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Structure of the Atom
Structure of the Atom
The Atomic Models of Thomson and
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
Atomic Excitation by Electrons
Niels Bohr (1885-1962)
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
Never express yourself more clearly than you are able to think.
Prediction is very difficult, especially about the future.
- Niels Bohr
• 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 given
temperature contain the same number of molecules per unit
• 1900 – Kinetic theory of gases.
Consequence – Great three quantization discoveries of XX
century: (1) electric charge: (2) light energy; (3) energy of
oscillating mechanical systems.
Historical Developments in Modern Physics
• 1895 – Discovery of x-rays by Wilhelm Röntgen.
• 1896 – Discovery of radioactivity of uranium by Henri
• 1897 – Discovery of electron by J.J.Thomson
• 1900 – Derivation of black-body radiation formula by Max
• 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,….
The Structure of Atoms
There are 112 chemical elements that have
been discovered, and there are a couple of
additional 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
Georgy Flerov, the founder of the Joint Institute for
Nuclear Research in Dubna, Russia, where the
element was discovered.
Georgi Flerov (1913-1990)
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 five
neighbouring 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 longestlived nuclei in the superheavy element region.
The Structure of Atoms
Each element is characterized by atom that
contains 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.
The Nuclear Atoms
Nearly all the mass of the atom is concentrated
in a tiny nucleus which contains the protons and
Typically, the nuclear radius is approximately
from 1 fm to 10 fm (1fm = 10-15m). The distance
between the nucleus and the electrons is
approximately 0.1 nm=100,000fm. This distance
determines the size of the atom.
Nuclear Structure
An atom consists of an extremely small, positively
charged nucleus surrounded by a cloud of negatively
charged electrons. Although typically the nucleus is less
than one ten-thousandth the size of the atom, the
nucleus contains more than 99.9% of the mass of the
The number of protons in the
nucleus, Z, is called the atomic
number. This determines what
chemical 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.
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 were
intimately 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 atomic
structure (valence).
4. The discoveries of radioactivity, x-rays, and the
electron (all seemed to involve atoms breaking apart in
some way).
The Nuclear Atoms
We will begin our study of atoms by
discussing some early models, developed in
beginning of 20 century to explain the spectra
emitted by hydrogen atoms.
Atomic Spectra
By the beginning of the 20th century a large body of data
has been collected on the emission of light by atoms of
individual elements in a flame or in a gas exited by electrical
Diagram of the spectrometer
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 all
the 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.
The source emits only two wavelengths, λ2>λ1. 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 of
the prism. Ordinary dispersion in the prism bends the shorter
wavelength through the lager total angel, separating the two
wavelength at the screen.
In this arrangement each wavelength appears as a narrow line,
which is the image of the slit. Such a spectrum was dubbed a
line spectrum for that reason. Prisms have been almost entirely
replaced in modern spectroscopes by diffraction gratings, which
have much higher resolving power.
When viewed through the
spectroscope, the characteristic radiation,
emitted by atoms of individual elements in
flame or in gas exited by electrical charge,
appears as a set of discrete lines, each of a
particular color or wavelength.
The positions and intensities of the lines
are a characteristic of the element. The
wavelength of these lines could be
determined with great precision.
Emission line spectrum of hydrogen in the visible and
near ultraviolet. The lines appear dark because the
spectrum was photographed. The names of the first five
lines are shown. As is the point beyond which no lines
appear, H∞ called the limits of the series.
Atomic Spectra
In 1885 a Swiss schoolteacher, Johann Balmer,
found that the wavelengths of the lines in the visible
spectrum of hydrogen can be represented by formula
 n 
  364.6 2
 nm
 (n  4) 
n  3,4,5,......
Balmer suggested that this might be a special
case of more general expression that would be
applicable to the spectra of other elements.
Atomic Spectra
Such an expression, found by J.R.Rydberg
and W. Ritz and known as the Rydberg-Ritz
formula, gives the reciprocal wavelengths as:
1 
 1
 R
 2 
n 
where m and n are integers with n>m, and R is
the Rydberg constant.
Atomic Spectra
The Rydberg constant is the same for all
spectral series.
For hydrogen the RH = 1.096776 x 107m-1.
For very heavy elements R approaches the
value R∞ = 1.097373 x 107m-1.
Such empirical expressions were successful
in predicting other spectra, such as other
hydrogen lines outside the visible spectrum.
Atomic Spectra
So, the hydrogen Balmer series wavelength are
those given by Rydberg equation
1 
 1
 R
 2 
n 
with m=2 and n=3,4,5,…
Other series of hydrogen spectral lines were found
for m=1 (by Lyman) and m=3 (by Paschen).
Hydrogen Spectral Series
Compute the wavelengths of the first lines
of the Lyman, Balmer, and Paschen series.
Emission line spectrum of hydrogen in the visible and
near ultraviolet. The lines appear dark because the
spectrum was photographed. The names of the first five
lines are shown, as is the point beyond which no lines
appear, H∞ called the limits of the series.
The Limits of Series
Find the  predicted by Rydberg-Ritz
formula for Lyman, Balmer, and Paschen
A portion of the emission spectrum of sodium. The two
very close bright lines at 589 nm are the D1 and D2 lines.
They are the principal radiation from sodium street
A portion of emission spectrum of mercury.
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 at
those points. Note that frequency increases toward the
right , wavelength toward the left in the spectra shown.
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 much
lighter than the atom, and that it was electrically
The most popular model was that of
J.J.Thomson, already quite successful in
explaining chemical reactions.
Knowledge of atoms in 1900
1897) carried the negative
Electrons were very light,
even compared to the atom.
Protons had not yet been
positive charge had to be
present to achieve charge
Thomson’s Atomic Model
Thomson’s “plum-pudding”
model of the atom had the
uniformly throughout a sphere
the size of the atom, with
electrons embedded in the
uniform background.
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
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 experiments
conducted by Ernest Rutherford and his student
Experiments of Geiger and Marsden
Rutherford, Geiger, and
Marsden conceived a new
technique for investigating
the structure of matter by
scattering a particles from
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
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
The 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
Gamma Decay
Gamma 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 a
gamma particle. Here's a diagram of gamma decay with
Gamma Decay of Helium-3
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.
Schematic diagram of the Rutherford apparatus. The beam
of α - particles is defined by the small hole D in the shield
surrounding the radioactive source R of 214Bi . The α beam
strikes an ultra thin gold foil F, and α particles are individually
scattered through various angels θ. The experiment consisted
of counting the number of scintillations on the screen S as a
function of θ.
A diagram of the original apparatus as it appear
in Geiger’s paper describing the results.
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 either
undeflected, or deflected through very small angles of the
order 10, however, a few α-particles were deflected through
angles of 900 and more.
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 900 and more.
Experiments of Geiger and Marsden
Geiger showed that many a particles were scattered from
thin gold-leaf targets at backward angles greater than 90°.
can’t backscatter a
Calculate the maximum scattering angle - corresponding to the
maximum momentum change.
It can be shown that the maximum
momentum transfer to the a particle is:
Dpmax  2me va
Determine qmax by letting
Dpmax be perpendicular
to the direction of motion:
q max
Dpa 2me va
M a va
too small!
Try multiple scattering from electrons
If an a particle is scattered by N electrons:
N = the number of atoms across the thin gold layer, t = 6 × 10−7 m:
The distance between atoms, d = n-1/3, is:
still too small!
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 scattering
deflection angle could be observed.
Such model could not possibly account for the
large angles scattering. The unexpected large
angles α-particles scattering was described by
Rutherford with these words:
It was quite incredible event that ever
happened to me in my life. It was as incredible
as if you fired a 15-inch shell at a piece of tissue
paper and it came back and hit you.
Rutherford’s Atomic Model
even if the α
particle is scattered from all 79
electrons in each atom of gold.
Experimental results were not
consistent with Thomson’s atomic
Rutherford proposed that an atom
has a positively charged core
(nucleus!) surrounded by the
negative electrons.
Geiger and Marsden confirmed the
idea in 1913.
Ernest Rutherford
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 for
the scattered α particles.
His predictions on the dependence of scattering
probability on angle, nuclear charge and kinetic
energy were completely verified in experiments.
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 kqαQ/r2. 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.
Force on a point charge versus distance r from the center
of a uniformly charged sphere of radius R. Outside the
sphere the force is proportional to Q/r2, inside the sphere
the force is proportional to qI/r2= Qr/R3, where qI = Q(r/R)3
is the charge within a sphere of radius r. The maximum
force occurs at r =R
Two α particles with equal kinetic energies
approach the positive charge Q = +Ze with
impact parameters b1 and b2, where b1<b2.
According to equation for impact parameter in
this case θ1 > θ2.
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
qa Q
ma v
The quantity πb2, which has the dimension of the
area , is called the cross section σ for scattering.
The cross section σ
is thus defined as the
number of particles
scattered per nucleus
per unit time divided
The total number of nuclei of foil atoms in the area covered
by 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.
The total number of particles scattered per second is
obtained by multiplying πb2I0 by the number of nuclei in the
scattering foil. Let n be the number of nuclei per unit volume:
 g   at 
 3 N A
mol  N A at
M cm3
 g 
 mol 
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 total
number scattered per second through angles grater than θ is
thus πb2I0ntA. If divide this by the number of α particles
incident per second I0A we get the fraction of α particles f
scattered through angles grater than θ:
f = πb2nt
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 :
DN 
I 0 Ant
 kZe
 2 Ek
  1
 sin
All this predictions was verified in Geiger
experiments, who observe several hundreds
thousands α particles.
(a)Geiger data for α scattering from thin Au and Ag foils. The
graph is in log-log plot to cover over several orders of
(b)Geiger also measured the dependence of ΔN on t for
different elements, that was also in good agreement with
Rutherford formula.
Data from Rutherford’s group showing observed α
scattering at a large fixed angle versus values of rd
qa Q for various kinetic energies.
computed from
rd  k
ma v 2
The Size of the Nucleus
This equation can be used to estimate the
size of the nucleus
qa Q
rd  k
ma v
For the case of 7.7-MeV α particles the
distance of closest approach for a head-on collision is
(2)(79)(1.44eV  nm)
rd 
7.7 10 eV
The Classical Atomic Model
Consider an atom as a planetary
The Newton’s 2nd Law force of attraction
on the electron by the nucleus is:
1 e
Fe 
4 0 r
where v is the tangential velocity of the electron:
4 0 mr
4 0 r
The total energy is then:
This is negative, so
the system is bound,
which is good.
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!!
into the
According to classical physics, a charge e moving with an
acceleration a radiates at a rate
1 e2a 2
 
6  0 c 3
(a) Show that an electron in a classical hydrogen atom spirals
into the nucleus at a rate
 
2 2
2 3
1 2  0 r m e c
(b) Find the time interval over
which the electron will reach r = 0,
starting from r0 = 2.00 × 10–10 m.
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 electron
does 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
 2. The photon frequency from energy conservation is
given by
f 
Ei  E f
where Ei and Ef are the energy of initial and final state, h is the
Plank’s constant.
Such a model is mechanically stable , because the Coulomb
U  k
provides the centripetal force
k Ze
F  2 
k Ze
mv 
The total energy for a such system can be written as the
sum of kinetic and potential energy:
k Ze k Ze
k Ze
The Bohr’s Postulates
Combining the second postulate with the
equation for the energy we obtain:
E1  E 2 1 k Ze  1 1 
  
f 
2 h  r2 r1 
where r1 and r2 are the radii of the initial and
final orbits.
The Bohr’s Postulates
E1  E 2 1 k Ze  1 1 
  
f 
2 h  r2 r1 
To obtain the frequencies implied by the
experimental Rydberg-Ritz formula,
 1
1 
f   cR  2  2 
 n 2 n1 
it is evident that the radii of the stable orbits must be
proportional to the squares of integers.
The Bohr’s Postulates
Bohr found that he could obtain this condition if he
postulates the angular momentum of the electron in a
stable 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:
mvr 
 n
where ħ=h/2π=1.055 x 10-34J·s=6.582x10 -16eV·s
The Bohr’s Postulates
The obtained equation mvr = nh/2π=nħ relates the
speed of electron v to the radius r. Since we had
v 
We can write
v n 2 2
m r
k Ze
n 2 2 
m r
The Bohr’s Postulates
Solving for r, we obtain
2 a0
m k Ze
where a0 is called the first Bohr’s radius
a0 
 0 . 0529 nm
Bohr’s Postulates
Substituting the expression for r in equation for
2 4
1 k Ze 1 1
2 mk e
    Z
f 
2 h  r2 r1 
4 
 1
1 
 2  2 
 n 2 n1 
If we will compare this expression with Z=1 for f=c/λ
with the empirical Rydberg-Ritz formula:
 1
1 
 R  2  2 
 n 2 n1 
2 4
we will obtain for the Rydberg constant R  mk e
4 c 
Bohr’s Postulates
2 4
mk e
4c 3
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
1 kZe2
2 r
Energy levels
If we will substitute the quantized value of r
as given by
2 a0
we obtain
2 2
2 4
1 kZe
1 kZ e
1 mk Z e
En  
2 2
2 r
2 n a0
2 n
Energy levels
1 kZe2
1 kZ 2e 2
1 mk 2 Z 2e 4
En  
2 r
2 n a0
2 n 2 2
En   Z 2
mk 2e 4 1 ke2
E0 
 13.6eV
2 a0
The energies En with Z=1 are the quantized allowed
energies for the hydrogen atom.
Energy level diagram for hydrogen showing the seven lowest
stationary states. The energies of infinite number of levels
are given by En = (-13.6/n2)eV, where n is an integer.
A hydrogen atom is in its first excited state (n = 2). Using the
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 the
electron, (e) the potential energy of the system, and (f) the
total energy of the system.
Energy levels
Transitions between this allowed energies
result in the emission or absorption of a
photon whose frequency is given by
f 
Ei  E f
and whose wavelength is
 
f Ei  E f
Energy levels
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 the
radiation emitted by the hydrogen atom are
quantized in agreement with the observed
line spectrum.
(a) In the classical orbital model, the electron orbits
about the nucleus and spirals into the center because of
the energy radiated.
(b) In the Bohr model, the electron orbits without
radiating until it jumps to another allowed radius of lower
energy, at which time radiation is emitted.
λ21=hc / (E1-E2)
Energy level diagram for hydrogen showing the seven lowest
stationary states and the four lowest energy transitions for the
Lyman, Balmer, and Pashen series. The energies of infinite
number of levels are given by En = (-13.6/n2)eV, where n is an
1240eV  nm
12 
 121.5nm
E1  E2 (13.6  3.4)eV
1 
 1
 R 2  2   R1  2   1.096776 107 m 1 (0.75)
m n 
 2 
12  1.2156 10 7 m  121.5nm
The spectral lines corresponding to the
transitions shown for the three series.
λ21=hc / (E1-E2)
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
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: Use
conservation of momentum.)
In a hot star, because of the high temperature, an atom can
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 spectral
lines 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 lines
nearest to the line of longest wavelength.
A stylized picture of Bohr
circular orbits for n=1,2,3,4.
The radii rn≈n2.
In high Z-elements
(elements with Z ≥12),
electrons are distributed over
all the orbits shown. If an
electron in the n=1 orbit is
knocked from the atom (e.g.,
by being hit by a fast electron
accelerated by the voltage
across an x-ray tube) the
vacancy 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 a
photon, whose wavelength will be in the x-ray region of the
spectrum, if Z is large enough.
Characteristic x-ray spectra. (a) Part the spectra of neodymium
(Z=60) and samarium (Z=62).The two pairs of bright lines are
the Kα and Kβ lines. (b) Part of the spectrum of the artificially
produced element promethium (Z=61), its Kα and Kβ lines fall
between those of Nd and Sm. (c) Part of the spectrum of all
three elements.
The Franck-Hertz Experiment
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 V0.
The Franck-Hertz Experiment
Schematic diagram of the
Franck-Hertz experiment.
Electrons ejected from the
heated cathode C at zero
potential are drawn to the
positive grid G. Those
passing through the holes
in the grid can reach the
plate P and contribute in
the current I, if they have
sufficient kinetic energy to
overcome the small back
potential ΔV. The tube
contains a low pressure
gas of the element being
The Franck-Hertz Experiment
As V0 increased from 0, the current
increases until a critical value (about 4.9 V for
mercury) is reached. At this point the current
suddenly decreases. As V0 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, (mAmp)
Current versus acceleration voltage in the Franck-Hertz
decreases because
many electrons lose
with mercury atoms
in the tube and
overcome the small
back potential.
Current, (mAmp)
The regular spacing of the
peaks in this curve
indicates that only a
certain quantity of energy,
4.9 eV, can be lost to the
interpretation is confirmed
by the observation of
radiation of photon energy
mercury atoms.
Suppose mercury atoms have an energy level
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 by
emitting a photon. The wavelength of the photon
should be
hc 1240 eV  nm
 253 . 06 nm  0 . 25  m
4 . 9 eV
This is equal to the measured wavelength,
confirming the existence of this energy level of the
mercury atom.
Similar experiments with other atoms yield the
same kind of evidence for atomic energy levels.
Lets consider an experimental tube filled by
hydrogen atoms instead of mercury. Electrons
accelerated by V0 that collide with hydrogen
electrons cannot transfer the energy to letter
electrons unless they have acquired kinetic
eV0=E2 – E1=10.2eV
If the incoming electron does not have
sufficient energy to transfer ΔE = E2 - E1 to the
hydrogen electron in the n=1 orbit (ground state),
than the scattering will be elastic.
If the incoming electron does have at least ΔE
kinetic energy, then an inelastic collision can occur in
which ΔE is transferred to the n=1 electron, moving it
to the n=2 orbit. The excited electron will typically
return to the ground state very quickly, emitting a
photon of energy ΔE.
Energy loss spectrum measurement. A well-defined
electron beam impinges upon the sample. Electrons
inelastically scattered at a convenient angle enter the slit of the
magnetic spectrometer, whose B field is directed out of the
page, and turn through radii R determined by their energy (Einc
– E1) via equation
2me ( Einc  E1 )
An energy-loss spectrum for a thin Al film.
Reduced mass correction
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:
m1 m 2
m1  m 2
Reduced mass correction
If the nucleus has the mass M its kinetic energy
will be ½Mv2 = p2/2M, where p = Mv is the
If we assume that the total momentum of the
atom is zero, from the conservation of momentum
we will have that momentum of electron and
momentum of nucleus are equal on the magnitude.
Reduced mass correction
The total kinetic energy is then:
p (M  m) p p  
M m 
Ek 
 
1  
2 M 2m
 M
The Rydberg constant equation than changed to:
R  R
 R
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
of the Earth’s orbit are given by
n 2 2
where MS is the mass of the Sun, ME 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 from
the Sun corresponding to the quantum number n + 1