More experimental evidence of the quantized nature of radiation and the photon picture
was provided by the famous Compton effect observed when monochromatic X-rays of wavelnegth λ were scattered from a target material like graphite. At high angles of scattering an extra peak at wavelength λ' greater than the original
wavelength was observed in the intensity vs wavelength plot.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compdat.html#c1
This was termed as the Compton shift Δλ=λ-λ' was dependent on the scattering angle θ
The classical picture of radiation as an EMW predicted that the free electron in the material would be set into forced oscillations by the oscillating electric field of the EMW and would re radiate at the same wavelength and frequency as the incident radiation. There was no explanation for a peak in the Intensity for a shifted wavelength λ' from Classical Theory of EMW.
Compton viewed this effect as Relativistic Collison between an energetic photon
with E=hν and a stationary electron with rest mass m0.
Relativistic collison theory is distinct from collisons in classical mechanics
due to the rest mass energy and energy and momentum are on the same footing.
(In fact the 3 mom and energy are the 4 components of the relativistic momentum 4-vector in 3 space+ 1 time dimensional Minkowski space)
The operative relation for energy is the famous equation of Einstein
E=[ m0c2]γ where
1/γ=(1-v2/c2)1/2
and m0 is the rest mass of the particle.
The energy momentum formula is
E2=c2p2+ m02C4
For photons with rest mass ZERO E=cp=hν and hence p=h/λ where p is the photon momentum
Now we apply momentum and energy conservation along x and y directions and obtain the two key formula
E0-E1=K where E are the energy of the incident and scattered photons and K is the kinetic energy of the electron
or
c(p0-p1=K ......(1) as E=Cp for the photon and
p2=p20+ p21 -2 p0 p1Cos θ.....(2)
We also have the formula
K2/C2 - 2Km0=p2=p2......(3)
we substitute eqn. (2) in (3) and using (1) we have
1/p1 - 1/p0 = (1/m0c) (1-Cos θ )........(4)
multiplying (4) by h and using p=h/λ we have
λ1 - λ0 = λcomp (1-Cos θ)
=Δλ
where λcomp= (h/m0c) is called the Compton Wavelength and is measured for the electron to be 0.0243 Angstorms.
This analysis explains the shifted wavelength in the Compton effect. The other peak at the original incident wavelength may be understood as scattering from tightly bound electrons. If the incident photon energy is low this situation is like that of the incident photon colliding with the atom as a whole ( electron is tightly bound to the atom) so the mass in the Compton wavelength in the scattering formula is
the mass of the atom M which is far larger than that of the electron. Hence the Compton wavelength is vanishingly small and the Compton shift goes to ZERO.
was provided by the famous Compton effect observed when monochromatic X-rays of wavelnegth λ were scattered from a target material like graphite. At high angles of scattering an extra peak at wavelength λ' greater than the original
wavelength was observed in the intensity vs wavelength plot.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compdat.html#c1
This was termed as the Compton shift Δλ=λ-λ' was dependent on the scattering angle θ
The classical picture of radiation as an EMW predicted that the free electron in the material would be set into forced oscillations by the oscillating electric field of the EMW and would re radiate at the same wavelength and frequency as the incident radiation. There was no explanation for a peak in the Intensity for a shifted wavelength λ' from Classical Theory of EMW.
Compton viewed this effect as Relativistic Collison between an energetic photon
with E=hν and a stationary electron with rest mass m0.
Relativistic collison theory is distinct from collisons in classical mechanics
due to the rest mass energy and energy and momentum are on the same footing.
(In fact the 3 mom and energy are the 4 components of the relativistic momentum 4-vector in 3 space+ 1 time dimensional Minkowski space)
The operative relation for energy is the famous equation of Einstein
E=[ m0c2]γ where
1/γ=(1-v2/c2)1/2
and m0 is the rest mass of the particle.
The energy momentum formula is
E2=c2p2+ m02C4
For photons with rest mass ZERO E=cp=hν and hence p=h/λ where p is the photon momentum
Now we apply momentum and energy conservation along x and y directions and obtain the two key formula
E0-E1=K where E are the energy of the incident and scattered photons and K is the kinetic energy of the electron
or
c(p0-p1=K ......(1) as E=Cp for the photon and
p2=p20+ p21 -2 p0 p1Cos θ.....(2)
We also have the formula
K2/C2 - 2Km0=p2=p2......(3)
we substitute eqn. (2) in (3) and using (1) we have
1/p1 - 1/p0 = (1/m0c) (1-Cos θ )........(4)
multiplying (4) by h and using p=h/λ we have
λ1 - λ0 = λcomp (1-Cos θ)
=Δλ
where λcomp= (h/m0c) is called the Compton Wavelength and is measured for the electron to be 0.0243 Angstorms.
This analysis explains the shifted wavelength in the Compton effect. The other peak at the original incident wavelength may be understood as scattering from tightly bound electrons. If the incident photon energy is low this situation is like that of the incident photon colliding with the atom as a whole ( electron is tightly bound to the atom) so the mass in the Compton wavelength in the scattering formula is
the mass of the atom M which is far larger than that of the electron. Hence the Compton wavelength is vanishingly small and the Compton shift goes to ZERO.
Combining The understanding of photoelectric effect with Black body experiment. What would be the significance of n in E=nhv since photon has energy hv and there were several states of energy levels of radiation in the boltzman distribution.
ReplyDeleteI want to know what this E*P(E) vs E graph represents. I have some idea but i wanted to ask the physical implication of the graph. Is it
ReplyDeleteY-axis :: E*P(E) The energy content of an oscillator of "the given" frequency at energy level E
X-axis :: E The energy level of the oscillators at the given frequency
For the same E*P(E) vs E
ReplyDeleteArea under the curve is the total energy content of all the oscillators of that frequency divided by the total number of oscillators (To convert N to P) oscillating at that frequency.
Is it that::: all the frequencies are possible in the spectrum but the energy of the oscillators at a particular frequency can take only a discrete value.
ReplyDeleteThat is :: Discreteness lies in the energy level of the waves at some frequency but all the frequencies are possible (continuously), fundamentally speaking.
What would be the meaning of the cloud concept that multiple number of wave packets occur at one point. Would there be any superposition and addition of any quantity like field etc
2>> sir, the textbook continues on the lines of the gamma ray microscope thought experiment and tries to derive the energy-time uncertainty from the same ideas... while doing so they take px/m = vx as the recoil velocity due to the incident illumination... however this was the velocity before illumination... the recoil velocity seems to my eyes more likely to be del-px/m
ReplyDeletethe quantity vx del-t they call uncertainty is simply the distance travelled in the original direction over the time of obsevation which can be which is precisely known if time of illumination(which is in our hands) is specified exactly. i am not able to find the quantum uncertainty in this... where am i going wrong?
posting this doubt here until the post on uncertainty principle appears... will repost it wherever it is finally most relevant...
ReplyDelete1>> in the microscope thought experiment the statemnt is "we can see the electron if only one scattered photon enters the objective lens of the microscope"... later we speak of a diffraction pattern due to which the position of the electron is uncertain... would a single scattered photon form a diffraction pattern?? isnt it expected to show particle-like dot in the measurement device (retina) that ABSORBS it?
i am a bit lost in the hand waving it seems... plz help...
can a finite del-lambda actually result in nonzero amplitude being completely restricted in a finite del-x?? i am somehow not able to convince myself that this can happen without seeing it in a graph...
ReplyDeleteany finite number of waves howsoever closely spaced would have auxiliary waves after a finite spatial period(=LCM of wavelengths)... to get rid of these auxiliaries we need to discard any possibility of discreteness in the wavelength available to the photon... but if any method of generating these photons has an inherent discreteness associated with it isnt our assumption of a continuum of freq available to the photon just a mathematical idealization? if there is no continuum then the particle may be in any of the auxiliary waves extending over all space...
where am i going wrong?
Could you please clarify this statement given in the Eisberg Resnick book
ReplyDeleteIn the Davison and Germer experiment
The interference involved is not between waves associated with one electron and wave associated with another. Instead it is an interferece between different parts of waves associated with a single electron that has been from various regions of of the crystal.
The interference is always between different parts
ReplyDeleteof a single wave so even a single electron would show a diffraction pattern when averagde over a long exposure time. That is single electrons at a time from a source.
Tamaghna:
ReplyDeleteI am unable to follow your arguments about aux waves. However its always true that delta k delta x ~ 1.
Tamaghna:
ReplyDeleteDiffraction is a wave phenomena and in principle even a single photon would show a diffraction pattern when averaged over a sufficiently long exposure time. That is a source which gives one photon at a time. You cant deal with both particle and wave aspect simultaneously. The wave behaviour is predominant at low frequencies and large photon numbers. Check the text.
Rahul:
ReplyDeleteAs far as black body spectrum is concerned the discreteness is in the energy of the atomic oscillators and delta epsilon=h nu in Plancks argument.
Tamaghna: All delta x and delta p for free particles is of the order of x and p respectively.
ReplyDeleteThese are order of magnitude estimates and are correct.
Rahul:
ReplyDeleteE is the energy of the oscillator and P(E) is the probability of finding the oscillator between E and E + dE.
Rahul:
ReplyDeleteIn the averaging process the denominator has value 1
and the numerator is the energy weighted by the probability that an oscillator will be found in E to E + dE. This is the average energy.
Sir, I am not referring to a source that gives one photon "at a time"... the quoted text if i understand it correctly claims that the detector is sensitive enough to register the absorption of a single electron "ever"... would this single electron not a train of electrons cause a particle like dot instead of a diffraction pattern on the detector??
ReplyDeleteSir, isnt the delta x taken in delx delk ~ 1 the separation between two consecutive minima... what I am referring to is the distance beyond which the amplitude is roughly zero till x=infinty
ReplyDelete