Transcript Slide 1

Definition of the symbols:
– Atom in its normal (non-excited) state
– Atom in excited state
 h 0 
Photon
0
Absorption of
the photon by
the atom produces:
Now, I want you to solve a simple problem:
● The typical lifetime of an electron on an excited orbit is ~ 1 ns;
● The linear speed of an electron on the n = 1 orbit in a H atom
1/137 of the speed of light (it was a homework problem, remember?);
● The radius of the n = 1 orbit in hydrogen (i.e., the “Bohr radius”)
is approximately a0 = 0.05 nm ;
I want you to find how many revolutions the electron on the
excited n = 2 orbit makes before it returns to the “ground
state” (i.e., the n=1 state)? – only the order of magnitude
Is of interest to us, don’t do very accurate calculations!
Answer :
T heradius of the n  2 orbit is 4a0  2 nm;
T heangularm om entumof theelectronon then  2 orbit
is twice thaton then  1 orbit,so that thelinear speed is
1 c
c
one- half of thaton then  1 orbit, i.e., it is 

;
2 137 274
speed time
T he# of revolutions 
orbit circumference
3 108 m/s 10-9 s
4


9

10
274 2  2 10-9 m
The electron makes nearly 100,000 revolutions, so it is
clear that it “does not remember” from which direction
came the photon which pushed it to the higher orbit.
The excited atom returns
To the ground state,
re-emitting the photon
which flies away
in a completely
random direction:
Re-emitted
photon
So, if a beam of photons
with frequency 0 is
incident on a container
with such atoms,they
will excitetheatoms...
Gas of
atoms
...and they will
be re - emittedin
all possible
directions.
Gas of
atoms
OK, why we needed to explain that will become clear
little later….
Please remember: this type of photon emission we
Considered is called a “spontaneous emission”.
Now, lets go back to the year 1900. Max Planck first
finds an empirical formula describing the spectrum of
the blackbody radiation.
Shortly afterwards, he derives the same formula
theoretically, based on an assumption that electromagnetic
radiation can be absorbed and emitted by the atoms
comprising the blackbody only in discrete “portions” that
he calls quanta. Quantum physics is born at this moment!
T hefamousPlanck's equation:

1
 c  8  hc 
R( )    4   hc / kT 
 1
 4      e
A few years later, Albert Einstein explains the photoelectric effect, based on the hypothesis that light consists of
particle-like photons of energy h .
Max Planck does not believe in “light quanta” and strongly
opposes the Einstein’s hypothesis. He insists that only
atoms has quantum properties, but electromagnetic
radiation is continuous.
But Einstein strongly believes in his own theory. Therefore,
he does not like the way Plank derived his famous formula.
He thinks of how it could be done in a way consistent with
his “photon theory”
Einstein reaches the conclusion that in addition to the
“spontaneous emission” of photons there must be
another emission process:
An atom that has
already been excited
by a photon with
frequency 0
Photon
0
is hit by another
photon of the
same frequency
Two photons
The second photon
“catalyzes” an
emission process.
The atom returns to
the ground state.
of frequency 0
Photon
0
Photon
0
fly away in the
same direction!
This process becomes known as a
“Stimulated Emission”
But at the time Einstein constructs this hypothesis, there is no
way of verifying it by an experiment… However, Einstein obtains
the proof in an indirect way…
In 1917, Einstein attacs the “blackbody problem” using a completely
different approach than that of Planck. He considers a container of
atoms, NA of which are in the non-excited state, and NB are in the
excited state.
The atoms
are “immersed” in a
“photon gas”.
(0) is the
density of
of photons
with freqency 0 in
this gas.
The gray is a “gas” of photons with frequency 0
Let A be therateof spontaneous emmision:
if N B atomsare in theexcitedstate,then
the number that will return totheground state
via a spontaneous emission in a timeunit is N B A.
Since thestimulatedemission is caused by photons,
the number of atomsthat willreturn totheground
statevia stimulatedemission in a timeunit must be
proportion
al to thedensityof photons  ( 0 ); this
number can be writtenas : N B B ( 0 ),
where B is an appropriate coefficient.
Now, consider the atomsthat
are in theground state.T hey
will absorb photonsand will
be transferred to theexcited
state.
T henumber of atomstransferred per unit timeis proportion
al to
N A and to thephotongas density,so it can be writtenas N AC ( 0 ),
where C is a coefficient.
Altogether, therateat which he
t populationN B of theexcited
atomschangetin timecan be writtenas :
dNB
  N B A  N B B ( 0 )  N AC ( 0 )
dt
Repeatedfrom theprecedingpage :
dNB
  N B A  N B B ( 0 )  N AC ( 0 )
dt
Now, there comes a moment for a very important step in
our reasoning. We assume that the system is in
EQUILIBRIUM. The proportion of the excited and nonexcited atoms remains constant in time.
Therefore, the above derivative is zero.
From that, we obtain in a very simple way:
 N B A  N B B ( 0 )  N AC ( 0 )  0
A
  ( 0 ) 
( N A / N B )C  B
A
From theprecedingslide :  ( 0 ) 
( N A / N B )C  B
Now comes the only moment when you have to accept
something I tell you without a proof.
Let’s call the energy of the atom in the ground
state as EA, and the energy of the excited atom as EB.
From thermal physics it is known that if a system
is in equilibrium at temperature T, then the proportion
of the # of particles with energy EA to the # of particles
with energy EB is given by the so-called “Boltzmann
Factor”:
 E A / kT
NA e
  EB / kT  e( EB  E A ) / kT  e h 0 / kT
NB e
(because in thepresentcase EB  E A  h 0 ).
A
So, we have:  ( 0 ) 
and
( N A / N B )C  B
NA
 e h 0 / kT
NB
And hence:
A
 ( 0 )  h 0 / kT
e
CB
When Einsteinobtainedtheabove formula,he comparedit with
the P lanck's equation expressednot in termsof wavelength,
but in termsof frequency- namely:
1
4h 3
 ( )  3  h 0 / kT
c e
1
Einstein noticed that his formula becomes essentially
identical with the Planck’s one, if B=C.
B=C: what does it mean? At the moment Einstein
reached that conclusion, he discovered something
that is of crucial importance in LASER physics!
Let’s recall:
N B B ( 0 ) : thenumber of stimulatedemission events;
N AC ( 0 ) : thenumber of " ordinary"absorptionevents;
B=C means that the transition rate for stimulated emission
is the same as for absorption.
Translating this into a somewhat more “human”
language: the above two expressions determine the
way a photon is going TO BE USED.
Let’s consider not a “gas” of photons, but a SINGLE photon incident
on the system:
Photon
0
If the number of the non-excited (“blue”) atoms is larger
than of these excited (“red”), then the photon most likely
will engage into an “ordinary absorption act”;
It will be scattered “out of the system”
Recall:
Gas of
atoms
Photon
0
Only if the number of excited (“red”) atoms is HIGHER
than the number of non-excited (“blue”) atoms,
THE PHOTON HAS A GREATER CHANCE “TO BE USED”
for triggering a STIMULATED EMISSION
So, if
N excited  N non -excited
(and onlyif),
a " CHAIN REACT ION"may occur
But “chain reaction” is a word from the nuclear weapon vocabulary.
And since we are PEACEFUL PEOPLE, we prefer the term
LASER ACTION
LASER =
Light Amplification through Stimulated Emission of Radiation
The most important message to remember:
Laser action may occur only if:
# of excited atoms  # of those non - excited
Number of excited atoms 1

OR:
Number of all atoms
2
T hereare more than50%
of excitedatoms
OR:
in thesystem
Such situation
is called
“INVERSE
POPULATION”
(or “inverted
population”)
THE GREATEST CHALLENGE IN LASER PHYSICS IS TO ACHIEVE
SUCH A SITUATION!!!!!