Lecture 29A

Atomic Physics
The study of quantum mechanics led to amazing theories as to how the subatomic world worked.
One of the first theories of how the atom was composed was the Plum Pudding Model by J.J.
Thomson (incorrect!!!).
In this model the atom was thought to be a large volume of positive charge with smaller electrons embedded throughout.
Almost like a watermelon with seeds.

Atomic Physics
But then in 1911, Ernest Rutherford performed an experiment where he shot a beam of positively charged particles (alphas) against a thin metal foil.
Most of the alpha particles passed directly through the foil.
A few alpha particles were deflected from their original paths (some even reversed direction).

Atomic Physics
This thin foil experiment led Rutherford to believe that positive charge is concentrated in the center of the atom, which he called the nucleus.
He then predicted that the electrons would orbit the nucleus like planets orbit the sun.
Centripetal acceleration should keep them from spiraling in (like the Moon).
Thus, his model was named the Planetary
Model .

Atomic Physics
But there were a few problems with the Planetary model of the atom. Such as, what happens when a charged particle (like the electron) is accelerated?
It gives off light of a particular frequency.
As light is given, the electron will lose energy and its radius should decrease.
The electron should eventually spiral into the nucleus.

Emission Spectra
Finally, the key to understanding atoms was to look at the light that was emitted from them.
When a low-pressure gas is subjected to an electric discharge, it will emit light characteristic of the gas.
When the emitted light is analyzed with a spectrometer, we observe a series of discrete lines.

Emission Spectra
This is known as emission spectra.
Each line has a different wavelength (color).
The elemental composition of the gas will tell what the resulting color lines will be.
Note that in general elements with a higher atomic number will have more lines.

Emission Spectra
The easiest gas to analyze is hydrogen gas .
Four prominent visible lines were observed, as well as several ultraviolet lines.
In 1885, Johann Balmer, found a simple functional form to describe all of the observed wavelengths: where R
H is known as the Rydberg constant
R
H
= 1.0973x10
7 m -1 .
n = 3, 4, 5, ....

Emission Spectra
Every value of n led to a different line in the spectrum.
For example, n = 3 led to a λ
3 led to a λ
4
= 486nm.
= 656nm and n = 4
The series of lines described by this equation is known as the Balmer Series.
Note how the spacing between the lines gets closer and closer the smaller the wavelength gets.

Absorption Spectra
In addition to emission spectra (lines emitted from a gas), there is also absorption spectra (lines absorbed by a gas).
An element can also absorb light at specific wavelengths.
An absorption spectrum can be obtained by passing a continuous radiation spectrum through a cloud of gas.
The elements in gas will absorb certain wavelengths.

Absorption Spectra
The absorption spectrum consists of a series of dark lines superimposed on an otherwise continuous spectrum.
The dark lines of the absorption spectrum coincide with the bright lines of the emission spectrum.
This is how the element of Helium was discovered.

Hydrogen Atom
In 1913, Neils Bohr explained atomic spectra by utilizing Rutherford’s Planetary model and quantization.
In Bohr’s theory for the hydrogen atom, the electron moves in circular orbit around the proton.
The Coulomb force provides the centripetal acceleration for continued motion.

Hydrogen Atom
Only certain electron orbits are stable.
In these orbits the atom does not emit energy in the form of electromagnetic radiation.
Radiation is only emitted by the atom when the electron
“jumps” between stable orbits.

Hydrogen Atom
The electron will move from a more energetic initial state to less energetic final state.
The frequency of the photon emitted in the
“jump” is related to the change in the atom’s energy:
If the electron is not
“jumping” between allowed orbitals, then the energy of the atom remains constant.

Hydrogen Atom
Bohr then turned to conservation of energy of the atom in order to determine the allowed electron orbitals.
The total energy of the atom will be:
But the electron is undergoing centripetal acceleration (Newton’s second law):

Angular Momentum
Recall from classical mechanics that there was this variable known as angular momentum, L .
Angular momentum, L , was defined as:
L = I ω where I was rotational inertia and ω was angular velocity.
For an electron orbiting a nucleus we have that:
Giving us:

Hydrogen Atom
Bohr postulated that the electron’s orbital angular momentum must be quantized as well: where ħ is defined to be h /2π.
This gives us a velocity of:
Substituting into the last equation from two slides before:

Hydrogen Atom
Solving for the radii of Bohr’s orbits gives us:
The integer values of n = 1, 2, 3, … give you the quantized Bohr orbits.
Electrons can only exist in certain allowed orbits determined by the integer n .
When n = 1, the orbit has the smallest radius, called the Bohr radius , a o
.
a o
= 0.0529nm


We know that the radii of the Bohr orbits in a hydrogen atom are quantized: r n
 m n 2 2 e k e e 2
We also know that when n = 1, the radius of that orbit is called the Bohr radius ( a o
= 0.0529nm).
So, in general we have:
 r n
n 2 a o
The total energy of the atom can be expressed as :
E tot
 KE  PE elec

1
2 m e v 2  k e e 2 r
(assuming the nucleus is at rest)


Hydrogen Atom
Plus, from centripetal acceleration we found that: m e v 2  k e e 2 r
Putting this back into energy we get:
E tot
 
1
2 m e v 2  k e e 2 r

1
2
 k e e 2 r

 k e e 2 r
E tot
 
1
2
 k e e 2 r


But we can go back to the result for the radius
( r n
= n 2 a o
) to get a numerical result.


Hydrogen Atom
E tot
 
1
2

 k e e 2 n 2 a o


 13.6
eV n 2
This is the energy of any quantum state (orbit).
Please note the negative sign in the equation.
 When n = 1, the total energy is –13.6eV.
This is the lowest energy state and it is called the ground state .
The ionization energy is the energy needed to completely remove the electron from the atom.
The ionization energy for hydrogen is 13.6eV.

Hydrogen Atom
So, a general expression for the radius of any orbit in a hydrogen atom is: r n
= n 2 a o
The energy of any orbit is:
If you would like to completely remove the electron from the atom it requires 13.6eV of energy.

Hydrogen Atom
What are the first four energy levels for the hydrogen atom?
When n = 1 => E
1
= –13.6eV.
When n = 2 => E
2
= – 13.6eV/2 2 = – 3.40eV.
When n = 3 => E
3
= – 13.6eV/3 2 = – 1.51eV.
When n = 4 => E
4
= – 13.6eV/4 2 = – 0.850eV.
Note that the energy levels get closer together as n increases (similar to how the wavelengths got closer in atomic spectra).
When the atom releases a photon it will experience a transition from an initial higher energy level ( n i
) to a final lower energy level ( n f
).

Hydrogen Atom
The energies can be compiled in an energy level diagram.
As the atom is in a higher energy state and moves to a lower energy state it will release energy (in the form of a photon).
The wavelength of this photon will be determined by the starting and ending energy levels.


Hydrogen Atom
The photon will have a wavelength λ and a frequency f : f 
E i
 E f h
To find the wavelengths for an arbitrary transition from one orbit with n f to another orbit with n i
, we can generalize Rydberg’s formula:
1


R
H


1 n f
2

1 n i
2



Hydrogen Atom
The wavelength will be represented by a different series depending on your final energy level ( n f
).
For n f
= 1 it is called the
Lyman series ( n i
= 2,3,4,...).
For n f
= 2 it is called the
Balmer series ( n i
= 3,4,5...).
For n f
= 3 it is called the
Paschen series ( n i
= 4,5,...).

Concept Question
When a cool gas is placed between a glowing wire filament source and a diffraction grating, the resultant spectrum from the grating is which one of the following?
A) line emission.
B) line absorption.
C) continuous.
D) monochromatic.
E) de Broglie.

Example
Atomic Spectra
What are the first four wavelengths for the
Lyman, Balmer, and Paschen series for the hydrogen atom?
Answer
The final energy level for either series will be n f
= 1 (Lyman), n f
= 2 (Balmer), and n f
= 3
(Paschen).


Atomic Spectra
Answer
Turn to the generalized Rydberg equation:
1


R
H



1 n f
2

1 n i
2 


For the Lyman series we have:
1

 R
H


1
1
 n
1 i
2


R
H

 n n i i
2
2

1 n i
2


1

 R

H
 n i
2 n i
2
 1

  n

1
R
H

 n i
2 n 2 i
 1


1
1.097
 10 7 m -1

 n i
2 n 2 i
 1




Atomic Spectra
Answer
Finally for the Lyman series:

1

1
1.097
 10 7 m -1

 2 2
2 2
 1


121 nm


3
2

1
1.097
 10 7 m -1

 3 2
3 2
 1


103 nm


1
1.097
 10 7 m -1

 4 2
4 2
 1


97.2 nm


4

1
1.097
 10 7 m -1

 5 2
5 2
 1


95.0 nm

For the Balmer series we have from before:
λ
1
= 656nm, λ
2
= 486nm, λ
3
= 434nm, λ
4
= 410nm.




Atomic Spectra
Answer
For the Paschen series we have:
1

 R
H


1
9
 n
1 i
2


R
H

 n
9 n i
2 i
2

9
9 n i
2


 n

1
1.097
 10 7 m -1

 n i
9 n
2 i
2
 9




1

3


1
1.097

10 7 m -1
   


4 2  9
1880 nm
1
1.097
 10 7 m -1
  

6 2  9


1090 nm


2

4

1
1.097
 10 7 m -1

9 5

5 2  9

 1280 nm

1
1.097
 10 7 m -1

9 7

7 2  9


1010 nm


Atomic Spectra
The only series that lies in the visible range (390 –
750nm) is the Balmer series.
The Lyman series lies in the ultraviolet range and the Paschen series lies in the infrared range.
We can extend the Bohr hydrogen atom to fully describe atoms that are “close” to hydrogen.
These hydrogen-like atoms are those that only contain one electron. Examples: He+, Li++, Be+++
In those cases, when you have Z as the atomic number of the element ( Z is the number of protons in the atom), you replace e 2 with Ze 2 in the hydrogen equations.

Concept Question
Consider a hydrogen atom, a singly-ionized helium atom, a doubly-ionized lithium atom, and a triplyionized beryllium atom. Which atom has the lowest ionization energy ?
A) hydrogen
B) helium
E tot
 
1
2
 k e e 2 r

 r n
 n 2 2 m e k e e 2
C) lithium
D) beryllium
 
E) the ionization energy is the same for all four

For Next Time (FNT)