Transcript Document

Compton effect
Another experiment revealing the
particle nature of X-ray (radiation,
with wavelength ~ 10-10 nm)
Compton, Arthur Holly (1892-1962),
American physicist and Nobel
laureate whose studies of X rays led
to his discovery in 1922 of the socalled Compton effect.
The Compton effect is the change
in wavelength of high energy
electromagnetic radiation when it
scatters off electrons. The discovery
of the Compton effect confirmed
that electromagnetic radiation has
both wave and particle properties,
a central principle of quantum
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theory.
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Beam of x-ray with sharp wavelength l
falls on graphite target. For various angle
q the scattered x-ray is measured as a
function of their wavelength
Although initially the incident
beam consists of only a single
well-defined wavelength (l) the
scattered x-rays have intensity
peaks at two wavelength (l’ in
addition), where l’ > l.
Wavelength
shift
Unexplained by classical
wave theory for radiation
No shift of wavelength is
predicted in wave theory
of light
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Compton (and independently by Debye)
explain this in terms of collision
between collections of (particle-like)
photon, each with energy E = hn, with
the free electrons in the target graphite
(imagine billard balls collision)
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Initial
photon,
E=hc/l,
p=h/l
Initial
electron, at
rest,
Eei=mec2,
pei=0
1: Conservation of E:
hc/l + mec2 = hc/l’ + Ee
2: Conservation of
momentum:p = p’ + pe
(vector sum)
Scattered
photon,
E’=hc/l’,
p’=h/l’
y
q
x
f
Scattered
electron, Ee,pe
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Conservation of momentum in 2-D
p = p’ + pe (vector sum) actually comprised
of two equation for both conservation of
momentum in x- and y- directions
Conservation
of l.mom in ydirection
p’sinq = pesinf
p = p’cosq + pecosf
Conservation of l.mom in x-direction
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Some algebra…
Mom conservation in y:
p’sinq = pesinf
(PY)
Mom conservation in y:
pecosf = p - p’cosq
(PX)
Conservation of total relativistic energy:
E + mec2 = E’ + Ee
(RE)
(PY)2 + (PX)2, substitute into (RE)2 to eliminate Ee and pe:
(E + mec2 - E’)2 = c2(p2 – 2pp’cosq +p’2)+ mec2,
leading to
Dl ≡ l’- l = (h/mec)(1 - cosq)
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Compton wavelength
le = h/mec = 0.0243 Angstrom, is the Compton
wavelength (for electron)
Note that the wavelength of the x-ray used in the
scattering is of the similar length scale to the
Compton wavelength of electron
The Compton scattering experiment can now be
perfectly explained by the Compton shift
relationship
Dl ≡ l’ - l = le(1 - cosq)
as a function of the photon scattered angle
Be reminded that the relationship is derived by
assuming light behave like particle (photon)
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Dl ≡ l’ - l = (h/mec)(1 - cosq)
Notice that Dl depend on q only,
not on the incident wavelength,
l.
Consider some limiting
behaviour of the Compton shift:
For q = 00  “grazing”
collision => Dl = 0
l
Dlmax
l  l’
q0
For q = 1800 “head on” collision
Dl= Dlmax = 2 lc =2( 0.00243nm)
(photon being reversed in direction), Dl is
180o maximum
l
l’=0.1795 nm
l’= l+Dlmax
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Example
X-rays of wavelength 0.2400 nm
are Compton scattered and the
scattered beam is observed at
an angle of 60 degree relative
to the incident beam.
Find (a) the wave length of the
scattered x-rays, (b) the
energy of the scattered x-ray
photons, (c) the kinetic energy
of the scattered electrons, and
(d) the direction of travel of
the scattered electrons
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solution
l’= l + lc (1 - cosq)
(1–
= 0.2400 nm + 0.00243 nm
cos 60o)
= 0.2412 nm
E’ = hc/l’
= 1240 eV∙nm /0.2412 nm
= 5141 eV
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p’g
pg
Initial
photon
Eg
me
E’g < Eg
q
f
K
pe
kinetic energy gained by the
scattered electron = energy
transferred by the incident photon
during the scattering:
K = hc/l - hc/l’ = (5167 – 5141)eV
= 26 eV
Note that we ignore SR effect here because K << rest mass of electron, me = 0.5
MeV
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p’g
pg
Initial
photon
Eg
me
E’g < Eg
q
f
K
pe
By conservation of momentum in the x- and y- direction:
pg= p’g cosq + pe cosf; p’g sinq = pe sin f;
tan f = pe sin f / pe cosf = (p’g sinq)/ (pg - p’g cosq)
= (E’g sinq)/ (Eg - E’g cosq) = (5141 sin 600 / [5167-5141 (cos
600] = 0.43 = 1.71
Hence, f = 59.7 degree
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Rayleigh scattering and
Compton scattering
There are two kinds of electrons in the graphite target:
free electron and electron that is bounded to individual
atom
The sifted peak corresponds to scattering of stationary
free electron with incident photon (for Ke >> biding
energy). This is just the Compton effect
The unshifted peak corresponds to collision of
photon with strongly bounded electron which is not
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free (incident photon energy << biding energy)
Photon scattering off a bounded
electron sees the mass of the
whole atom instead of only that
of a free electron. Hence the
Compton shift, instead being
given by
Dle = (h/mec)(1 - cosq)
the mass term is now give by me
 Matom (Matom >> me) instead, such
that Dl (due to M atom) << Dl(due to
me). Hence, the scattered
l’appears unmodified
This is called Rayleigh scattering 15
explainable in classical theory
mass seen by
photon = me
l
le
Compton
scattering:
Dl1 = (h/mec)(1 - cosq)
mass seen by
photon = Matom
l
le
Rayleigh
scattering:
Dl2 = (h/Matomc)(1 - cosq)
Dl1 << Dl2 because me << Matom
(or equivalently, l ~ le >> latom
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In the limit l infinity (low incident
photon energy), l >> lc , Rayleigh
scattering dominates (The effective
mass is large as seen by the incident
photons)
As l  lc Compton scattering
dominates
Note that Compton scattering occurs at
energy scale which is much higher, ~
0.1 MeV (typical energy of the x-ray
photon, with the wavelength ~ Compton
wavelength of the electron
In comparison, photoelectricity occurs
at a much lower energy scale, ~ eV.
Different mode of interaction happen at different
energy scale (or length scale)
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Ref: A general reminder
1) The next tutorial will happen in 26 Dec
2003, friday, 5.00 pm. Please make sure to
pass up the solution to your tutors latest by
12.00 pm on that day (Friday). You can
check out the course schedule for the rest
of the semester in
http://www.fizik.usm.my/tlyoon/teaching/
calander.htm
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tutor Mr. Sia Chen How (Kumpulan
12), please collect back your tutorial
solutions from the box outside his room
(room no. 256).
3) I have instructed all tutors to provide the
password to the tutorial 1 full solution,
which u can download from our zct 104
course web page. If you dunt have the
password, pls obtain it from your frens or
contact me via e-mail.
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4) Please check that your SR lecture notes (in pdf. format)
is the latest version. I discover that the version of notes
some students has is an old one. It is advised that you
should download the latest version (updated on 2 Dec
2003).
5) I strongly encourage students to discuss with me any
problem they face with the course (e.g. poor english,
problem to obtain lecture notes, difficulties in learning or
comprehending the course material etc.). This, of course,
does not include your love affair problem lah.=b You may
discuss with me in person after lectures or in my room
(which i prefer); call me up or write to me. Please step
forward and dont remain silent. I am always willing to
listen and to help.
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1.
2.
3.
4.
5.
6.
4) Please take some time to read the extra
material I have put up in the course web page,
e.g.
Extra reading material
Michelson-Morley experiment:
http://www.phys.virginia.edu/classes/109N/lect
ures/michelson.html
Ether
Relativity
Time
photon
photoelectric effect
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Even God could not help those
who don’t help themselves
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