Upload
mary-dunham
View
218
Download
0
Embed Size (px)
Citation preview
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 2/102
▪ The Atomic Models of Thomson andRutherford
▪ Rutherford Scattering
▪ The Classic Atomic Model
▪ The Bohr Model of the Hydrogen Atom
▪ Successes & Failures of the Bohr Model
▪ Characteristic X-Ray Spectra and Atomic
Number
▪ Atomic Excitation by Electrons
CHAPTER 4
Structure of the Atom
The opposite of a correct statement is a false statement. But the opposite of a profound
truth may well be another profound truth.
An expert is a person who has made all the mistakes that can be made in a very narrow
field.
Never express yourself more clearly than you are able to think.
Prediction is very difficult, especially about the future. - Niels Bohr
Niels Bohr (1885-1962)
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 3/102
History
• 450 BC, Democritus – The idea that matter is composed of
tiny particles, or atoms.
• XVII-th century, Pierre Cassendi, Robert Hook – explained
states of matter and transactions between them with a model
of tiny indestructible solid objects.
• 1811 – Avogadro’s hypothesis that all gases at giventemperature contain the same number of molecules per unit
volume.
• 1900 – Kinetic theory of gases.
Consequence – Great three quantization discoveries of XXcentury: (1) electric charge: (2) light energy; (3) energy of
oscillating mechanical systems.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 4/102
Historical Developments in Modern Physics • 1895 – Discovery of x-rays by Wilhelm Röntgen.
• 1896 – Discovery of radioactivity of uranium by HenriBecquerel
• 1897 – Discovery of electron by J.J.Thomson
• 1900 – Derivation of black-body radiation formula by MaxPlank.
• 1905 – Development of special relativity by Albert Einstein,and interpretation of the photoelectric effect.
• 1911 – Determination of electron charge by Robert Millikan.
• 1911 – Proposal of the atomic nucleus by Ernest Rutherford .
• 1913 – Development of atomic theory by Niels Bohr .• 1915 – Development of general relativity by Albert Einstein.
• 1924+ - Development of Quantum Mechanics by deBroglie,Pauli, Schrödinger, Born, Heisenberg, Dirac,….
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 5/102
There are 112 chemical elements that have
been discovered, and there are a couple ofadditional chemical elements that recently have
been reported.
Flerovium is the radioactive chemical element
with the symbol Fl and atomic number 114. The
element is named after Russian physicist GeorgyFlerov , the founder of the Joint Institute for Nuclear
Research in Dubna, Russia, where the element was
discovered.
Georgi Flerov (1913-1990)
The Structure of Atoms
The name was adopted by IUPAC on May 30, 2012. About 80 decays of atoms of
flerovium have been observed to date. All decays have been assigned to the fiveneighbouring isotopes with mass numbers 285–289. The longest-lived isotope
currently known is 289Fl with a half-life of ~2.6 s, although there is evidence for a
nuclear isomer, 289bFl, with a half-life of ~66 s, that would be one of the longest-
lived nuclei in the superheavy element region.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 6/102
The Structure of Atoms
Each element is characterized by atom thatcontains a number of protons Z , and equal number
of electrons, and a number of neutrons N . The
number of protons Z is called the atomic number .The lightest atom, hydrogen (H ), has Z=1; the
next lightest atom, helium (He), has Z=2 ; the third
lightest, lithium (Li ), has Z=3 and so forth.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 7/102
The Nuclear Atoms
Nearly all the mass of the atom is concentrated
in a tiny nucleus which contains the protons andneutrons.
Typically, the nuclear radius is approximately
from 1 fm to 10 fm (1fm = 10
-15
m). The distancebetween the nucleus and the electrons is
approximately 0.1 nm=100,000fm. This distance
determines the size of the atom.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 8/102
Nuclear Structure
An atom consists of an extremely small, positivelycharged nucleus surrounded by a cloud of negativelycharged electrons. Although typically the nucleus is lessthan one ten-thousandth the size of the atom, thenucleus contains more than 99.9% of the mass of theatom!
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 9/102
The number of protons in the
nucleus, Z , is called the atomic
number . This determines whatchemical element the atom is. The
number of neutrons in the nucleus
is denoted by N . The atomic mass
of the nucleus, A, is equal to Z + N .
A given element can have
many different isotopes, which
differ from one another by the
number of neutrons contained in
the nuclei. In a neutral atom, the
number of electrons orbiting the
nucleus equals the number of
protons in the nucleus.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 10/102
Structure of the Atom
Evidence in 1900 indicated that
the atom was not a fundamental unit:
1. There seemed to be too many kinds
of atoms, each belonging to a distinct chemical
element (way more than earth, air, water, and fire!).
2. Atoms and electromagnetic phenomena wereintimately related (magnetic materials; insulators vs.
conductors; different emission spectra).
3. Elements combine with some elements but not with
others, a characteristic that hinted at an internal atomicstructure (valence).
4. The discoveries of radioactivity, x-rays, and the
electron (all seemed to involve atoms breaking apart in
some way).
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 11/102
The Nuclear Atoms
We will begin our study of atoms bydiscussing some early models, developed in
beginning of 20 century to explain the spectra
emitted by hydrogen atoms.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 12/102
Atomic Spectra
By the beginning of the 20th century a large body of datahas been collected on the emission of light by atoms of
individual elements in a flame or in a gas exited by electricaldischarge.
Diagram of the spectrometer
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 13/102
Atomic Spectra
Light from the source passed through a narrow slit before
falling on the prism. The purpose of this slit is to ensure that allthe incident light strikes the prism face at the same angle so that
the dispersion by the prism caused the various frequencies that
may be present to strike the screen at different places with
minimum overlap.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 14/102
The source emits only two wavelengths, λ2>λ
1. The source is
located at the focal point of the lens so that parallel light passes
through the narrow slit, projecting a narrow line onto the face ofthe prism. Ordinary dispersion in the prism bends the shorter
wavelength through the lager total angel, separating the two
wavelength at the screen.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 15/102
In this arrangement each wavelength appears as a narrow line,
which is the image of the slit. Such a spectrum was dubbed aline spectrum for that reason. Prisms have been almost entirely
replaced in modern spectroscopes by diffraction gratings, which
have much higher resolving power.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 16/102
When viewed through the spectroscope,
the characteristic radiation, emitted byatoms of individual elements in flame or ingas exited by electrical charge, appears asa set of discrete lines, each of a particular
color or wavelength.
The positions and intensities of the lines
are a characteristic of the element. Thewavelength of these lines could bedetermined with great precision.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 17/102
Emission line spectrum of hydrogen in the visible and
near ultraviolet. The lines appear dark because thespectrum was photographed. The names of the first fivelines are shown. As is the point beyond which no linesappear, H
∞ called the limits of the series.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 18/102
Atomic Spectra
In 1885 a Swiss schoolteacher, Johann Balmer,
found that the wavelengths of the lines in the visiblespectrum of hydrogen can be represented by formula
Balmer suggested that this might be a specialcase of more general expression that would be
applicable to the spectra of other elements.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 19/102
Atomic Spectra
Such an expression, found by J.R.Rydberg
and W. Ritz and known as the Rydberg-Ritzformula, gives the reciprocal wavelengths as:
where m and n are integers with n>m, and R is
the Rydberg constant .
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 20/102
Atomic Spectra
The Rydberg constant is the same for all
spectral series.
For hydrogen the R H
= 1.096776 x 10 7 m-1.
For very heavy elements R approaches the
value R ∞ = 1.097373 x 10 7 m-1. Such empirical expressions were successful
in predicting other spectra, such as other
hydrogen lines outside the visible spectrum.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 21/102
Atomic Spectra
So, the hydrogen Balmer series wavelength are
those given by Rydberg equation
with m=2 and n=3,4,5,…
Other series of hydrogen spectral lines were foundfor m=1 (by Lyman) and m=3 (by Paschen).
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 22/102
Hydrogen Spectral Series
Compute the wavelengths of the first lines
of the Lyman, Balmer , and Paschen series.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 23/102
Emission line spectrum of hydrogen in the visible and
near ultraviolet. The lines appear dark because thespectrum was photographed. The names of the first fivelines are shown, as is the point beyond which no linesappear, H
∞ called the limits of the series.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 24/102
The Limits of Series
Find the predicted by Rydberg-Ritz
formula for Lyman, Balmer, and Paschenseries.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 25/102
A portion of the emission spectrum of sodium. The two
very close bright lines at 589 nm are the D1 and D
2 lines.
They are the principal radiation from sodium street
lighting.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 26/102
A portion of emission spectrum of mercury.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 27/102
Part of the dark line (absorption) spectrum of sodium.White light shining through sodium vapor is absorbed at
certain wavelength, resulting in no exposure of the film atthose points. Note that frequency increases toward theright , wavelength toward the left in the spectra shown.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 28/102
Nuclear Models
Many attempts were made to construct a
model of the atom that yielded the Balmer and
Rydberg-Ritz formulas.
It was known that an atom was about 10-10m
in diameter, that it contained electrons muchlighter than the atom, and that it was electrically
neutral.
The most popular model was that of J.J.
Thomson, already quite successful in explainingchemical reactions.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 29/102
Knowledge of atoms in 1900
Electrons (discovered in1897) carried the negative
charge.
Electrons were very light,
even compared to the atom.Protons had not yet been
discovered, but clearly
positive charge had to be
present to achieve chargeneutrality.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 30/102
In Thomson’s view, when the atom was heated, the
electrons could vibrate about their equilibrium positions,thus producing electromagnetic radiation.
Unfortunately, Thomson couldn’t explain spectra with this
model.
Thomson’s Atomic Model
Thomson’s “plum-pudding”
model of the atom had the
positive charges spread
uniformly throughout a sphere
the size of the atom, with
electrons embedded in theuniform background.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 31/102
The difficulty with all such models was that
electrostatic forces alone cannot produce stable
equilibrium. Thus the charges were required to move
and, if they stayed within the atom, to accelerate.However, the acceleration would result in continuous
radiation, which is not observed.
Thomson was unable to obtain from his model a set of
frequencies that corresponded with the frequencies of
observed spectra.
The Thomson model of the atom was replaced by
one based on results of a set of experimentsconducted by Ernest Rutherford and his student H.W.
Geiger .
E i t f G i d M d
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 32/102
Experiments of Geiger and Marsden
Rutherford, Geiger, andMarsden conceived a new
technique for investigating
the structure of matter by
scattering α particles from
atoms.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 33/102
Rutherford was investigating radioactivity
and had shown that the radiations from uranium
consist of at least two types, which he labeled α and β.
He showed, by an experiment similar to that
of Thompson, that q /m for the α - particles was
half that of the proton.Suspecting that the α particles were double
ionized helium, Rutherford in his classical
experiment let a radioactive substance decay
and then, by spectroscopy, detected the spectra
line of ordinary helium.
Beta decay
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 34/102
Beta decay Beta decay occurs when the neutron to proton ratio is too
great in the nucleus and causes instability. In basic beta
decay, a neutron is turned into a proton and an electron.The electron is then emitted. Here's a diagram of beta
decay with hydrogen-3:
Beta Decay of Hydrogen-3 to Helium-3.
Alpha Decay
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 35/102
Alpha DecayThe reason alpha decay occurs is because the nucleus has too many
protons which cause excessive repulsion. In an attempt to reduce the
repulsion, a Helium nucleus is emitted. The way it works is that the Helium
nuclei are in constant collision with the walls of the nucleus and because
of its energy and mass, there exists a nonzero probability of transmission.
That is, an alpha particle (Helium nucleus) will tunnel out of the nucleus.
Here is an example of alpha emission with americium-241:
Alpha Decay of Americium-241 to Neptunium-237
G D
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 36/102
Gamma DecayGamma decay occurs because the nucleus is at too high
an energy. The nucleus falls down to a lower energy state
and, in the process, emits a high energy photon known as agamma particle. Here's a diagram of gamma decay with
helium-3:
Gamma Decay of Helium-3
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 37/102
Rutherford was investigating radioactivity
and had shown that the radiations from uranium
consist of at least two types, which he labeled α and β.
He showed, by an experiment similar to that
of Thompson, that q /m for the α - particles was
half that of the proton.Suspecting that the α particles were double
ionized helium, Rutherford in his classical
experiment let a radioactive substance decay
and then, by spectroscopy, detected the spectra
line of ordinary helium.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 38/102
Schematic diagram of the Rutherford apparatus. The beam
of α - particles is defined by the small hole D in the shieldsurrounding the radioactive source R of 214Bi . The α beamstrikes an ultra thin gold foil F , and α particles are individuallyscattered through various angels θ . The experiment consistedof counting the number of scintillations on the screen S as a
function of θ .
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 39/102
A diagram of the original apparatus as it appear
in Geiger’s paper describing the results.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 40/102
An α -particle by such an atom (Thompson model) would have
a scattering angle θ much smaller than 10 . In the Rutherford’s
scattering experiment most of the α -particles were eitherundeflected, or deflected through very small angles of the
order 10 , however, a few α -particles were deflected through
angles of 90 0 and more.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 41/102
An α -particle by such an atom (Thompson model) would have
a scattering angle θ much smaller than 10 . In the Rutherford’s
scattering experiment a few α -particles were deflected through
angles of 90 0 and more.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 42/102
Experiments of Geiger and Marsden
Geiger showed that many α particles were scattered from
thin gold-leaf targets at backward angles greater than 90°.
B f Aft
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 43/102
Electrons
can’t back-
scatter α
particles.
Calculate the maximum scattering angle - corresponding to the
maximum momentum change.
It can be shown that the maximummomentum transfer to the α particle is:
Determine θ max
by letting
Δ pmax
be perpendicular
to the direction of motion:
Before After
too small!
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 44/102
If an α particle is scattered by N electrons:
Try multiple scattering from electrons
The distance between atoms, d = n-1/3
, is:
N = the number of atoms across the thin gold layer , t = 6 × 10−7 m:
still too small!
n =
N = t / d
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 45/102
If the atom consisted of a positively charged
sphere of radius 10 -10 m, containing electrons as in
the Thomson model, only a very small scatteringdeflection angle could be observed.
Such model could not possibly account for the
large angles scattering. The unexpected large
angles α -particles scattering was described byRutherford with these words:
It was quite incredible event that ever
happened to me in my life. It was as incredibleas if you fired a 15-inch shell at a piece of tissue
paper and it came back and hit you.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 46/102
even if the α
particle is scattered from all 79electrons in each atom of gold.
Experimental results were not
consistent with Thomson’s atomic
model.
Rutherford proposed that an atom
has a positively charged core
(nucleus!) surrounded by the
negative electrons.
Geiger and Marsden confirmed theidea in 1913.
Rutherford’s Atomic Model
Ernest Rutherford
(1871-1937)
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 47/102
Rutherford concluded that the large angle
scattering could result only from a single encounter
of the α particle with a massive charge with volume
much smaller than the whole atom.
Assuming this “nucleus” to be a point charge,
he calculated the expected angular distribution forthe scattered α particles.
His predictions on the dependence of scatteringprobability on angle, nuclear charge and kinetic
energy were completely verified in experiments.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 48/102
Rutherford Scattering geometry. The nucleus is assumed to
be a point charge Q at the origin O . At any distance r the α particle experiences a repulsive force kq
α Q/r 2 . The α particle
travel along a hyperbolic path that is initially parallel to line
OA a distance b from it and finally parallel to OB, which
makes an angle θ with OA.
Rutherford Scattering
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 49/102
There’s a relationship
between the impact
parameter b and thescattering angle θ .
Rutherford Scattering
When b is small,
r is small.
the Coulomb force is large.
θ can be large and the particle can be
repelled backward.
where
cot(θ /2)
θ π 0
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 50/102
Any particle inside
the circle of area π
b0
2 will be similarly (or
more) scattered.
Rutherford
Scattering
The cross section σ = π b2 is related to
the probability for a particle being
scattered by a nucleus (t = foil thickness):
The fraction of incident
particles scattered is:
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 51/102
Force on a point charge versus distance r from the center
of a uniformly charged sphere of radius R . Outside thesphere the force is proportional to Q/r 2 , inside the sphere
the force is proportional to q I /r 2 = Qr/R 3, where q I = Q(r/R)3
is the charge within a sphere of radius r . The maximum
force occurs at r =R
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 52/102
Two α particles with equal kinetic energiesapproach the positive charge Q = +Ze withimpact parameters b
1 and b
2 , where b
1<b
2 .
According to equation for impact parameter inthis case θ
1
> θ 2
.
The path of α particle can be shown to be a
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 53/102
The path of α particle can be shown to be a
hyperbola, and the scattering angle θ can be relate to
the impact parameter b from the laws of classical
mechanics.
The quantity πb2 , which has the dimension of the
area , is called the cross section σ for scattering.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 54/102
The cross section σ
is thus defined as the
number of particles
scattered per nucleusper unit time divided
by the incident
intensity.
The total number of nuclei of foil atoms in the area coveredby the beam is nAt , where n is the number of foil atoms per
unit volume. A is the area of the beam, and t is the thickness
of the foil.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 55/102
The total number of particles scattered per second isobtained by multiplying πb2 I
0 by the number of nuclei in the
scattering foil. Let n be the number of nuclei per unit volume:
For a foil of thickness t , the total number of nuclei as “seen”
by the beam is nAt , where A is the area of the beam. The totalnumber scattered per second through angles grater than θ isthus πb2 I
0 ntA. If divide this by the number of α particles
incident per second I 0 A we get the fraction of α particles f
scattered through angles grater than θ :
f = πb2 nt
On the base of that nuclear model Rutherford
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 56/102
On the base of that nuclear model Rutherford
derived an expression for the number of α particles
ΔN that would be scattered at any angle θ :
All this predictions was verified in Geiger
experiments, who observe several hundredsthousands α particles.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 57/102
(a)Geiger data for α scattering from thin Au and Ag foils.
The graph is in log-log plot to cover over several orders ofmagnitudes.
(b)Geiger also measured the dependence of ΔN on t fordifferent elements, that was also in good agreement with
Rutherford formula.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 58/102
Data from Rutherford’s group showing observed α
scattering at a large fixed angle versus values of r d
computed from for various kinetic energies.
Th Si f th N l
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 59/102
The Size of the Nucleus
This equation can be used to estimate the
size of the nucleus
For the case of 7.7-MeV α particles the
distance of closest approach for a head-on collision is
The Classical Atomic Model
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 60/102
Consider an atom as a planetary system.
The Newton’s 2nd Law force of attraction
on the electron by the nucleus is:
where v is the tangential velocity of the electron:
The total energy is then:
This is negative, so
the system is bound,
which is good.
The Planetary Model is Doomed
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 61/102
The Planetary Model is Doomed
From classical E&M theory, an accelerated electric
charge radiates energy (electromagnetic radiation),which means the total energy must decrease. So the
radius r must decrease!!
Electron
crashes
into the
nucleus!?
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 62/102
According to classical physics, a charge e moving with anacceleration a radiates at a rate
(a) Show that an electron in a classical hydrogen atom spiralsinto the nucleus at a rate
(b) Find the time interval over
which the electron will reach r = 0,starting from r 0= 2.00 × 10 –10 m.
The Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 63/102
The Bohr s Postulates
Bohr overcome the difficulty of the collapsing atom by
postulating that only certain orbits, called stationary
states, are allowed, and that in these orbits the electrondoes not radiate. An atom radiates only when the electron
makes a transition from one allowed orbit (stationary state)
to another:
▪ 1. The electron in the hydrogen atom can move only in
certain nonradiating, circular orbits called stationary
states.
▪ 2. The photon frequency from energy conservation is
given by
where E i and E
f are the energy of initial and final state, h is the
Plank’s constant.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 64/102
Such a model is mechanically stable , because the
Coulomb potential
provides the centripetal force
The total energy for a such system can be written as the
sum of kinetic and potential energy:
The Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 65/102
The Bohr s Postulates
Combining the second postulate with the
equation for the energy we obtain:
where r 1 and r
2 are the radii of the initial and
final orbits.
The Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 66/102
The Bohr s Postulates
To obtain the frequencies implied by the
perimental Rydberg-Ritz formula,
it is evident that the radii of the stable orbits must be
proportional to the squares of integers.
The Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 67/102
The Bohr s Postulates
Bohr found that he could obtain this condition if he
postulates the angular momentum of the electron in astable orbit equals an integer times ħ=h/2π . Since the
angular momentum of a circular orbit is just mvr , the
third Bohr’s postulate is:
3. n=1,2,3……….
where ħ=h/2π=1.055 x 10 -34J·s=6.582x10 -16 eV·s
The Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 68/102
The Bohr s Postulates
The obtained equation mvr = nh/2π=nħ relates the
speed of electron v to the radius r . Since we had
or
We can write
or
The Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 69/102
The Bohr s Postulates
Solving for r , we obtain
where a0 is called the first Bohr’s radius
Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 70/102
Substituting the expression for r in equation for
frequency:
If we will compare this expression with Z=1 for f=c/λ
with the empirical Rydberg-Ritz formula:
we will obtain for the Rydberg constant
Bohr’s Postulates
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 71/102
Using the known values of m, e, and ħ, Bohr
calculated R and found his results to agree with the
spectroscopy data.The total mechanical energy of the electron in the
hydrogen atom is related to the radius of the circular
orbit
Energy levels
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 72/102
Energy levels
If we will substitute the quantized value of r
as given by
we obtain
Energy levels
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 73/102
or
where
The energies E n with Z=1 are the quantized allowed
energies for the hydrogen atom.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 74/102
Energy level diagram for hydrogen showing the seven
lowest stationary states. The energies of infinite number of
levels are given by E n
= (-13.6/n2 )eV , where n is an integer.
A hydrogen atom is in its first excited state (n = 2). Using the
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 75/102
y g ( ) g
Bohr theory of the atom, calculate (a) the radius of the orbit,
(b) the linear momentum of the electron, (c) the angular
momentum of the electron, (d) the kinetic energy of theelectron, (e) the potential energy of the system, and (f) the
total energy of the system.
Energy levels
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 76/102
gy
Transitions between this allowed energies
result in the emission or absorption of aphoton whose frequency is given by
and whose wavelength is
Energy levels
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 77/102
gy
Therefore is convenient to have the value
of hc in electronvolt nanometers! hc = 1240 eV∙nm
Since the energies are quantized, the
frequencies and the wavelengths of theradiation emitted by the hydrogen atom are
quantized in agreement with the observed
line spectrum.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 78/102
(a) In the classical orbital model , the electron orbits
about the nucleus and spirals into the center because ofthe energy radiated.(b) In the Bohr model, the electron orbits without
radiating until it jumps to another allowed radius of lowerenergy, at which time radiation is emitted.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 79/102
Energy level diagram for hydrogen showing the seven loweststationary states and the four lowest energy transitions for theLyman, Balmer, and Pashen series. The energies of infinitenumber of levels are given by E
n = (-13.6/n2 )eV , where n is an
integer.
λ21
=hc / (E1-E
2)
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 81/102
The spectral lines corresponding to the
transitions shown for the three series.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 82/102
λ21
=hc / (E1-E
2)
Compute the wavelength of the Hβ
spectral line of the
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 83/102
Compute the wavelength of the Hβ
spectral line of the
Balmer series predicted by Bohr model.
A hydrogen atom at rest in the laboratory emits a photon of
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 84/102
yd oge ato at est t e abo ato y e ts a p oto o
the Lyman α radiation. (a) Compute the recoil kinetic energy
of the atom. (b) What fraction of the excitation energy of the
n = 2 state is carried by the recoiling atom? (Hint: Useconservation of momentum.)
In a hot star, because of the high temperature, an atom can
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 85/102
absorb sufficient energy to remove several electrons from the
atom. Consider such a multiply ionized atom with a single
remaining electron. The ion produces a series of spectrallines as described by the Bohr model. The series
corresponds to electronic transitions that terminate in the
same final state. The longest and shortest wavelengths of the
series are 63.3 nm and 22.8 nm, respectively. (a) What is the
ion? (b) Find the wavelengths of the next three spectral linesnearest to the line of longest wavelength.
A stylized picture of Bohrf 1 2 3
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 86/102
circular orbits for n=1,2,3,4.The radii r
n≈n2 .
In high Z -elements(elements with Z ≥12 ),electrons are distributed overall the orbits shown. If anelectron in the n=1 orbit is
knocked from the atom (e.g.,by being hit by a fast electronaccelerated by the voltageacross an x-ray tube) thevacancy that produced is filed
by an electron of higher energy(i.e., n=2 or higher).
The difference in energy between the two orbits is emitted as aphoton, whose wavelength will be in the x-ray region of thespectrum, if Z is large enough.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 87/102
Characteristic x-ray spectra. (a) Part the spectra of neodymium
(Z=60) and samarium (Z=62).The two pairs of bright lines arethe Kα and K β lines. (b) Part of the spectrum of the artificiallyproduced element promethium (Z=61), its Kα and K β lines fallbetween those of Nd and Sm. (c) Part of the spectrum of allthree elements.
(a)
(b)
(c)
The Franck-Hertz Experiment
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 88/102
While investigating the inelastic scattering of
electrons, J.Frank and G.Hertz in 1914 performed
an important experiment that confirmed by direct
measurement Bohr’s hypothesis of energy
quantization in atoms.
The experiment involved measuring the plate
current as a function of V 0 .
The Franck-Hertz Experiment
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 89/102
Schematic diagram of the
Franck-Hertz experiment.
Electrons ejected from theheated cathode C at zero
potential are drawn to the
positive grid G . Those
passing through the holes
in the grid can reach theplate P and contribute in
the current I , if they have
sufficient kinetic energy to
overcome the small backpotential ΔV . The tube
contains a low pressure
gas of the element being
studied.
The Franck-Hertz Experiment
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 90/102
p
As V 0 increased from 0, the currentincreases until a critical value (about 4.9 V for
mercury) is reached. At this point the current
suddenly decreases. As V 0 is increased further,
the current rises again. They found that when the electron’s
kinetic energy was 4.9 eV or greater, the vapor
of mercury emitted ultraviolet light of wavelength
0.25 μm.
Current versus acceleration voltage in the Franck-Hertz
experiment
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 91/102
The currentdecreases becausemany electrons lose
energy due toinelastic collisionswith mercury atomsin the tube andtherefore cannot
overcome the smallback potential.
C u r r e n t , ( m A m p )
experiment.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 92/102
The regular spacing ofthe peaks in this curve
indicates that only acertain quantity of energy,4.9 eV, can be lost to themercury atoms. Thisinterpretation is confirmed
by the observation ofradiation of photon energy4.9 eV, emitted bymercury atoms.
C u r r e n t , ( m A m p )
Suppose mercury atoms have an energy level
4 9 V b h l l l A
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 93/102
4.9 eV above the lowest energy level. An atom can
be raised to this level by collision with an electron; it
later decays back to the lowest energy level byemitting a photon. The wavelength of the photon
should be
This is equal to the measured wavelength,
confirming the existence of this energy level of themercury atom.
Similar experiments with other atoms yield the
same kind of evidence for atomic energy levels.
Lets consider an experimental tube filled by
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 94/102
p y
hydrogen atoms instead of mercury. Electrons
accelerated by V 0
that collide with hydrogen
electrons cannot transfer the energy to letter
electrons unless they have acquired kinetic
energy
eV0=E
2 – E
1=10.2eV
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 95/102
If the incoming electron does not have
sufficient energy to transfer ΔE = E 2 - E 1 to thehydrogen electron in the n=1 orbit (ground state),
than the scattering will be elastic.
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 96/102
If the incoming electron does have at least ΔE
kinetic energy, then an inelastic collision can occur inwhich ΔE is transferred to the n=1 electron, moving itto the n=2 orbit. The excited electron will typicallyreturn to the ground state very quickly, emitting a
photon of energy ΔE .
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 97/102
Energy loss spectrum measurement. A well-definedelectron beam impinges upon the sample. Electronsinelastically scattered at a convenient angle enter the slit of the
magnetic spectrometer, whose B field is directed out of thepage, and turn through radii R determined by their energy (E
inc
– E 1 ) via equation
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 98/102
An energy-loss spectrum for a thin Al film.
Reduced mass correction
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 99/102
The assumption by Bohr that the nucleus is fixed is
equivalent the assumption that it has an infinity mass.
If instead we will assume that proton and electron both
revolve in circular orbits about their common center of mass we
will receive even better agreement for the values of the
Rydberg constant R and ionization energy for the hydrogen.
We can take in account the motion of the nucleus (the
proton) very simply by using in Bohr’s equation not the electron
rest mass m but a quantity called the reduce mass μ of the
system. For a system composed from two masses m1
and m2
the reduced mass is defined as:
Reduced mass correction
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 100/102
If the nucleus has the mass M its kinetic energy
will be ½Mv 2 = p2 /2M , where p = Mv is themomentum.
If we assume that the total momentum of the
atom is zero, from the conservation of momentumwe will have that momentum of electron and
momentum of nucleus are equal on the magnitude.
Reduced mass correction
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 101/102
The total kinetic energy is then:
The Rydberg constant equation than changed to:
The factor μ was called mass correction factor .
As the Earth moves around the Sun, its orbits are quantized. (a) Follow the
steps of Bohr’s analysis of the hydrogen atom to show that the allowed radii
8/10/2019 Nuclear Atom
http://slidepdf.com/reader/full/nuclear-atom 102/102
steps of Bohr s analysis of the hydrogen atom to show that the allowed radii
of the Earth’s orbit are given by
where MS is the mass of the Sun, M
E is the mass of the Earth, and n is an
integer quantum number. (b) Calculate the numerical value of n. (c) Find the
distance between the orbit for quantum number n and the next orbit out fromthe Sun corresponding to the quantum number n + 1