Transcript Slide 1

We get a
planetary view
of the atom
Nucleus
1/10,000 atoms
diameter
99.9% of atoms
mass is in the
nucleus
BUT.....!!!
An orbiting electron must be
accelerating…..WHY???
Changing direction
So, it must radiate
energy….WHY?
Accelerating charges
cause EM radiation
KE & Momentum should be
lost due to E-M radiation.
Electron should spiral
inward to nucleus
62 52
42
n=32
n=6
n=5
n=4
n=3 (2nd excited state)
n=2 (1st excited state)
Larger Jump = More Energy = Bluer Wavelength
n=1 (Ground State)
26 25
24
n=23
n=6
n=5
n=4
n=3 (2nd excited state)
n=2 (1st excited state)
Photons of all other energies (wavelengths) are
ignored and pass on by unabsorbed.
n=1 (Ground State)
Hydrogen
Helium
Oxygen
Neon
Iron

Nebula NGC 2363



This nebula is a glowing
gas cloud about
10,000,000 LY from Earth.
The hot stars in the Nebula
emit high energy photons
that are absorbed by the
gas.
The heated gases produce
an emission spectrum and
the particular wavelength of
the red light of the nebula
is 656nm. The exact
wavelength of Hydrogen.
The Suns Absorption Spectrum
The Suns Absorption Spectrum from 420 – 430 nm. (TOP)
The emission spectrum of Iron (Bottom)
In the late 1800’s astronomers were trying
to organize and make sense of all the data
they were collecting.
 At the time, spectra studies were the most
reliable, but there is a huge diversity of
stellar spectra.
 In 1870’s stars were classified into various
letters based upon their spectral patterns.

Hydrogen Balmer Lines Very Weak due to
extreme temps
Hydrogen Balmer Lines Strongest
Hydrogen Balmer Lines fading out and trace
amounts of heavier elements starting to appear
Stars containing heavier metals such as
Calcium and Iron (Including the Sun, a G2 star)
Stars containing Titanium Oxide

In short, the
OBAFGKM system
allows us to


Identify the surface
temperature of the
stars
Chemical composition
of the stars
2.9 x103
max ( meters ) 
T ( Kelvin)
So we can determine the
distances and characteristics
of Stars
We need a better
classification scheme
Where 4πR2 =
area of a sphere