Chapter 27
Early Quantum
theory and models of
the atom
Photons
Electromagnetic waves are composed of particle-like
entities called photons.
E = hf
h = 6.626 ×10
−34
J • s ( Planck 's const.)
v = c (in vacuum )
What are the mass and momentum of a photon?
Mass:
2
2
E = mc = γ m0 c =
m0 c 2
v2
1− 2
c
E
v2 E
c2 E
E
m0 = 2 1− 2 = 2 1− 2 = 2 1−1 = 2 0 = 0
c
c
c
c
c
c
Momentum:
E = p 2 c 2 + m02 c 4 = p 2 c 2 + 0 = pc
E hf h ( c λ ) h
∴p= =
=
=
c c
c
λ
true if E is
finite,
i.e. not infinity
Example. What are the a) momentum and b) energy
of a blue photon with wavelength 470 nm?
h 6.63×10 −34
m
−27
a) p = =
= 1.41×10 kg
−9
λ 470 ×10
s
b) E = pc = (1.41×10 −27 ) (3.00 ×108 )
= 4.23×10 −19 J = 2.64 eV
Example: Solar Sails and the Propulsion of Spaceships
One propulsion method that is currently being studied for interstellar
travel uses a large sail. The intent is that sunlight striking the sail
creates a force that pushes the ship away from the sun, much as wind
propels a sailboat. Assume such a sail has a mass per area of
3.00 x 10-3 kg/m2 and the intensity of sunlight with an average wavelength
of 500 nm is 1.01 x 106 W/m2 (at a distance of 8 solar radii from the Sun).
Find the acceleration of the sail and the speed it would have after 1 year
of this acceleration. Assume the mass of the spacecraft is small compared
with that of the sail.
sail mass
kg
W
= σ = 3.00 ×10 −3 2 I Sun = 1.01×10 6 2 λ = 500 nm
area
m
m
h 6.63×10 −34
m
−27
p photon = =
= 1.33×10 kg
−9
λ 500 ×10
s
# photons
I Sun
I Sun
1.01×10 6
=N=
=
=
2
s•m
hf h ( c / λ ) 6.63×10 −34 (3.00 ×108 / 500 ×10 −9 )
= 2.54 ×10 24
photons
s • m2
Calculate the total force on the sail, Fsail, for sail area Asail. Assume the sail is a
black body which absorbs all of the photon momenta.
Δp
= Np photon Asail = ( 2.54 ×10 24 ) (1.33×10 −27 ) Asail = 3.38 ×10 −3 Asail
Δt
Fsail 3.38 ×10 −3 Asail 3.38 ×10 −3
m
=
=
=
= 1.13 2
−3
independent of
msail
σ Asail
3.00 ×10
s
Fsail =
asail
sail area!
v = asail Δt1 year
m
= (1.13) (3.15 ×10 ) = 3.56 ×10
= 0.12 c à  Almost relativistic;
s
à  If reflective à x 2
7
7
(For example, the mass of a 5 km x 5 km sail would be 75,000 kg)
Scattering of a photon off an electron
The Compton effect (c. 1923)
The scattered photon and
the recoil electron depart the
collision in different directions.
Due to conservation of energy,
the scattered photon must have
a smaller frequency.
This is called the Compton effect.
Momentum and energy are
conserved in the collision.

pe, Ee
2
e
hc hc p
E = E ! + Ee ⇒
= +
λ λ ! 2m0
  
h
h
p = p! + pe p =
p! =
λ
λ!

p, E

p!, E !
h
⇒ λ" − λ =
(1− cosθ )
m0 c
m0 = electron mass
h
= Compton wavelength
m0 c
Example. A photon with a wavelength of 0.1140 nm (X-ray) scatters off
an electron at rest. Find a) the Compton wavelength for an electron
and b) the wavelength of the scattered photon if the scattering
angle of the photon with respect to the incident photon direction is 30o,
and c) if the scattering angle is 180o.
λ! − λ =
h
(1− cosθ )
m0 c
h
6.626 ×10 −34
−12
a)
=
=
2.426
×10
m = 0.002426 nm
−31
8
m0 c ( 9.109 ×10 ) ( 2.998 ×10 )
b) λ ! − λ = ( 2.43×10 −12 ) (1− cos30 o ) = 3.26 ×10 −13 m = 0.000326 nm
λ ! = λ + 0.000326 nm = 0.1140 + 0.000326 = 0.1143 nm
c) λ ! − λ = ( 2.43×10 −12 ) (1− cos180 o ) = ( 2.43×10 −12 ) ( 2 ) = 4.86 ×10 −12 m
λ ! = λ + 0.00486 nm = 0.1140 + 0.00486 = 0.1189 nm
Wave – particle duality
We’ve seen examples of how light can act like a wave or like
a particle:
Wave behavior:
Double-slit interference, single-slit diffraction, c = fλ ……
Particle behavior:
Photons (Compton scattering, photoelectric effect, p = h/λ ….)
It is found that particles with mass also exhibit
wave-particle duality
If electrons are used in a double-slit experiment
we expect to see the image of the double slits
on the screen
Instead, a fringe pattern appears indicating
interference effects.
The wavelength of a particle is given
by the same relation that applies to a
photon:
h
λ=
p
de Broglie wavelength
X-ray diffraction on NaCl
Neutron diffraction on NaCl
Neutron diffraction is a manifestation of the wave-like properties
of particles.
Example: The de Broglie Wavelength of an Electron and a Baseball
Determine the de Broglie wavelength of (a) an electron moving at a speed
of 6.0 x 106 m/s and (b) a baseball (mass = 0.15 kg) moving at a speed
of 13 m/s.
6.63×10 J s)
(
h
λ= =
p ( 9.1×10 kg) ( 6.0 ×10
−34
−31
6
m s)
= 1.2 ×10 −10 m
At the scale of atomic
spacing in crystals
interference observable!
−34
h ( 6.63×10 J s)
λ= =
= 3.3×10 −34 m
p ( 0.15 kg) (13m s)
Much smaller than the
scale of the nucleus
interference not observable!
Example. What kinetic energy should neutrons (m = 1.67 x 10-27 kg) have to
exhibit diffraction effects in a crystal with interatomic spacing of 1.00 x 10-10 m?
Set the neutron wavelength to be the interatomic spacing: λ = 1.00 x 10-10 m
h
h 6.63×10 −34
m
−24
λ= ⇒ p= =
= 6.63×10 kg
−10
p
λ 1.00 ×10
s
−24 2
6.63×10 )
(
p
−20
KE =
=
=
1.32
×10
J = 0.0825 eV
−27
2m 2 (1.67 ×10 )
e.g. “thermal” neutrons from a
2
nuclear reactor corresponding
to a temperature of 640oK
3
KE = kT
2
k = Boltzman const.
Example. What is the de Broglie wavelength of a proton with kinetic
Energy of 7 TeV produced from the Large Hadron Collider?
λ=
h
p
E2
2 2
E = p c +m c ⇒ p=
−
m
0c
2
c
2 2
2 4
0
2
12
E = KE + E0 = KE + m0 c = 7.00 ×10
8 2
(1.6 ×10 ) + (1.67 ×10 ) (3.00 ×10 )
−19
−27
= 1.12 ×10 −6 J
−6 2
p=
(1.12 ×10 ) − 1.67 ×10 3.00 ×10
(
)(
)
(3.00 ×10 )
8 2
−27 2
8 2
= 3.73×10 −15 kg
m
s
h 6.63×10 −34
−19
∴λ = =
=
1.78
×10
m à  much smaller than the size of a
−15
p 3.73×10
proton,~10-15 m à good for probing the
proton’s substructure, i.e. quarks & gluons
Particles are waves of probability.
Building up a double-slit interference pattern
electron by electron.