × −
=
2
7
2
0.9134
10 m
( 1)
i
i
n
n
−
×
−
= 913.4 n
i
2
/(n
i
2
–1) Å
Substituting n
i
= 2,3,4,5, we get λ
21
= 1218 Å, λ
31
= 1028 Å, λ
41
= 974.3 Å,
and λ
51
= 951.4 Å.
12.6 DE BROGLIE’S EXPLANATION OF BOHR’S
SECOND POSTULATE OF QUANTISATION
Of all the postulates, Bohr made in his model of the atom, perhaps the
most puzzling is his second postulate. It states that the angular
momentum of the electron orbiting around the nucleus is quantised (that
is, L
n
= nh/2π; n = 1, 2, 3 …). Why should the angular momentum have
only those values that are integral multiples of h/2π? The French physicist
Louis de Broglie explained this puzzle in 1923, ten years after Bohr
proposed his model.
We studied, in Chapter 11, about the de Broglie’s hypothesis that
material particles, such as electrons, also have a wave nature. C. J. Davisson
and L. H. Germer later experimentally verified the wave nature of electrons
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Atoms
in 1927. Louis de Broglie argued that the electron in its
circular orbit, as proposed by Bohr, must be seen as a particle
wave. In analogy to waves travelling on a string, particle waves
too can lead to standing waves under resonant conditions.
From Chapter 15 of Class XI Physics textbook, we know that
when a string is plucked, a vast number of wavelengths are
excited. However only those wavelengths survive which have
nodes at the ends and form the standing wave in the string. It
means that in a string, standing waves are formed when the
total distance travelled by a wave down the string and back is
one wavelength, two wavelengths, or any integral number of
wavelengths. Waves with other wavelengths interfere with
themselves upon reflection and their amplitudes quickly drop
to zero. For an electron moving in n
th
circular orbit of radius
r
n
, the total distance is the circumference of the orbit, 2πr
n
.
Thus
2π r
n
= n
λ
, n = 1, 2, 3... (12.24)
Figure 12.10 illustrates a standing particle wave on a
circular orbit for n = 4, i.e., 2πr
n
= 4
λ
, where
λ
is the de Broglie
wavelength of the electron moving in n
th
orbit. From Chapter
11, we have
λ
= h/p, where p is the magnitude of the electron’s
momentum. If the speed of the electron is much less than the speed of
light, the momentum is mv
n
. Thus,
λ
= h/mv
n
. From Eq. (12.24), we have
2π r
n
= n h/mv
n
or m v
n
r
n
= nh/2π
This is the quantum condition proposed by Bohr for the angular
momentum of the electron [Eq. (12.13)]. In Section 12.5, we saw that
this equation is the basis of explaining the discrete orbits and energy
levels in hydrogen atom. Thus de Broglie hypothesis provided an
explanation for Bohr’s second postulate for the quantisation of angular
momentum of the orbiting electron. The quantised electron orbits and
energy states are due to the wave nature of the electron and only resonant
standing waves can persist.
Bohr’s model, involving classical trajectory picture (planet-like electron
orbiting the nucleus), correctly predicts the gross features of the
hydrogenic atoms*, in particular, the frequencies of the radiation emitted
or selectively absorbed. This model however has many limitations.
Some are:
(i) The Bohr model is applicable to hydrogenic atoms. It cannot be
extended even to mere two electron atoms such as helium. The analysis
of atoms with more than one electron was attempted on the lines of
Bohr’s model for hydrogenic atoms but did not meet with any success.
Difficulty lies in the fact that each electron interacts not only with the
positively charged nucleus but also with all other electrons.
FIGURE 12.10 A standing
wave is shown on a circular
orbit where four de Broglie
wavelengths fit into the
circumference of the orbit.
* Hydrogenic atoms are the atoms consisting of a nucleus with positive charge
+Ze and a single electron, where Z is the proton number. Examples are hydrogen
atom, singly ionised helium, doubly ionised lithium, and so forth. In these
atoms more complex electron-electron interactions are nonexistent.
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The formulation of Bohr model involves electrical force between
positively charged nucleus and electron. It does not include the
electrical forces between electrons which necessarily appear in
multi-electron atoms.
(ii) While the Bohr’s model correctly predicts the frequencies of the light
emitted by hydrogenic atoms, the model is unable to explain the
relative intensities of the frequencies in the spectrum. In emission
spectrum of hydrogen, some of the visible frequencies have weak
intensity, others strong. Why? Experimental observations depict that
some transitions are more favoured than others. Bohr’s model is
unable to account for the intensity variations.
Bohr’s model presents an elegant picture of an atom and cannot be
generalised to complex atoms. For complex atoms we have to use a new
and radical theory based on Quantum Mechanics, which provides a more
complete picture of the atomic structure.
LASER LIGHT
Imagine a crowded market place or a railway platform with people entering a gate and
going towards all directions. Their footsteps are random and there is no phase correlation
between them. On the other hand, think of a large number of soldiers in a regulated march.
Their footsteps are very well correlated. See figure here.
This is similar to the difference between light emitted by
an ordinary source like a candle or a bulb and that emitted
by a laser. The acronym LASER stands for Light Amplification
by Stimulated Emission of Radiation. Since its development
in 1960, it has entered into all areas of science and technology.
It has found applications in physics, chemistry, biology,
medicine, surgery, engineering, etc. There are low power
lasers, with a power of 0.5 mW, called pencil lasers, which
serve as pointers. There are also lasers of different power,
suitable for delicate surgery of eye or glands in the stomach.
Finally, there are lasers which can cut or weld steel.
Light is emitted from a source in the form of packets of
waves. Light coming out from an ordinary source contains a mixture of many wavelengths.
There is also no phase relation between the various waves. Therefore, such light, even if it is
passed through an aperture, spreads very fast and the beam size increases rapidly with
distance. In the case of laser light, the wavelength of each packet is almost the same. Also
the average length of the packet of waves is much larger. This means that there is better
phase correlation over a longer duration of time. This results in reducing the divergence of
a laser beam substantially.
If there are N atoms in a source, each emitting light with intensity I, then the total
intensity produced by an ordinary source is proportional to NI, whereas in a laser source,
it is proportional to N
2
I. Considering that N is very large, we see that the light from a laser
can be much stronger than that from an ordinary source.
When astronauts of the Apollo missions visited the moon, they placed a mirror on its
surface, facing the earth. Then scientists on the earth sent a strong laser beam, which was
reflected by the mirror on the moon and received back on the earth. The size of the reflected
laser beam and the time taken for the round trip were measured. This allowed a very
accurate determination of (a) the extremely small divergence of a laser beam and (b) the
distance of the moon from the earth.
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SUMMARY
1. Atom, as a whole, is electrically neutral and therefore contains equal
amount of positive and negative charges.
2. In Thomson’s model, an atom is a spherical cloud of positive charges
with electrons embedded in it.
3. In Rutherford’s model, most of the mass of the atom and all its positive
charge are concentrated in a tiny nucleus (typically one by ten thousand
the size of an atom), and the electrons revolve around it.
4. Rutherford nuclear model has two main difficulties in explaining the
structure of atom: (a) It predicts that atoms are unstable because the
accelerated electrons revolving around the nucleus must spiral into
the nucleus. This contradicts the stability of matter. (b) It cannot
explain the characteristic line spectra of atoms of different elements.
5. Atoms of most of the elements are stable and emit characteristic
spectrum. The spectrum consists of a set of isolated parallel lines
termed as line spectrum. It provides useful information about the
atomic structure.
6. The atomic hydrogen emits a line spectrum consisting of various series.
The frequency of any line in a series can be expressed as a difference
of two terms;
Lyman series:
ν = −
Rc
n
1
1
1
2 2
; n = 2, 3, 4,...
Balmer series:
ν= −
Rc
n
1
2
1
2 2
; n = 3, 4, 5,...
Paschen series:
ν = −
Rc
n
1
3
1
2 2
; n = 4, 5, 6,...
Brackett series:
ν= −
Rc
n
1
4
1
2 2
; n = 5, 6, 7,...
Pfund series:
ν = −
Rc
n
1
5
1
2 2
; n = 6, 7, 8,...
7. To explain the line spectra emitted by atoms, as well as the stability
of atoms, Niel’s Bohr proposed a model for hydrogenic (single elctron)
atoms. He introduced three postulates and laid the foundations of
quantum mechanics:
(a) In a hydrogen atom, an electron revolves in certain stable orbits
(called stationary orbits) without the emission of radiant energy.
(b) The stationary orbits are those for which the angular momentum
is some integral multiple of h/2π. (Bohr’s quantisation condition.)
That is L = nh/2π, where n is an integer called the principal
quantum number.
(c) The third postulate states that an electron might make a transition
from one of its specified non-radiating orbits to another of lower
energy. When it does so, a photon is emitted having energy equal
to the energy difference between the initial and final states. The
frequency (
ν
) of the emitted photon is then given by
h
ν
= E
i
– E
f
An atom absorbs radiation of the same frequency the atom emits,
in which case the electron is transferred to an orbit with a higher
value of n.
E
i
+ h
ν
= E
f
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8. As a result of the quantisation condition of angular momentum, the
electron orbits the nucleus at only specific radii. For a hydrogen atom
it is given by
2
2
0
2
4
2
n
n h
r
m
e
ε
π
=
π
The total energy is also quantised:
4
2 2 2
0
8
n
me
E
n h
ε
= −
= –13.6 eV/n
2
The n = 1 state is called ground state. In hydrogen atom the ground
state energy is –13.6 eV. Higher values of n correspond to excited
states (n > 1). Atoms are excited to these higher states by collisions
with other atoms or electrons or by absorption of a photon of right
frequency.
9. de Broglie’s hypothesis that electrons have a wavelength
λ =
h/mv gave
an explanation for Bohr’s quantised orbits by bringing in the wave-
particle duality. The orbits correspond to circular standing waves in
which the circumference of the orbit equals a whole number of
wavelengths.
10. Bohr’s model is applicable only to hydrogenic (single electron) atoms.
It cannot be extended to even two electron atoms such as helium.
This model is also unable to explain for the relative intensities of the
frequencies emitted even by hydrogenic atoms.
POINTS TO PONDER
1. Both the Thomson’s as well as the Rutherford’s models constitute an
unstable system. Thomson’s model is unstable electrostatically, while
Rutherford’s model is unstable because of electromagnetic radiation
of orbiting electrons.
2. What made Bohr quantise angular momentum (second postulate) and
not some other quantity? Note, h has dimensions of angular
momentum, and for circular orbits, angular momentum is a very
relevant quantity. The second postulate is then so natural!
3. The orbital picture in Bohr’s model of the hydrogen atom was
inconsistent with the uncertainty principle. It was replaced by modern
quantum mechanics in which Bohr’s orbits are regions where the
electron may be found with large probability.
4. Unlike the situation in the solar system, where planet-planet
gravitational forces are very small as compared to the gravitational
force of the sun on each planet (because the mass of the sun is so
much greater than the mass of any of the planets), the electron-electron
electric force interaction is comparable in magnitude to the electron-
nucleus electrical force, because the charges and distances are of the
same order of magnitude. This is the reason why the Bohr’s model
with its planet-like electron is not applicable to many electron atoms.
5. Bohr laid the foundation of the quantum theory by postulating specific
orbits in which electrons do not radiate. Bohr’s model include only
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one quantum number n. The new theory called quantum mechanics
supportes Bohr’s postulate. However in quantum mechanics (more
generally accepted), a given energy level may not correspond to just
one quantum state. For example, a state is characterised by four
quantum numbers (n, l, m, and s), but for a pure Coulomb potential
(as in hydrogen atom) the energy depends only on n.
6. In Bohr model, contrary to ordinary classical expectation, the
frequency of revolution of an electron in its orbit is not connected to
the frequency of spectral line. The later is the difference between two
orbital energies divided by h. For transitions between large quantum
numbers (n to n – 1, n very large), however, the two coincide as expected.
7. Bohr’s semiclassical model based on some aspects of classical physics
and some aspects of modern physics also does not provide a true
picture of the simplest hydrogenic atoms. The true picture is quantum
mechanical affair which differs from Bohr model in a number of
fundamental ways. But then if the Bohr model is not strictly correct,
why do we bother about it? The reasons which make Bohr’s model
still useful are:
(i) The model is based on just three postulates but accounts for almost
all the general features of the hydrogen spectrum.
(ii) The model incorporates many of the concepts we have learnt in
classical physics.
(iii) The model demonstrates how a theoretical physicist occasionally
must quite literally ignore certain problems of approach in hopes
of being able to make some predictions. If the predictions of the
theory or model agree with experiment, a theoretician then must
somehow hope to explain away or rationalise the problems that
were ignored along the way.
EXERCISES
12.1 Choose the correct alternative from the clues given at the end of
the each statement:
(a) The size of the atom in Thomson’s model is .......... the atomic
size in Rutherford’s model. (much greater than/no different
from/much less than.)
(b) In the ground state of .......... electrons are in stable equilibrium,
while in .......... electrons always experience a net force.
(Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on .......... is doomed to collapse.
(Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a ..........
but has a highly non-uniform mass distribution in ..........
(Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the
mass in .......... (Rutherford’s model/both the models.)
12.2 Suppose you are given a chance to repeat the alpha-particle
scattering experiment using a thin sheet of solid hydrogen in place
of the gold foil. (Hydrogen is a solid at temperatures below 14 K.)
What results do you expect?
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12.3 What is the shortest wavelength present in the Paschen series of
spectral lines?
12.4 A difference of 2.3 eV separates two energy levels in an atom. What
is the frequency of radiation emitted when the atom make a
transition from the upper level to the lower level?
12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the
kinetic and potential energies of the electron in this state?
12.6 A hydrogen atom initially in the ground level absorbs a photon,
which excites it to the n = 4 level. Determine the wavelength and
frequency of photon.
12.7 (a) Using the Bohr’s model calculate the speed of the electron in a
hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital
period in each of these levels.
12.8 The radius of the innermost electron orbit of a hydrogen atom is
5.3×10
–11
m. What are the radii of the n = 2 and n =3 orbits?
12.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at
room temperature. What series of wavelengths will be emitted?
12.10 In accordance with the Bohr’s model, find the quantum number
that characterises the earth’s revolution around the sun in an orbit
of radius 1.5 × 10
11
m with orbital speed 3 × 10
4
m/s. (Mass of earth
= 6.0 × 10
24
kg.)
ADDITIONAL EXERCISES
12.11 Answer the following questions, which help you understand the
difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of
α
-particles by a thin gold foil
predicted by Thomson’s model much less, about the same, or
much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of
α
-particles at angles greater than 90°) predicted by Thomson’s
model much less, about the same, or much greater than that
predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for
small thickness t, the number of
α
-particles scattered at
moderate angles is proportional to t. What clue does this linear
dependence on t provide?
(d) In which model is it completely wrong to ignore multiple
scattering for the calculation of average angle of scattering of
α
-particles by a thin foil?
12.12 The gravitational attraction between electron and proton in a
hydrogen atom is weaker than the coulomb attraction by a factor of
about 10
–40
. An alternative way of looking at this fact is to estimate
the radius of the first Bohr orbit of a hydrogen atom if the electron
and proton were bound by gravitational attraction. You will find the
answer interesting.
12.13 Obtain an expression for the frequency of radiation emitted when a
hydrogen atom de-excites from level n to level (n–1). For large n,
show that this frequency equals the classical frequency of revolution
of the electron in the orbit.
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Atoms
12.14 Classically, an electron can be in any orbit around the nucleus of
an atom. Then what determines the typical atomic size? Why is an
atom not, say, thousand times bigger than its typical size? The
question had greatly puzzled Bohr before he arrived at his famous
model of the atom that you have learnt in the text. To simulate what
he might well have done before his discovery, let us play as follows
with the basic constants of nature and see if we can get a quantity
with the dimensions of length that is roughly equal to the known
size of an atom (~ 10
–10
m).
(a) Construct a quantity with the dimensions of length from the
fundamental constants e
, m
e
, and c. Determine its numerical
value.
(b) You will find that the length obtained in (a) is many orders of
magnitude smaller than the atomic dimensions. Further, it
involves c. But energies of atoms are mostly in non-relativistic
domain where c is not expected to play any role. This is what
may have suggested Bohr to discard c and look for ‘something
else’ to get the right atomic size. Now, the Planck’s constant h
had already made its appearance elsewhere. Bohr’s great insight
lay in recognising that h, m
e
, and e will yield the right atomic
size. Construct a quantity with the dimension of length from h,
me, and e and confirm that its numerical value has indeed the
correct order of magnitude.
12.15 The total energy of an electron in the first excited state of the
hydrogen atom is about –3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the
zero of potential energy is changed?
12.16 If Bohr’s quantisation postulate (angular momentum = nh/2π) is a
basic law of nature, it should be equally valid for the case of planetary
motion also. Why then do we never speak of quantisation of orbits
of planets around the sun?
12.17 Obtain the first Bohr’s radius and the ground state energy of a
muonic hydrogen atom [i.e., an atom in which a negatively charged
muon (µ
–
) of mass about 207m
e
orbits around a proton].
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