simon greaves · 2020. 10. 16. · compton studied higher energy x-rays and found a wavelength...
TRANSCRIPT
Compton scattering
Simon Greaves
Research Institute of Electrical Communication, Tohoku University, Sendai
QM3
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Compton scattering I
Scattering of (low energy) X-rays first observed by Barkla in 1909.
The scattered X-rays had the same wavelength as the incident
X-rays.
Thomson’s theory (1900): Incident X-rays cause electrons to
oscillate at the same frequency. Electrons also emit
electromagnetic radiation at this frequency.
Compton studied higher energy X-rays and found a wavelength
shift for some scattered X-rays.
Compton scattering occurs from free, or weakly bound electrons,
i.e. electron binding energy ≪ X-ray energy. It is independent of
the type of material used.
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Compton scattering II
First measured by Arthur Compton in 1922. A photon loses part of
its energy to an electron during a collision. The lower photon
energy results in a lower frequency through the E = hf relationship.
The change in frequency is dependent upon the angle of incidence.
h/mc = 2.426×10−12 m for an electron.
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Compton scattering III
As the scattering angle increases,
two distinct peaks appear.
The Compton effect suggests
that light has a particulate
nature.
Compton won the Nobel prize
in 1927 for the discovery of this
effect.
The shift in wavelength
becomes greater as the
scattering angle increases.
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X-rays I
X-rays are generated by accelerating electrons emitted from a
cathode.
X-rays are emitted when the accelerated electrons collide with the
target material, e.g. Cu, W, Mo etc.
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X-rays II
The incident electron ejects an electron from a target atom.
An electron in the target atom fills the vacancy, losing energy as it
does so.
The energy lost is emitted as an X-ray photon.
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X-rays III
X-rays have fixed wavelengths that depend on the target material.
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Relativistic equations I
Relativistic energy
E =mc2
(
1 −v2
c2
)1
2
(as v → 0 , E → mc2)
Relativistic momentum
p =mv
(
1 −v2
c2
)1
2
(m is the electron mass)
Somv
p=
mc2
Eand v =
pc2
E
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Relativistic equations II
Substitute into E to get
E
(
1 −p2c4
E2c2
)
1
2
= mc2
So
E2− p2c2 = m2c4 and E2 = m2c4 + p2c2
This is the energy of a moving particle with mass m and momentum
p.
For a stationary particle (p = 0) E = mc2.
For a photon (m = 0) E = cp.
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Conservation of momentum I
The incident photon has energy E0 and momentum p0. The
scattered photon has energy E1 and momentum p1 and is scattered
through an angle θ.
The recoiling electron has momentum p2. It recoils at an angle φ.
From conservation of momentum we have
p0 = p1 cos θ + p2 cosφ along the direction of incidence and
p1 sin θ + p2 sinφ = 0 perpendicular to the direction of incidence.
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Conservation of momentum II
Square and collect p2 terms to get
p2
2 sin2 φ+ p2
2 cos2 φ = p2
1 sin2 θ + (p0 − p1 cos θ)
2
Then
p2
2= p2
1sin2 θ + p2
0− 2p0p1 cos θ + p2
1cos2 θ
So
p2
2 = p2
0 + p2
1 − 2p0p1 cos θ
or
p2
2= (p0 − p1)
2 + 2p0p1(1 − cos θ)
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Conservation of energy I
Next, we deal with the energy. Assuming the electron is at rest
before the collision (E = mc2) we have
E0 + mc2 = E1 + (m2c4 + p2
2c2)1/2
Therefore
E0 − E1 = (m2c4 + p2
2c2)1/2
− mc2 = c(p0 − p1)
where c(p0 − p1) is the change in photon energy (E = cp).
So
(m2c4 + p2
2c2)1/2 = c(p0 − p1) + mc2
or
m2c4 + p2
2c2 = m2c4 + c2(p0 − p1)
2 + 2mc3(p0 − p1)
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Conservation of energy II
Divide by c2 to get
p2
2= (p0 − p1)
2 + 2mc(p0 − p1)
We can combine the expressions for p2 obtained from the
conservations of energy and momentum to give
mc(p0 − p1) = p0p1(1 − cos θ)
Now, for the photon
E = cp = hf = hc/λ so p = h/λ
Then
p0 − p1 =h
λ0
−h
λ1
=h(λ1 − λ0)
λ0λ1
and p0p1 =h2
λ0λ1
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Final result
We havemch(λ0 − λ1)
λ0λ1
=h2(1 − cos θ)
λ0λ1
Finally
(λ0 − λ1) = ∆λ =h(1 − cos θ)
mc
The quantity h/mc is known as the Compton wavelength of the
electron. It is independent of the material and incident photon
wavelength.
∆λ increases for larger θ. This experiment also demonstrates the
particulate nature of light.
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