Quantum PhysicsAdapted from the 2007 AP Institute

Objectives: After completing this module, you should be able to:

• Discuss the meaning of Discuss the meaning of quantum quantum physicsphysics and and Planck’s constantPlanck’s constant for the for the description of matter in terms of description of matter in terms of waves or particles.waves or particles.

• Demonstrate your understanding of the Demonstrate your understanding of the photoelectric effectphotoelectric effect, the , the stopping stopping potentialpotential, and the , and the deBroglie deBroglie wavelengthwavelength..

• Explain and solve problems similar to Explain and solve problems similar to those presented in this unit.those presented in this unit.

Planck’s ConstantIn his studies of black-body radiation, In his studies of black-body radiation, Maxwell Planck discovered that Maxwell Planck discovered that electromagnetic energy is emitted or electromagnetic energy is emitted or absorbed in discrete quantities.absorbed in discrete quantities.

Planck’s Equation:

E = hf (h = 6.626 x 10-34 J s)

Apparently, light consists Apparently, light consists of tiny bundles of energy of tiny bundles of energy called called photonsphotons, each , each having a well-defined having a well-defined quantumquantum of energy. of energy.

Apparently, light consists Apparently, light consists of tiny bundles of energy of tiny bundles of energy called called photonsphotons, each , each having a well-defined having a well-defined quantumquantum of energy. of energy.

E = hf

Photon

Energy in Electron-voltsPhoton energies are so small that the Photon energies are so small that the

energy is better expressed in terms of the energy is better expressed in terms of the electron-voltelectron-volt..

One One electron-volt (eV) is the energy of is the energy of an electron when accelerated through a an electron when accelerated through a potential difference of one volt.potential difference of one volt.

One One electron-volt (eV) is the energy of is the energy of an electron when accelerated through a an electron when accelerated through a potential difference of one volt.potential difference of one volt.

1 eV = 1.60 x 10-

19 J 1 keV = 1.6 x 10-16 J

1 MeV = 1.6 x 10-13

J

Example 1: What is the energy of a photon of yellow-green light ( = 555 nm)?

First we find First we find f f from wave equation: from wave equation: c = f

; c hc

f E hf

34 8

-9

(6.626 x 10 J s)(3 x 10 m/s)

555 x 10 mE

E = 3.58 x 10-19 JE = 3.58 x 10-19 J E = 2.24 eVE = 2.24 eVOrOr

Since 1 eV = 1.60 x 10Since 1 eV = 1.60 x 10-19-19 JJ

Useful Energy ConversionSince light is often described by its Since light is often described by its wavelength in wavelength in nanometers (nm)nanometers (nm) and its and its energy energy E E is given in is given in eVeV, a conversion formula , a conversion formula is useful. (1 nm = 1 x 10is useful. (1 nm = 1 x 10-9-9 m) m)

-19(in Joules) ; 1 eV 1.60 x 10 Jhc

E

9

-19

(1 x 10 nm/m)(in eV)

(1.6 x 10 J/eV)

hcE

If If is inis in nm nm, the energy in, the energy in eV eV is found is found from:from:

1240E

1240E

Verify the Verify the

answer in answer in Example 1 . . .Example 1 . . .

The Photo-Electric Effect

When light shines on When light shines on the cathode the cathode CC of a of a photocell, electrons photocell, electrons are ejected from are ejected from CC and attracted by the and attracted by the positive potential positive potential due to battery.due to battery.

Cathode

Anode

Incident light

Ammeter++-- A

AC

There is a certain There is a certain thresholdthreshold energy, energy, called the called the work function Wwork function W, that must be , that must be overcome beforeovercome before any any electrons can be electrons can be emitted.emitted.

There is a certain There is a certain thresholdthreshold energy, energy, called the called the work function Wwork function W, that must be , that must be overcome beforeovercome before any any electrons can be electrons can be emitted.emitted.

Photo-Electric Equation

Cathode

Anode

Incident light

Ammeter++-- A

AC

The The conservation of energyconservation of energy demands that demands that the energy of the incoming light the energy of the incoming light hc/hc/ be be equal to the work function equal to the work function W W of the surface of the surface plus the kinetic energy plus the kinetic energy ½mv2 of the emitted of the emitted electrons.electrons.

The The conservation of energyconservation of energy demands that demands that the energy of the incoming light the energy of the incoming light hc/hc/ be be equal to the work function equal to the work function W W of the surface of the surface plus the kinetic energy plus the kinetic energy ½mv2 of the emitted of the emitted electrons.electrons.

212

hcE W mv

212

hcE W mv

0

hcW

Threshold

wavelength

Example 2: The threshold wavelength of light for a given surface is 450 nm. What is the kinetic

energy of emitted electrons if light of wavelength 600 nm shines on the metal?

A

= 600 nmhcW K

0

hc hcK

0

1240 1240

450 nm 600 nm

hc hcK

; K; K = 2.76 eV – 2.07 eV = 2.76 eV – 2.07 eV

K = 0.690 eVK = 0.690 eV OrOr K = 1.10 x 10-19 JK = 1.10 x 10-19 J

450

1240

600

1240K

Stopping Potential

A

Cathode

Anode

Incident light

Potentiometer

++ --

V

A potentiometer is A potentiometer is used to vary to the used to vary to the voltage voltage V V between between the electrodes.the electrodes.

KKmaxmax = eV = eVoo

0E hf W eV Photoelectric Photoelectric

equation:equation:

The stopping The stopping potential is that potential is that voltage voltage VVoo that just that just stops the emission of stops the emission of electrons, and thus electrons, and thus equals their original equals their original K.E.K.E.

0

h WV f

e e

0

h WV f

e e

Slope of a Straight Line (Review)The general equation The general equation for a straight line is:for a straight line is:

y = mx + by = mx + b

The The x-interceptx-intercept xxoo

occurs when line occurs when line crosses crosses xx axis or axis or when when y = 0y = 0. . The slope of the line is The slope of the line is the rise over the run:the rise over the run:

ySlope m

x

ySlope m

x

xo x

y

The slope of a line:

y

x

Slope

Finding Planck’s Constant, hUsing the apparatus on the previous slide, Using the apparatus on the previous slide, we determine the stopping potential for a we determine the stopping potential for a number of incident light frequencies, then number of incident light frequencies, then plot a graph.plot a graph.

Note that the x-intercept Note that the x-intercept ffoo is the is the threshold threshold

frequency.frequency.

0

h WV f

e e

0

h WV f

e e

hSlope

e

hSlope

e

fo

Stopping potential

Frequency

V

Finding h constant

y

x

Slope

Example 3: In an experiment to determine Planck’s constant, a plot of stopping potential

versus frequency is made. The slope of the curve is 4.13 x 10-15 V/Hz. What is Planck’s constant?

fo

Stopping potential

Frequency

V

yx

Slope0

h WV f

e e

0

h WV f

e e

-154.13 x 10 V/Hzh

Slopee

h = eh = e(slope) = (1.6 x 10(slope) = (1.6 x 10-19-19C)(4.13 x 10C)(4.13 x 10-15-15 V/Hz) V/Hz)

Experimental Planck’s h = 6.61 x 10-34

J/HzExperimental Planck’s h = 6.61 x 10-34

J/Hz

Example 4: The threshold frequency for a given surface is 1.09 x 1015 Hz. What is the stopping

potential for incident light whose photon energy is 8.48 x 10-19 J?

0E hf W eV Photoelectric Photoelectric

Equation:Equation:

0 0; eV E W W hf

WW = (6.63 x 10 = (6.63 x 10-34 -34 Js)(1.09 x 10Js)(1.09 x 1015 15 Hz) =7.20 x Hz) =7.20 x 1010-19 -19 JJ -19 -19 -19

0 8.48 x 10 J 7.20 x 10 J 1.28 x 10 JeV -19

0 -19

1.28 x 10 J

1.6 x 10 JV Stoppin

g potential:

Vo = 0.800 V

A

Cathode

AnodeIncident light

++ --

V

Total Relativistic EnergyRecall that the formula for the relativistic Recall that the formula for the relativistic total energy was given by:total energy was given by:

Total Energy, E

For a particle with For a particle with zero momentumzero momentum p p = = 00::A light photon has A light photon has mmoo = 0, but it = 0, but it does have does have momentum momentum pp::

E = moc2

E = pc

22222 cpmcE

Compton Effect

• Arthur Holly Compton showed the x-ray photons have a momentum of hf/c.

• Recall the unit for the above, is it dimensionally consistent?

• He showed the wavelength shift of a scattered x-ray only depends on the scattering angle.

• So it gives up some of its energy to the scattered electron.

Waves and ParticlesWe know that light behaves as both a wave We know that light behaves as both a wave and a particle. The rest mass of a photon is and a particle. The rest mass of a photon is zero, and its wavelength can be found from zero, and its wavelength can be found from momentum.momentum.

hcE pc

h

p Wavelengt

h of a photon:

All objectsAll objects, not just EM waves, have , not just EM waves, have wavelengths which can be found from their wavelengths which can be found from their momentummomentum

de Broglie Wavelengt

h:

h

mv

Finding Momentum from K.E.

In working with particles of momentum In working with particles of momentum p = p = mv, mv, it is often necessary to find the it is often necessary to find the momentum from the given kinetic energy momentum from the given kinetic energy K. Recall the formulas:K. Recall the formulas:

K = K = ½mv2 ; p = mv

mK =mK = ½m2v2 = ½p2

Multiply first Multiply first Equation by Equation by mm::

Momentum from K:

2p mK

Example 5: What is the de Broglie wavelength of a 90-eV electron? (me = 9.1 x 10-31 kg.)

-ee-- 90 eV90 eV

Next, we find momentum Next, we find momentum from the kinetic energy:from the kinetic energy: 2p mK

-31 -172(9.1 x 10 kg)(1.44 x 10 J)p

-19-171.6 x 10 J

90 eV 1.44 x 10 J1 eV

K

p = p = 5.125.12 x 10x 10-24-24 kg kg m/sm/s

h h

p mv

h h

p mv

-34

-24

6.23 x 10 J

5.12 x 10 kg m/s

h

p = 0.122

nm = 0.122 nm

Summary

Planck’s Equation:

E = hf (h = 6.626 x 10-34 J s)

Apparently, light consists Apparently, light consists of tiny bundles of energy of tiny bundles of energy called called photonsphotons, each , each having a well-defined having a well-defined quantumquantum of energy. of energy.

Apparently, light consists Apparently, light consists of tiny bundles of energy of tiny bundles of energy called called photonsphotons, each , each having a well-defined having a well-defined quantumquantum of energy. of energy.

E = hf

Photon

1 eV = 1.60 x 10-

19 J 1 keV = 1.6 x 10-16 J 1 MeV = 1.6 x 10-13

J

The Electron-volt:

Summary (Cont.)

If If is inis in nm nm, the energy in, the energy in eV eV is found is found from:from:

1240E

1240E

Wavelength in Wavelength in nm; Energy in eVnm; Energy in eV

Cathode

Anode

Incident light

Ammeter++-- A

AC

212

hcE W mv

212

hcE W mv

0

hcW

Threshold

wavelength

Summary (Cont.)

A

Cathode

Anode

Incident light

Potentiometer

++ --

V

KKmaxmax = eV = eVoo

0

h WV f

e e

0

h WV f

e e

hSlope

e

hSlope

e

Planck’s Planck’s Experiment:Experiment:

fo

Stopping potential

Frequency

V

yx

Slope

Summary (Cont.)

For a particle with For a particle with zero momentumzero momentum p = p = 0:0:A light photon has A light photon has mmoo = 0 = 0, but it , but it does have does have momentum momentum pp::

E = moc2

E = pc

Quantum physics works for waves or Quantum physics works for waves or particles:particles:

Quantum physics works for waves or Quantum physics works for waves or particles:particles:

h

p Wavelengt

h of a photon:

de Broglie Wavelengt

h:

h

mv

CONCLUSION: Quantum Physics Rocks!