Chapter 27

Early Quantum theory and models of

the atom

Photons

Electromagnetic waves are composed of particle-like entities called photons.

E = hf

v = c in vacuum( )

h = 6.626×10−34 J •s Planck 's const.( )

What are the mass and momentum of a photon?

E =mc2 = γm0c2 =

m0c2

1− v2

c2

m0 =Ec2

1− v2

c2=Ec2

1− c2

c2=Ec2

1−1 = Ec2

0 = 0true if E is finite, i.e. not infinity

Mass:

Momentum:

E = p2c2 +m02c4 = p2c2 + 0 = pc

∴ p = Ec=hfc=h c λ( )c

=hλ

Example. What are the a) momentum and b) energy of a blue photon with wavelength 470 nm?

a) p = hλ=6.63×10−34

470×10−9=1.41×10−27kg m

sb) 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 massarea

=σ = 3.00×10−3 kgm2 ISun =1.01×10

6 Wm2 λ = 500 nm

pphoton =hλ=6.63×10−34

500×10−9=1.33×10−27kg m

s# photonss•m2 = N =

ISunhf

=ISun

h c / λ( )=

1.01×106

6.63×10−34 3.00×108 / 500×10−9( )= 2.54×1024 photons

s•m2

Fsail =ΔpΔt

= NpphotonAsail = 2.54×1024( ) 1.33×10−27( )Asail = 3.38×10−3Asail

asail =Fsailmsail

=3.38×10−3Asail

σ Asail=3.38×10−3

3.00×10−3=1.13m

s2

v = asailΔt1 year = 1.13( ) 3.15×107( ) = 3.56×107 ms = 0.12 c

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.

independent of sail area!

(For example, the mass of a 5 km x 5 km sail would be 75,000 kg)

à Almost relativistic; à  If reflective à x 2

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.

Scattering of a photon off an electron The Compton effect (c. 1923)

Momentum and energy are conserved in the collision.

⇒ "λ −λ =hm0c

1− cosθ( )

m0 = electron masshm0c

=Compton wavelength

p,E

!p , !E

pe,Ee E = !E +Ee ⇒hcλ=hc!λ+pe2

2m0

p = !p + pe p = hλ

!p = h!λ

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 =hm0c

1− cosθ( )

a) hm0c

=6.626×10−34

9.109×10−31( ) 2.998×108( )= 2.426×10−12m = 0.002426 nm

b) !λ −λ = 2.43×10−12( ) 1− cos30o( ) = 3.26×10−13m = 0.000326 nm

!λ = λ + 0.000326 nm = 0.1140+ 0.000326 = 0.1143 nm

c) !λ −λ = 2.43×10−12( ) 1− cos180o( ) = 2.43×10−12( ) 2( ) = 4.86×10−12m!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:

λ =hp

de Broglie wavelength

Neutron diffraction is a manifestation of the wave-like properties of particles.

X-ray diffraction on NaCl

Neutron diffraction on NaCl

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.

λ =hp=

6.63×10−34 J s( )9.1×10−31kg( ) 6.0×106 m s( )

=1.2×10−10m

λ =hp=

6.63×10−34 J s( )0.15 kg( ) 13m s( )

= 3.3×10−34 m

At the scale of atomic spacing in crystals interference observable!

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?

λ =hp⇒ p = h

λ=6.63×10−34

1.00×10−10= 6.63×10−24kg m

s

KE = p2

2m=6.63×10−24( )

2

2 1.67×10−27( )=1.32×10−20 J = 0.0825 eV

e.g. “thermal” neutrons from a nuclear reactor corresponding to a temperature of 640oK

KE = 32kT

k = Boltzman const.

Set the neutron wavelength to be the interatomic spacing: λ = 1.00 x 10-10 m

Example. What is the de Broglie wavelength of a proton with kinetic Energy of 7 TeV produced from the Large Hadron Collider?

λ =hp

E = p2c2 +m02c4 ⇒ p = E 2

c2−m0

2c2

E = KE +E0 = KE +m0c2 = 7.00×1012 1.6×10−19( )+ 1.67×10−27( ) 3.00×108( )

2

=1.12×10−6 J

p =1.12×10−6( )

2

3.00×108( )2 − 1.67×10

−27( )23.00×108( )

2= 3.73×10−15kg m

s

∴λ =hp=6.63×10−34

3.73×10−15=1.78×10−19m à much smaller than the size of a

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.