Transcript The Franck-Hertz Experiment

Atomic and Molecular
Spectra
Atoms are very small,
 Atoms are stable
 Atoms contain negatively charge electrons,
but are neutral(because matters are neutral)
 Atoms emit and absorb electromagnetic
radiation(Black body radiation, photoelectric
effect)

• Blackbody Radiation and Photoelectric
Effect gave the impulse to understand the
nature of light
• However, there was no understanding on
how electrons “exist” inside matter
–
It was known that atoms that form matter contain
electrons (Thompson, 1897) and that atoms are
neutral
• Atomic structure?
Thomson Model
• Model of J.J. Thomson:
– Electrons (Corpuscular) are embedded within positive
charge sphere (so-called “plum pudding” model)
– Both the positive charge and the mass of the atom would
be more or less uniformly distributed over its size
• Approximately 10-10 meters across = 1 Å = 0.1 nm
Challenges to Thomson’s Model
• How does the atom emit radiation?
• This model soon came into conflict with
experiments by Rutherford
The Rutherford Experiments
• Rutherford discovered α (alpha)-particles
– α-particles are the nuclei of helium atoms, which were
products of nuclear disintegration
– He used α-particles in various studies
• In 1909, he with co-workers (Hans Geiger and Ernest
Marsden) experimented with streams of alpha
particles passing through a thin gold foil
Rutherford Experiment
The Rutherford Experiments
Observations:
1.
Almost all α-particles went
through the gold foil with very
slight deflections
2.
A very, few α-particles were
reflected – scattered by more than
of 90°
1.
2.
3.
Conclusions:
Most α-particles must have passed
through the atoms without being
deflected. An atom must be mostly
empty space
The α-particles occasionally strick a
small strongly scattering region. This
region could be a concentrated
positive charge
Observed trajectory suggested the
latter is the case!
1. Can calculate atomic volume
and area using Avagadro’s
number
2. Can calculate cross sectional
area of nucleus relative to
atomic cross section based on
fraction that scatterer and the
distribution of scattering angles
which is consistent with
trajectories of the charged aparticle scattering from a
positive core
Rutherford’s Experiments
•
Since atoms have no overall electrical charge (atoms are
neutral), there must be just enough negatively-charged
particles outside the nucleus to just balance the positivelycharged nucleus
Therefore,
The Rutherford's experiment suggested that the atom consists
of mostly empty space with a very small positivelycharged nucleus, outside which are just enough negative
charges to equal the positive charge in the nucleus
Since the electrons do not crash into the nucleus but exist in
stable atoms, this suggests a planetary model with
negatively charged electrons circulating around a positive
core
Estimate of Nuclear Cross Section I
• Approximately one in 104  particles with energies of ~
107 eV are scattered backward from a one micron thick
sheet of metal foil
• If atomic spacing is ~ 1 A, then there are 104 atomic
layers so the probability of scattering by a single layers is
10-8
• This implies the nuclear cross section is one part in 108
of the atomic cross section
• Since the area goes as the square of the radius the
nuclear radius is smaller than atomic radius by a factor
of 10-4
• Since the atomic diameter is ~ 1 A, the nuclear diameter
is ~ 10-4 A
Estimate of Nuclear Cross Section II
• The model of a small positively charge nuclear core was
confirmed by a comparison with the angular distribution
to scattered -particles
• The incident -particles are assumed to have a uniform
distribution in space. Initial trajectories that are headed
towards the nuclei of atoms in the foil with random
separations from the nearest nuclear core
• Full trajectory calculated for each distance of the
trajectory line from the center of the nearest nucleus to
give the scattering angle
• From the distribution of distances of the -particles from
the positive core calculate distribution of scattering
angles
• Agreement found with measured scattering angle
distribution
The Rutherford Model
• “Planetary Model”
– Positive charge in the center of
the atom with almost all mass
concentrated within this
positive charge - nuclei
– Electrons - negative charge- are
attracted to the nucleus about
which they orbit (just as
planets orbit the sun due to
attractive 1/r2 force)
– Sizes
• nuclei ~ 10-14 m (calculated from
fraction of -particles that scatter
more than 900 in a foil of given
thickness)
• atom ~ 10-10 m (from the mass
density and number of atoms in a
mole – Avogadro’s number)
Difficulties with the Rutherford
Model
• Since electron travels in a
circular orbit, it is
constantly accelerated
(even though its speed is
constant.) Thus, the
electron emits EM
radiation, which carriers
away energy. The energy
of the atoms is reduced.
Thus the atom has a lower
potential energy and
moves closer to the
nucleus
( Ze)e
Etotal  
80 r
Thus, classically, the
Rutherford Atom is Unstable
Difficulties of the Rutherford Model
• Another problem is that the spectrum of the
emitted EM radiation would be continuous.
Classical approach gives the following expression
1/ 2
 e c



3
 16  0 m 
2 2
1
r
3/ 2

1
r 3/ 2
• As r decrease, the emission wavelength changes
continuously, so this model predicts that the
emission spectrum of atoms is broad
• But sharp spectral lines are observed, not a
continuum
The Spectrum of Hydrogen
• At room temperature, hydrogen gas does not emit
light
• When heated to high temperatures, hydrogen
emits visible radiation
– distinct spectral lines are observed rather than the
continuous radiation spectrum expected classically
• Example of visible part of the spectrum
The Spectrum of Hydrogen
• Several families of such lines were observed
• They can be fit empirically by the Rydberg-Ritz
Formula
1
1
7
1
 RH  2  2 , RH  1.0973732 10 m

n 
k
1
• RH is known as the Rydberg constant
• k and n are integers, and n > k
• The visible hydrogen emission is known as the
Balmer series with k = 2
The Bohr’s Atom
Assumption:
• Electron has, for some yet unknown reason, only
certain energies in the hydrogen atom
• Bohr called these allowed energy levels
• The atom can occasionally jump between energy
levels, emitting a photon when it makes a transition
to a lower energy state and absorbing a photon
when jumping to a higher energy level
• The difference in the energy between the two levels
is E = h
• So have discrete lines observed in hydrogen
spectrum
Bohr’s Postulates
• Bohr started from the assumption
that the electron moves in circular
orbits around the proton under
the influence of the attractive
electric field.
• Postulate 1: Only certain orbits are stable.
These are stationary or more precisely
quasi-stationary states. An electron does
not emit EM radiation when in one of these
states (orbits)
Bohr’s Postulates
• Postulate 2. If the electron is initially in an
allowed orbit (stationary state), i, having the
energy, Ei, goes into another allowed orbit, f,
having energy, Ef (< Ei), EM radiation is
emitted, with energy and frequency,
h  Ei  E f

Ei  E f
h
Bohr’s Postulates
• Postulate 3. The electron can only have an
orbit for which the angular momentum of the
electron, L, takes on discrete values (the
orbits are quantized):
L  me vr  n
h

2
Orbits characterized by angular momentum since this
depends on both the distance of the electron from nucleus
and its velocity
Electron Orbits in the Bohr Atom
L quantized  E quantized
L  mvr  n
n
*v 
mr
The Coulomb force is the radial force keeping the electron in orbit :
e2
mv 2
**

40 r 2
r
mv 2
e2
So, the kinetic energy is,

2
80 r
But the total energy is the sum of kinetic and potential energy :
mv 2
e2
e2
e2
e2
E




2
40 r 80 r 40 r
80 r
n 2
)
mr
e2
2
From the equation for radial force * * have: v 
,
40 rm
From quantizati on considerat ions * have : v 2  (
while
2
n 2 2
2  40
So, v 
 2 2 rn
 n 2 a0 , a0  0.53 A
2
40 rm m r
me
2
So, En  
e2
e2
80 r

e2
E0
1


, E0  13.6 eV , n  1,2,3...
80 a0 n 2
n2
Electron Orbits in The Bohr’s Atom
• The radii (orbits) are said to be discrete
(‘quantized’)
rn  a0 n  rmin  r1  a0
2
• rmin= a0 = 0.53×10-10 m = 0.53 Å, is called the
Bohr radius
• In this model, the electron and the nucleus in
the hydrogen atom cannot get closer than the
distance a0
The Wavelength of the Emitted
Electromagnetic radiation
• Following Postulate 2:
E0  E0 
1
1

 En  Ek   2    2   E0  2  2 
n  k 
n 
k
h nk ,n k
1


 nk
c
Enk E0  1
1


 2  2
ch
hc  k
n 
E0
7
1
 1.09774 10 m  RH
hc
The Spectrum of Hydrogen
 jk
E0  1 1 


 2  2
 c hc  k
j 
1
E0
 1.09774  10 7 m 1  Ry
hc
Ionization Energy
• For hydrogen atom, with only one electron, the
ionization energy has a clear meaning:
– This is the energy required to remove the electron from a
hydrogen atom:
• H0 + 13.6 eV  H+ + e• An atom is now ionized, since now have H+ instead of H0
• Similarly for any other atom, we can introduce the
ionization energy
– The minimum energy required to remove the most energetic
electron from the atom in its lowest energy state
• The energy required to remove the second electron is the
“second ionization energy”
– Question: Is it larger or smaller than “first” ionization energy?
Discrete Spectrum
• Bohr’s Postulates Predicted a Discrete Spectrum
• Consistent with the spectrum of hydrogen
• Direct proof found in measurements by James Frank and
Gustav Hertz in 1914
• Results consistent with spectra
The Franck-Hertz Experiment
• The heated filament ejects electrons into the tube, which can be
either evacuated (vacuum) or filled with Hg vapor
• A variable (positive) accelerating
voltage, VA, is applied to the grid
(a wire mesh)
– The electrons acquire kinetic energy
K = eVA
• There is (negative) retarding
voltage (Vr) between grid and
collector
• Only electrons having enough
[kinetic] energy will overcome this
potential and reach the Collector
and contribute to current measured
with an Electrometer
The Franck-Hertz Experiment
• If Vr > VA no electrons can reach the Collector, so no
current would be measured.
• If Vr < VA, then, if the tube is highly evacuated,
most of the electrons would reach the Collector,
and have energy |e|(VA - Vr)
The Franck-Hertz Experiment
• If the tube contains some gas, the electrons can loose
energy via collisions with the gas atoms
– Such collisions are inelastic, i.e. electrons lose energy, which is
transferred to internal energy of atoms in the gas
• Thus, even in the case when
Vr < VA, it is possible that the
electrons would not be able
to reach Collector, and
contribute to the current
The Franck-Hertz Experiment
• Franck and Hertz observed
the Collector current as a
function of VA (>Vr) when
tube was filled with various
gases (result for mercury gas
is shown here)
• At first, the current increased
as was expected for a typical
vacuum tubes, but at ~4.9 V
current suddenly dropped
• Then, the increase resumed
until 9.8 V, an so on
VA
The Franck-Hertz Experiment
• The current drops because fewer electrons reach the
Collector
• This occurs only if the electrons
undergo inelastic collisions
• Thus, when VA = 4.9 × n Volts
(n = 1, 2, 3, …) the electrons
undergo inelastic collisions
with Hg atoms!
• In inelastic collision kinetic
energy of electrons becomes
internal energy of Hg atoms –
Hg atoms absorb the energy of
electrons!
VA
The Franck-Hertz Experiment
• Why do we see the drop only at specific voltages?
• If distribution of energy levels of Hg atom is
continuous, then kinetic energy should be
transferred to Hg atoms regardless of the energy of
electrons
• However, If we assume that Hg spectrum is
discrete, then only when electrons reach certain
energies do they undergo inelastic collision with
Hg atoms
– Thus energy spectrum of Hg atoms is such that an
electron energy level lies ~4.9 eV above the ground state
The Franck-Hertz Experiment Explained
• Why did Frank and Hertz not
observe dips in the current at
other voltage?
– Their experiment was not
sufficiently sensitive
– Since as soon as electrons gain
energy of 4.9 eV they transfer it to
Hg atoms, only small fraction of the
electrons could have higher
energies, (e.g., 6.65 eV), making it
difficult to observe current dips
associated with other (higher)
voltages