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Physics 1C
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 RH is known as
the Rydberg constant
RH = 1.0973x107m-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 λ3 = 656nm and n = 4
led to a λ4 = 486nm.
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=Iω
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, ao.
ao = 0.0529nm
Hydrogen Atom: Bohr’s Theory
We know that the radii of the Bohr orbits in a hydrogen
atom are quantized:
2 2
n
rn 
2
me kee
We also know that when n = 1, the radius of that orbit
is called the Bohr radius (ao = 0.0529nm).
So, in general we have:
rn = n2ao

The total energy of the atom can be expressed as :
E tot  KE  PE elec
2
1
e (assuming the
2
 me v  ke
nucleus is at rest)
2
r
Hydrogen Atom
Plus, from centripetal acceleration we found that:
2
e
mev  ke
r
2
Putting this back into energy we get:
2 
2

1
e
1
e
e
2
E tot  mev  ke  ke  ke
2
r 2  r 
r
2
2 

1
e
E tot   ke 
2  r 
But we can go back to the result for the radius
(rn = n2 ao) to get a numerical result.
Hydrogen Atom
2 

1
e
13.6 eV
E tot   ke 2 
2
2  n ao 
n
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:
rn = n2ao
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 => E1 = –13.6eV.
When n = 2 => E2 = – 13.6eV/22 = – 3.40eV.
When n = 3 => E3 = – 13.6eV/32 = – 1.51eV.
When n = 4 => E4 = – 13.6eV/42 = – 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 (ni) to a final
lower energy level (nf).
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:
Ei  E f
f 
h
To find the wavelengths
for an arbitrary transition
from one orbit with nf to
another orbit with ni, we
can generalize Rydberg’s
formula:
 1
1
1
 RH  2  2 

 nf ni 
Hydrogen Atom
The wavelength will be
represented by a different
series depending on your
final energy level (nf).
For nf = 1 it is called the
Lyman series (ni = 2,3,4,...).
For nf = 2 it is called the
Balmer series (ni = 3,4,5...).
For nf = 3 it is called the
Paschen series (ni = 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.
Atomic Spectra
Example
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
nf = 1 (Lyman), nf = 2 (Balmer), and nf = 3
(Paschen).
Atomic Spectra
Answer
Turn to the generalized Rydberg equation:
 1 1
1
 RH  2  2 

 nf ni 
For the Lyman series we have:
2



1
1 1
ni
1 
 RH   2  RH  2  2 

1 ni 
n i n i 
2

1
n i 1
 RH  2 
  n i 
2 
2 


1
ni
1
ni
n 
 2 

7
-1  2
RH n i 1 1.097 10 m n i 1
Atomic Spectra
Answer
Finally for the Lyman series:
 2 2 
1
1 
 121 nm
7
-1  2
1.097 10 m 2 1
 32 
1
2 
 103 nm
7
-1  2
1.097 10 m 3 1

 4 2 
1
3 
 97.2 nm
7
-1  2


1.097 10 m 4 1
 5 2 
1
4 
 95.0 nm
7
-1  2
1.097 10 m 5 1
For the Balmer series we have from before:
λ1 = 656nm, λ2 = 486nm, λ3 = 434nm, λ4 = 410nm.

Atomic Spectra
Answer
For the Paschen series we have:
1 1 
 n i2
9 
 RH   2  RH  2  2 

9 n i 
9n i 9n i 
1

 9ni2 
1
n 

7
-1  2
1.097 10 m n i  9 
2 
2 


9
4
9
5




1
1



 1280 nm
1 

1880
nm


2
7
-1  2
7
-1
2
5  9 
1.097 10 m 4  9 
1.097
10
m




2 

9
6


1

 1090 nm
3 
7
-1  2
1.097 10 m 6  9 
 
2 

9
7


1

 1010 nm
4 
7
-1  2
1.097 10 m 7  9 

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 e2 with Ze2 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
2 

1
e
E tot   ke 
2  r 


2
n
rn 
2
me kee
C) lithium
D) beryllium
2
E) the ionization energy is the same for all four
For Next Time (FNT)
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