Download Bohr`s model of the atom

The attractive Coulomb force between the positive nucleus and the orbiting electron
could provide the attractive force which keeps the electron in it’s orbit, much as the
planets orbit the sun with gravity providing the centripetal force.
2
Fcentripetal 
me v0
 me02 r
r
What’s wrong with
this picture?
Accelerating charges radiate.
Could this electromagnetic
radiation be the source of
the spectral lines?
No. This radiation must come
at the expense of the kinetic
energy of the orbiting
electron!
It will eventually spiral into
the nucleus. The atom would
be unstable!
Lyman series :
 n2 
 (cm)  C1  2 2  n  2, 3, 4...
 n 1 
Balmer series :
 n2 
 (cm)  C2  2 2  n  3, 4, 5...
n 2 
Paschen series :
It also turns out that
C1=C2=C3…
 n2 
 (cm)  C4  2 2  n  4, 5, 6...
 n 3 
But what does all of this mean??
•Classical electromagnetism does not
hold for atom sized systems.
•Used Planck’s energy quantization
ideas to postulate that electrons orbit
in fixed, stable, nonradiating states,
given by
mevr  n n 1,2,3...
•Used Einstein’s concept of a photon
to define the frequency of radiation
emitted
 when an electron jumps from
one state to another. The photon
energy is just the energy difference
between states, i.e., E f  Ei  hv
•Used classical mechanics to
calculate the orbit of the electron.
Coulomb force keeps electrons in orbit:
F  ma 
2
2
mv
e
 2
r
r
1
KE  me v 2
2
e2
U  k
r
But we know that the allowed values of the angular
momentum are:
mevr  n
mevr  n
(Are you wondering why these are the allowed
values? You should be.)
n
v 
me r

2
ke2 ke2
ke2
E  KE  U 


2r
r
2r
1
ke
KE  me v 2 
2
2r
1  n  ke2
KE  me   
2 me r  2r
2
n2 2
 rn 
me ke2
For n=1, this gives the “Bohr radius”:
a0 
n 1,2,3...
2
2
me ke
 0.0529 nm
which was in good agreement with experimental values.
For other values of n: rn  n 2 a0
ke2  1 
En  
 2  n  1, 2, 3,...
2a0  n 
13.6
En   2 eV
n
energy quantization therefore follows from
angular momentum conservation.
therefore the ionization energy is 13.6 eV. It takes 13.6
eV to liberate an electron in the ground state. At
n=infinity, it is removed.
This also agreed with experiment.
Independent of the orbital angular
momentum of the electron, the
frequency of a photon emitted is:
E f  Ei  hv

Ei  E f
h
ke2  1
1 


2a0 h  n 2f ni2 
ke2  1
1 
 
 2
2

 c 2a0 hc  n f ni 
1

 1
1 

R 2  2
n


 f ni 
1
1. The laser in its nonlasing state
2. The flash tube fires and
injects light into the ruby rod.
The light excites atoms in the
ruby.
4. Some of these photons run in a direction
parallel to the ruby's axis, so they bounce
back and forth off the mirrors. As they pass
through the crystal, they stimulate emission
in other atoms.
3. Some of these atoms
emit photons.
5. Monochromatic, single-phase,
collimated light leaves the ruby through
the half-silvered mirror -- laser light!
Note that the difference in energies of allowed orbits becomes smaller as n
becomes larger.
In the limit of large quantum numbers, the frequencies
and the intensities of radiation calculated from classical
theory must agree with quantum theory.
At the beginning of class, I said
that Bohr was mostly right…so
where did he go wrong?
•
Failed to account for why
some spectral lines are
stronger than others. (To
determine transition
probabilities, you need
QUANTUM MECHANICS!)
Auugh!
•
Treats an electron like a
miniature planet…but is an
electron a particle…or a
wave?