THE COMPTON EFFECT
Energy and
momentum are
conserved when
a photon collides
with an electron.
KE=1/2mv2
X-ray
E1
E2
E1 = E2 + KE(electron)
EXAMPLE
A) A Photon having an energy of 3.8eV collides
with an electron at rest. The electron has a gain
in kinetic energy of 1.2eV, what will be the
energy of the emitted photon in eV and joules?
E1 = E2 + KE(electron)
3.8eV = E2 + 1.2eV
E2= 2.6eV
E2 = 4.16x10-19J
*(1 eV = 1.6 x 10-19J)*
B) What is the frequency of the scattered
light wave and what color is it?
E2 =hf
4.16x10-19J= 6.63 x 10-34Js (f)
f = 6.23 x 1014Hz
color = blue
c) What is the velocity of the electron?
KE= 1/2mv2
1.2eV(1.6 x 10-19J/eV)= ½ (9.11x10-31kg)v2
1.92 x 10-19J = 4.56x10-31(v2)
4.2 x 1011 = v2
v= 6.49 x 105m/s
DeBroglie Wavelength
If light has particle properties then particles
must have wave properties
λ= h = h
mv
ρ
λ- wavelength h- Planks constant
m- mass v-velocity ρ- momentum
EXAMPLE 1
• What is the DeBroglie wavelength of a
0.25kg ball thrown at 20m/s?
λ=
h
mv
= 6.63 x 10-34 Js
(0.25kg)(20m/s)
λ =1.33 x 10-34m
*This is too small to observe.*
EXAMPLE 2
What is the DeBroglie wavelength of an
electron traveling at 1x106m/s?
λ=
h
mv
=
6.63 x 10-34Js
(9.11x10-31kg)(1x106m/s)
λ= 7.29 x 10-10m
* Suitable for interference and diffraction and
can be observed*
EXAMPLE 3
• What is the momentum of a proton having
a DeBroglie wavelength of 2.81x10-11m?
λ=h/ρ
2.81x10-11m = 6.63 x10-34Js
ρ
ρ = 2.36x10-23kgm/s
Heisenberg Uncertainty Principle
• “To study the nature and motion of a
moving electrons by bombarding them
with photons would change the motion and
position of the electron and lead to
uncertainty”
• You can’t accurately measure both
position and momentum of a moving
electron at any given time.