Transcript Blackbody Radiation

Blackbody Radiation
• All bodies emit EM radiation because of their
temperature (thermal radiaiton) but they are not
usually in equilibrium.
(but radiation emitted: thermal radiation+reflected)
• However consider an opaque enclosure with walls
at constant temperature.
Radiation and walls reach thermal equilibrium.
Under these conditions the radiation has quite
definite properties.
•How can we study this radiation?
If we cut a tiny hole in the enclosure it will not
disturb the eqbm. And the emitted radiation will
have the same properties as the Cavity Radiation.
•Emitted spectrum = spectrum from a perfect black
body at the same T as the enclosure.
For this reason cavity radiation is also called
black-body radiation.
Blackbody Radiation
•Thermal Radiation:-All heated objects emit EM radiation
e.g.Hot coals on a fire glow.
Stars shine-we see the radiation from the hot surface
Incandescent filament
•Classical Physics was unable to describe these spectra
Blackbody Radiation
Experimental results:
Stefan-Boltzmann law
PA =  T4 , where
= 5.67 x 10 -8 Wm -2 K -4
PA is the power emitted per unit surface area.
Wien’s displacement law
λmax.T = const
= 2.9 x 10 -3 mK
Origin of Blackbody Radiation
The origin of the em. waves in the cavity must relate
to the thermal motion of charged constituents of
atoms in the material (we know they are electrons)
Basic idea = radiation incident on a surface is reflected
because it is quickly absorbed and re-emitted.
Rayleigh and Jeans worked out the spectral shape
(classical theory).
Each wavelength is emitted by atoms oscillating at a
frequency  = c/λ.
The predicted spectrum is then the product of two
functions - N() the number of oscillators with freq.
, which is proportional to  2
and
The average energy of an oscillator with freq. ,
which is equal to kT
Result S prop. to 2,
which is infinite at large , small λ
Ultraviolet catastrophe
Rayleigh-Jeans Formula
The assumptions made by Rayleigh and Jeans
lead to the spectrum for a blackbody marked
“Classical Theory” below.
Note:-Here the spectrum is plotted as a function
of frequency.
Rayleigh-Jeans Formula
Here it is plotted as the more familiar function of
Wavelength.
Physical Picture of Planck’s quantum hypothesis
●Assume atomic oscillators behave like simple
harmonic oscillators with V = ½ kx2 parabola
Sun-Yellow
Red
• Spectrum of Blackbody Radiation
as a function of wavelength.
• Energy emitted by four blackbodies with
equal surface areas.
• Note that they are plotted on log-log scale.
• Area is proportional to total power per unit surface
area (PA)
• Stefan-Boltzmann Law: PA = .T4
• PA is in Wm-2 and  = 5.67 x 10 -8 Wm-2 K-4
Blackbody Radiation
• General question:-What is the spectrum of
EM radiation emitted by an object of arbitrary
temperature T in thermal equilibrum.
We assume that this “blackbody”
reflects no radiation at any 
•Max Planck showed that the spectrum is given by
ud  = 8hc -5.d
[exp(hc/kT) - 1]
ud  is the energy density
=energy/unit volume
• Although no perfect blackbody exists
solids and stars follow Planck’s Law very closely.
Note that picture is on log-log scale.
Wien’s Displacement Law
• Doubling T increases
P by 16 since
PA = .T4
• Note that maximum
wavelength max shifts with
T.This can be quantified in
Wien’s Displacement Law.
max.T = const.
= 2.9 x 10-3 mK
• This quantifies the observation
that an object changes colour
with Temperature e.g.At room
temp. spectrum peaks in infra-red.
• Very important since it allows us to obtain a
measure of the SURFACE TEMPERATURE of a
star from max.For the Sun max is in blue with a lot
of radiation in red so it looks yellow.For stars with
T = 3000k max is in infrared but significant amount
in red.Red Giants are at this T.
Planck’s Solution
●Planck decided that the problem lay in the
assumption that the average energy of the oscillators
is a constant and not in the number of modes of
oscillation.
●He looked for a way of reducing the number of high
frequency oscillators in the cavity. In other words
there is too much radiation at small λ.
●He suggested the idea that an oscillating atom can
absorb or re-emit energy only in discrete bundles or
quanta.
●If the energy of the quanta are proportional to the
frequency() then as  becomes large then  becomes
large. Since no standing wave can have E>kT, no
standing wave can exist with an energy quantum
larger than kT. This effectively limits the high
frequency (small wavelength) radiant intensity and
removes the “Ultraviolet catastrophe”.
[Max Planck Nobel Prize 1918
Blackbody radiation: ear thermometer
Planck’s Solution
● Planck had two new ideas.
1)he assumed that the energy allowed per oscillator
is not continuous.
In other words the oscillators can only have
certain discrete values of energy.
Energy is quantised.
2)He assumed that the gaps between allowed values
of energy are greater for high frequency
oscillators.
He made the simple guess that
E = nh,
where h = const., n = 1,2,3,4,----and the energy of each quantum is h.
The average energy per osc. then tends to zero as
 increases.
•Max Planck showed that the spectrum is given by
u d  =
8hc -5.d
[exp(hc/kT) - 1]
Wm-2
This fits perfectly if h = 6.63 x 10 -34 Js
Note:-Planck was slow to accept this-like others.
The Photoelectric Effect
• Many people, including Planck, were very sceptical
about his solution to the Black-body radiation
problem.The drawback was the idea of quantisation
of the energies of the oscillators.
• The thing which convinced them was Einstein’s
explanation of the Photoelectric Effect.
• The phenomenon was first observed by Hertz in 1887.
He noticed that the air in a spark gap became a better
conductor when illuminated by UV light.
• It was then studied in detail by Hallwachs, Lenard,
Millikan and others.
Basically they showed that electrons are emitted from
a metal surface when illuminated by light of
sufficiently high frequency. In other words there is a
threshold in frequency before anything happens.
•It needs a systematic study to highlight the important
features.
Vstop = kinetic energy of fastest electrons.
Photoelectric effect-Results to be explained
1) The number of electrons ejected per unit time
is proportional to the intensity of the Light.
2)The electrons are emitted with velocities up to a
maximum velocity(VSTOP < 0 and # 0)
3)The maximum kinetic energy does not depend on
intensity.
4)There is a THRESHOLD FREQUENCY 0 such
that there is no emission for  < 0
The threshold frequency depends on the metal.
5)There is no measurable delay between the light
striking the metal and the emission of the electrons
no matter the incident light intensity,i.e.no matter
how low the intensity.
Wave Theory and Photoelectric Effect.
• Wave theory would suggest that the maximum kinetic
energy should depend on the intensity of light.
•The wave theory predicts no threshold effect. It
would be simply a matter of collecting enough
energy.
•A measurable time delay would be expected. Enough
energy has to be accumulated before an electron can
be emitted.
So the Wave Theory has failed but Einstein had a
simple theory based on quantisation.
Einstein’s Theory of Photoelectric Effect.
•Einstein proposed that light (EM radiation) consists
of particle-like packets of energy called photons
•Each photon carries an energy
E = h
where h = Planck’s constant,  = wave frequency
•This extends Planck’s ideas regarding emission and
absorption of radiation so that they apply when the
radiation is transmitted.
•The emission of electrons is caused by single photons
being completely absorbed by individual electrons.
•
Free electrons
E

Bound
Electrons
•The energy  is called the Work Function.the
maximum kinetic energy for electrons is then
Kmax = h - 
Note:-We can have K < Kmax for tightly bound
electrons or because of collisons during emission.
Photoelectric Effect-Einstein’s Hypothesis
● Einstein proposed that radiation is quantised not just
in absorption and emission but in all situations.
•The energy  is called the Work Function
•The max. kinetic energy for electrons is
Kmax. = h - 
we can have K < Kmax for tightly bound electrons
or collisions during the emission of the electrons
•Note that Kmax = VSTOP x e
where e = charge on electron
•So VSTOP plotted vs.  has slope h/e and we can
measure h and compare it with value from BB
radiation formula.
•They agree with
h = 6.626 x 10-34 Js
= 4.136 x 10-15 eV.s
•The Einstein model accounts for all 5 experimental
features of the photoelectric effect.
Photoelectric effect-Results to be explained
1)The no.of electrons ejected per unit time is propnl.
to the intensity of the Light.
[ It is proportional to the no. of photons]
2)The electrons are emitted with velocities up to a
maximum velocity(VSTOP < 0 and # 0)
[Kmax = h -  ]
3)The maximum kinetic energy does not depend on
intensity.
[ It depends instead on the energies of the photons]
4)There is a THRESHOLD FREQUENCY 0 such
that there is no emission for  < 0
The threshold frequency depends on the metal.
[ if h <  then we cannot release an electron]
5)There is no measurable delay between the light
striking the metal and the emission of the electrons
no matter the incident light intensity,i.e.no matter
how low the intensity.
The Compton Effect
•Arthur Compton(1923). His experiments showed
definite particle-like(photon) behaviour for X-rays.
•Simply the result is () function of (0)
' =  + 
•Compton scattering occurs in addition to the classical
process of Thomson scattering where
() = (0)
and we have absorption and re-radiation.
The Compton shift  has
a definite ang. dependence.
•It does not depend on the
precise material used as the
scatterer.
•This suggests that the
photons collide with
something which is in all
materials,namely electrons.
Compton Effect
•Compare X-ray photon energy with electron energies
-Binding energies are  few eV
•X-ray photon with  = 0.07nm,the wavelength used
by Compton.
E = h = hc/
=6.63.10-34.3.108
0.07.10-9 1.6.10-19
= 17.7 keV
»electron binding energy
•Consider a collision between a photon and a free
unbound electron at rest.
For a photon, E = pc
[A photon must have m0 =0 since  =1 ]
•We can calculate the results for such a collision
assumimg
1)Conservation of momentum
2)Conservation of mass-energy
3)Energy-momentum relationship
Compton Scattering
•Collision can be analysed in
two dimensional lab. Frame
•Photon momentum is given
by p = E/c = h/c
•We can resolve momentum
in the horizontal(pL) and
transverse(pT) directions
pL:
h/c = h'/c x cos +pecos
pT:
0
E:
h + m0c2 = h' + sqrt(pe2c2 + m02c4 )
= h'/c x sin  =pesin
What we want is an expression for the final
frequency ' as a function of the initial frequency 
and the scattering angle  and must eliminate the
unknowns  and pe
Compton Scattering
Schematic of Compton
Scattering process.
We can also solve for the energy of the scattered
Photon.
h / =
h0
[ 1 + h0/ m0c2( 1 – cos θ )]
a) When θ = 1800 this gives an energy for the
scattered photon of ~ m0c2/2 if the energy of the
incident photon is much larger than the rest mass
of the electron.
b) When θ = 900 this gives an energy of ≈ m0c2 for
the scattered photon when the incident photon
energy is large.
Photons
The ideas of Planck and Einstein and the
experiments of Compton seem to show conclusively
that light is composed of photons.
However, earlier , we had decided with the 19th C.
physicists that light is a waveform since it describes
very well the phenomena of interference and
diffraction.
How can we reconcile the two. Let us first look at
what happens as we allow only a tiny amount of
light through an optical spectrum. This gives an
insight into the situation.
1AMQ, Part II Quantum Mechanics
4 Lectures
• Wave Particle Duality, the Heisenberg
Uncertainty Principle, de Broglie
(wavelength) hypothesis
• Schrodinger Wave Equation.
• Wavefunctions and the Free Particle.
• Electrons in a box, energy quantisation.
•K.Krane, Modern Physics, Chapters 4 and 5
• Eisberg and Resnick, Quantum Physics,
Chapters 3 (and 5 & 6)
Wave-Particle Duality.
Light behaves like: (a) a wave (diffraction,
interference) and/or (b) a stream of massless
particles or photons, (black body spectrum,
photoelectric effect, Compton scattering).
Consider diffraction
from a slit, one photon at
a time.
The experiment shows
that individual particles
of light gradually build
up the diffraction pattern
predicted by the classical
wave theory.
The wave pattern
describes the probability
of detecting a photon at
that point.
The wave pattern measures the probability
detecting a photon at that point.
Individual photons unpredictable in detail,
BUT their average behaviour is predictable!
At any point P in the interference pattern, the
intensity gives the probability: i.e.
I(P) is prop. to prob. that a photon hits pos. P.
For waves, the intensity = (amplitude)2 and thus
the wave and particle pictures are connected,
Probability
a (amplitude of wave)2
This is called the Born Interpretation (Max
Born, Nobel Prize, 1954).
The question then arises, if waves can act like
particles, is the reverse true ?
The de Broglie Hypothesis
Louis de Broglie (for his PhD thesis!) postulated
that in some phenomena, particles can act like
waves!
(Nobel Prize, 1929)
He guessed that for particles, the momentum, p,
could be related to a wavelength,  , by
p = h/ where  is the de Broglie wavelength.
Note this is consistent with the momentum for
photons, ie p = E/c = h/ = h/
This predicts, for example, that electrons should
show diffraction effects (like photons), if they
pass through a slit with width ~ dB.
Calculating the de Broglie Wavelength.
Example.
e-s in a TV Ke-=20 keV, me=9.1x10-31kg, gives
dB=0.009nm. (Note that the typical atomic
spacing is ~0.05nm).
dB=0.05nm corresponds to Ke-=601 eV for
electrons. [from Ke=p2/2me=(1/2m) . (h/)2]
ie. e-s accelerated across 600 V should be
diffracted exactly like x-rays with =0.05nm.
X-ray diffraction
pattern
Electron diffraction
pattern
Note how (2 slit)
e- interference
pattern builds up
with increasing
number of e-s.
Electron and neutron diffraction
are widely used in crystallography
studies.
The y-position for any transmitted photon is
specified to an accuracy (uncertainty)  y=a.
Making the slit smaller (ie. reducing  y)
increases the uncertainty in the angle, .
Thus, ypy~ h (for single slit diffraction).
This relates to the physical limits we can put
on determining either the position or
momentum of the individual photons as they
pass through the slit. These are a subset of the
Heisenberg Uncertainty Principle.
In general, if p = momentum, t = time and
E= energy, h = Planck’s constant ,then,
xp x  h
xp x  h
2
, xp x  h
2
, Et  h
2
2
The division by 2 comes about due to the
use of wave number (k) rather than
wavelength (=2/k) in the definition of
momentum (see Krane p114).
Basis for Schrodinger’s equation
• Light can be described
- as a wave
- as a stream of particles
}in some
circumstances
• Following De Broglie we can say that Matter also
has this dual wave-particle character.
• Probability  | Amplitude of wave | 2
- the Born Interpretation.
•Uncertainty Principle - Heisenberg
p.x  h/2
E.t  h/2
The Schrödinger Equation.
For particles, forces change momentum (p=mv)
and each force has an associated potential energy.
Newton’s 2nd Law relates the potential to the
change in momentum via the relations
2
dp
d x
dV
F = = ma = m 2 and for 1 - D, F = 
dt
dt
dx
Solving the Schrodinger equation specifies Y
(x,t) completely, except for a constant, ie. if Y
is a solution, so is A xY .
From the Born interpretation we have ||2.dx as
the probability of finding the particle at position
x.Since the particle must be somewhere the
integral of this quantity from - to + must
equal 1.0.This allows us to normalise the wavefn.
We will solve this equation for the wavefunction
with an amplitude  as a function of position 
and time t
Time Independent Schrodinger Equation
In many cases, (eg. Coulomb potential in H
atom), V=V(x), ie. not time dependent, then
Y( x, t ) =  ( x)e
2
i t
where (x) satisifies
2
 d
[
 V ( x)] = E or, H = E
2
2m dx
Properties of the T.I.S.E.
•Applies when V=V(x), ie. V is independent of
time.Eg. an e- wave acted on by a fixed nucleus,
but NOT an atom in an oscillating magnetic field.
•Applies only in the non-relativistic limit
(ie. assumes K=(1/2)mv2=p2/2m).
•Assumes p=h/ and E=h.
•Is linear in , ie no terms like 2, 3 etc. Laws
of superposition are valid. If a and b are
two possible solutions, then =caa+cbb
is a solution for all constants ca and cb.