SME
University Physics III: Waves, Optics & Modern
Photons and quantum basics
Use this for “X-rays of 50 pm scatter from the loosely held electrons in a block of graphite. What wavelength comes off at 120°, and how much energy does the electron carry away?”
Diagram
Δλ
=
(
h
mₑc
)(1
−
cos
θ)
Δλ = 2.426 × (1 − cos 90°) = 2.426 pm
λ′ = λ + Δλ = 71 + 2.426
λ′ = 73.43 pm
E
=
hc
λ
=
1,240
71
=
17.46
keV
E′
=
hc
λ′
=
1,240
73.43
=
16.89
keV
K = E − E′ = 17.46 − 16.89
K = 0.577 keV
The electron takes K at φ = 44.04° below the axis; the dashed pₑ closes p = p′ + pₑ.
Six waves of each, to one scale: the bar is 6Δλ = 14.56 pm.
This is an example. Type your own values to start a new problem.
λ′ (pm)
−
λ (pm)
=
(h
÷
mₑc)(1
−
cos(
θ
°))
Δλ
Wavelength shift
calculated · pm
E
Incoming photon energy
calculated · keV
E′
Scattered photon energy
calculated · keV
K
Electron’s kinetic energy
calculated · keV
Formulas
Δλ
=
2.426
×
(1
−
cos(
θ
))
2.426 = 2.426 × (1 − cos(90°))
λ
′
=
λ
+
Δλ
73.426 = 71 + 2.426
E
=
1,240
λ
17.4648
=
1,240
71
E
′
=
1,240
λ
′
16.8878
=
1,240
73.426
K
=
E
−
E
′
0.577 = 17.4648 − 16.8878
Assumptions
•
The electron starts free and at rest, and the photon bounces off it like a ball: energy and momentum are both kept.
•
Keeping both gives the shift Δλ = (h ÷ mₑc)(1 − cos θ), with h ÷ mₑc = 2.426 pm. It does not depend on λ.
•
A photon carries E = hc ÷ λ, with hc = 1240 keV·pm (the same number as 1240 eV·nm).
•
The electron leaves with what the photon lost: K = E − E′.
Step-by-step
WE KNOW
λ = 71 pm, θ = 90°
FIND
wavelength shift (Δλ), scattered wavelength (λ′), incoming photon energy (E), scattered photon energy (E′), electron’s kinetic energy (K)
Step 1 · Find wavelength shift (Δλ)
Use Δλ = 2.426 × (1 − cos(θ))
The
Compton
shift:
h
mₑc
=
2.426
pm
times
1
−
cos
θ
.
It
is
0
straight
ahead
and
4.852
pm
straight
back.
Δλ = 2.426 × (1 − cos(90°))
Δλ = 2.426 pm
Step 2 · Find scattered wavelength (λ′)
Use λ′ = λ + Δλ
The scattered photon’s wave is longer by the shift.
λ
′
=
71
+
2.426
λ
′
=
73.426
pm
Step 3 · Find incoming photon energy (E)
Use E = 1,240 ÷ λ
hc = 1240 keV·pm over the wavelength in pm gives the energy in keV.
E
=
1,240
71
E
=
17.4648
keV
Step 4 · Find scattered photon energy (E′)
Use E′ = 1,240 ÷ λ′
The same for the scattered photon: its longer wave carries less energy.
E
′
=
1,240
73.426
E
′
=
16.8878
keV
Step 5 · Find electron’s kinetic energy (K)
Use K = E − E′
Energy is kept: what the photon lost, the electron carries off.
K
=
17.4648
−
16.8878
K
=
0.577
keV
Check
Put the numbers back in. Both sides should match.
2.426 = 2.426 × (1 − cos(90°)) ✓
73.426 = 71 + 2.426 ✓
17.4648
=
1,240
71
✓
16.8878
=
1,240
73.426
✓
0.577 = 17.4648 − 16.8878 ✓
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The Lorentz transformation
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The photoelectric effect and the stopping potential
Topic check
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