Module 7 – The Atom
I. Atomic Emission and Absorption Spectrum
A. Experimental Facts
1. ATOMS emit and absorb ONLY certain FREQUENCIES
( WAVELENGTHS) of electromagnetic radiation .
2. The emission and absorption spectrum of each atom are unique !
The EMMISION and ABSORPTION LINE wavelengths are the SAME !! 3.
B. Explanations
1. The first condition is a consequence of the fact that bound . electrons in the atom are
When BOUNDARY CONDITIONS are imposed upon a differential equation,
ONLY CERTAIN SOLUTIONS are ALLOWED .
These solutions are called EIGEN (German for "proper") FUNCTIONS .
The parameter value the wavelength (or equivalently energy) is called the
Example: Wave on a string
that distinguishes a particular solution in our case
EIGEN VALUE
L
.
Unbounded String x = 0
Bounded String x = L

Any wavelength is possible for the unbounded string.
For the bounded string, the node conditions at the wall limit the wavelength to
2L/1, 2L/2, 2L/3, 2L/4, etc.. Thus, a particular solution to our bounded wave problem in the language of differential equations can be stated as follows
"Nth" Eigenvalue:
λ n
 2L n where n

1, 2, 3, .....
"Nth" Eigenfunction:
 n

A n
Sin 

2
π
λ n x

 for x

0 to L
Notice that the eigenfunction has an "associated" eigenvalue and range . The eigenfunction in our case the wavefunction tells us the displacement of our standing wave for any value of x in the range. In physics, the associated eigenvalue usually provides information about the value of an important physics property in our case the wavelength .
2.
3.
Notice also that our allowed eigenvalues are
. This is common for
.
We say that such eigenvalues are
.
From Einstein's photon hypothesus, we know that our first experimental fact implied that not only the wavelength but also the energy of emitted and absorbed electromagnetic radiation is quantized!
E n
 h
λ n c  1240 eV
λ n
 nm
The second experimental fact has both great practical application and theoretical implications .

4.
First it implies that emission and absorption spectroscopy can be used as an analytical technique to determine the composition of materials
EXAMPLES: a.
Particle Induced X-ray Emission - PIXE is used to analyze minerals, artifacts, medical samples, etc.
In this technique charged particle (electron from an electron microscope or proton or alpha particle from an accelerator) excites an atom by knocking out an electron. The remaining electron de-excites by giving off characteristic photons. (Along with RBS, it was used to explore the moon before Apollo 11, determined the cause of legionnaires disease, authentication of rare books and paintings, determining environmental pollutants, art restorations, finding rare gems, and other uses). b. X-ray Fluorescence Spectroscopy – A photon from an x-ray tube or radioactive sources excites an atom. XRF was used to analyze the surface of Mars. c. Optical spectroscopy- Look at the photons emitted from heated gasses.
This is used in Chemistry (Flame Spectroscopy) to determine unkown chemical compositions. It has also allows us to identify the composition of the sun and stars as well. The element Helium was first discovered in the spectrum of the stars (Hence its name). It was only found much later on
Earth when a Professor at the University of Kansas was analyzing an odorless gas which appeared while a group was drilling for oil.
Secondly, the fact that the spectrum is unique suggested that the energy of the emitted photon is connected to the structure of the atom (ie the energy levels of the electrons).
The third experimental fact lends further support to the fact that energy of the emitted photon is connected to the energy levels of the atom. Absorption is the transfer of the photon's energy to an atom's electron. If the electron falls back down to its original energy level then it must give off the same amount of energy that it absorbed.
E
2
E
2
E
1
E
1

E
2
E
1
2.
1.
Since a photon can not be partially absorbed, only a photon whose energy matches the energy difference between two electron energy levels would be absorbed.
C. Problems
The electrons in Rutherford's planetary model have kinetic and potential energy that depends upon their orbital distance from the nucleus. The electron should be allowed to have any energy value according to classical physics since any orbital distance is allowed.
E

K

U
 L 2
2I
 k
Ze r
2
  1
2 k
Ze where we have neglected electron-electron interactions. r
2
Unlike the planets, the electron has charge so it should be radiating a continuous spectrum as it spirals into the nucleus.
3. The spectral lines have very specific shapes
"diffuse."
. For instance some lines are very bright and are called the "principle" lines. Other lines are "fine", "sharp", or
Rutherford's model didn't explain the cause of these observations .
Note: These spectral notations still remain in chemistry (p,f, s, and d orbitals) today even though they don't shown up in the Quantum Mechanical formulation.
E
2
E
1

II. Hydrogen Spectrum
Scientists had compiled a large amount of data concerning the spectral lines of many elements before the turn of the century. Although a compilation of wavelengths and intensities was useful, it was difficult to analyze unknown samples since you had no computers. Also, there was no scheme to explain existing known spectral lines or to predict new ones.
A. Balmer
In 1885, Johann Balmer, a high school teacher, provided the following formula that fits the visible spectral lines of hydrogen :
λ n


364.56
nm
 n 2 n 2

4 where n

3, 4, 5, ...
B. Rydberg-Ritz Formula
W. Ritz and J. R. Rydberg extended Balmer's work by producing a formula that predicted additional lines of hydrogen as well as heavier elements.
1
λ

R
1 m 2
 1 n 2 where m is a positive constant n is a positive constant greater than m
R is the Rydberg constant
The Rydberg constant was actually a parameter in the Rydberg-Ritz formula and had the values of
R
H
= 1.09677576 x 10 7 m -1 for hydrogen
R  = 1.09737315 x 10 7 m -1 for heavy elements
This was a powerful formula for analytical work. However, it is still just a curve fit through empirical data and provides no deep understanding into the nature of the atom. Furthermore, some lines didn't fit the formula!!

C. Other Spectral Series of Hydrogen
In addition to the visible Balmer series, hydrogen has other spectral series:
Lyman Series (m = 1) is in the ultra-violet region
Paschen Series (m = 3) is in the infrared region
Brackett Series (m = 4)
Pfund Series (m = 5)
III. Bohr Atom
The great brilliance of Bohr’s work is how few assumptions are required to explain so many experimental details. Bohr knew that his model of the hydrogen atom wasn’t a complete or systematic system for solving problems, but it the model and Bohr’s insights were invaluable stepping stones on the way to true theory of Quantum Mechanics.
A. Bohr's Postulates
1. An electron moves in a circular orbit around the nucleus due to the
Coulomb force.

F
  k
  r 2
  rˆ

ke
2
Z r
2 rˆ
2. Electrons can only exist in orbits where the magnitude of its angular momentum is a positive integer of Planck's constant divided by 2

.
L
 m v r
 n 
3. The electrons do not radiate electromagnetic energy while they travel in their circular orbits. Thus, each electron orbit is a state of constant energy.
Violates Classical Electromagnetism

4. Electromagnetic radiation is emitted (absorbed) when an electron jumps from one energy state to a lower (higher) energy state. The energy of the emitted (absorbed) photon is equal to the energy lost by the electron.
E i

E f

 h
ν  hc
λ
Bohr's postulate are strange and difficult to accept. However, it is more amazing that with just these four postulates Bohr was able to reproduce the Ritz-Rydberg equation and determine the size of the hydrogen atom!
B. Size of the Atom
We will now use classical physics and Bohr's postulates to find the size of an atom. We will assume that the nucleus is stationary. To correct for the finite mass of the nucleus, you simply replace the mass of the electron by the reduced mass of the electron-nucleus system.
Ze r
The electron is in uniform circular motion due to the Coulomb force between the electron and the nucleus.
F
 k
   
 r 2 m v 2 r r
 m v 2 k r
Z e 2
2

r

 m v r

2 ke 2 Z m r
 ke 2
L 2 c 2
Z m c 2 r
 n 2 ke 2
 2 c 2
Z E o



 n
2
Z



4
π 2
 
2 ke 2 E o
For hydrogen (Z=1) ground state (n = 1), we have a o

4
π 2
 
2 ke 2 E o

4
π 2

1.44

1240 eV
 eV
 nm
 nm

2
5.11
x 10
5 eV

0 .
0529 nm
Thus, the nth orbit of an atom of atomic number Z is given by r n
 n
2 a o
Z
C. Energy Levels
We can now us our previous results to calculate the energy levels of the atom. The electron has rotational kinetic energy and electric potential energy.
E

K

U
 L 2
2 I
 k Z e 2 r

Using the moment of inertia for a point charge, we have
E
 L 2
 k Z e 2 r
2 m r 2
Using
Z k e 2 r
 m v 2

2 K that we found in the previous section, we have
E
 L 2
2 m r 2
 m v 2
L 2

2 m r 2
 L 2 m r 2
L 2
E
 
2 m r 2
E
 n 2  2

2 m

 a o n
2
Z


2
 n 2  2 Z 2

2 m n 4 a o
2
E
 
Z n
2
1


2 E o



 h c
2
π a o


For the ground state of hydrogen (n = 1, Z=1), we have
2
E
H
 


 2
1 x 5.11
x 10
5 eV





1240
2
π x eV
 nm
0.0529
nm


2
 
13.6
eV

Thus, the nth energy level of an atom with atomic number Z is given by
E n

Z 2 E
H n 2
IV. Bohr Model and the Ritz-Rydberg Formula
We can now use our results from the Bohr atom to develop the Ritz-Rydberg formula. If an electron drops from the mth energy level in the hydrogen atom to the lower energy nth level, it will emit a photon of energy E given by
E
 hc
λ
 
E n

E m
Substituting in the Bohr energy state relationship for the hydrogen atom, we have hc
λ
 
E
H


1 n
2

1 m
2


1
λ

E
H h c


1 n
2

1 m
2


1
λ



13.6
eV
1240 eV
 nm




1 n
2

1 m
2


1
λ

0 .
019677 nm

1


1 n
2

1 m
2


1
λ

1 .
9677 m

7


1 n
2

1 m
2



R
H


1 n
2

1 m
2



For an an atom of atomic number Z, the Bohr atom predicts that the spectral lines will be given by
1
λ

Z 2 R
H


1 n
2

1 m
2


V. Reduced Mass
Spectroscopic data is so precise that one must account for the finite mass of the nucleus . In the hydrogen atom, the nucleus and electron orbit around their center of mass and not around the center of nucleus since the nucleus has a finite mass. However, we can convert this two body problem into a single body problem by reducing the mass of the electron as shown below: r
X m
C.M.
M r

Two Bodies Rotating
About The Center of Mass
Equivalent One Body
Problem
Reduced Mass Formula
μ  m m

M
M
Since the mass of the electron is much less than the mass of the nucleus, the reduced mass is approximately the mass of the electron although slightly lower . Furthermore, the reduced mass increases slightly for higher Z atoms which changes the Rydberg constant for higher Z atoms.

V c
VI. Franck and Hertz Experiment
A. Experimental Setup
Mercury
Gas
B. Results
V acc
V r
4.9 eV
-
A
+
4.9 eV
I
V acc
Data indicates quantization of energy levels of mercury. This was further confirmed by observing a photon emission from mercury that matched the energy lost by the electrons!