Quantum Physics

Objectives: After completing this module, you should be

able to:• Discuss the meaning of quantum

physics and Planck’s constant for the description of matter in terms of waves or particles.

• Demonstrate your understanding of the photoelectric effect, the stopping potential, and the deBroglie wavelength.

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

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

Planck’s Equation: E = hf (h = 6.626 x 10-34 J

s)Apparently, light consists of tiny bundles of energy called photons, each having a well-defined quantum of energy.

E = hf

Photon

Energy in Electron-voltsPhoton energies are so small that the

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

One electron-volt (eV) is the energy of an electron when accelerated through a 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 (l = 555 nm)?

First we find f from wave equation: c = fl

; c hcf E hfl l

34 8

-9

(6.626 x 10 J s)(3 x 10 m/s)555 x 10 m

E

E = 3.58 x 10-19 J E = 2.24 eVOr

Since 1 eV = 1.60 x 10-19 J

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

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

9

-19

(1 x 10 nm/m)(in eV)(1.6 x 10 J/eV)hcE

l

If l is in nm, the energy in eV is found from:1240E

l Verify the

answer in Example 1 . . .

The Photo-Electric EffectWhen light shines on the cathode C of a photocell, electrons are ejected from A and attracted by the positive potential due to battery.

Cathode

AnodeIncident light

Ammeter+- A

AC

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

Photo-Electric EquationCathod

eAnode

Incident light

Ammeter+- A

AC

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

212

hcE W mvl

0

hcWl

Threshold wavelength

lo

Example 2: The threshold wavelength of light for a given surface is 600 nm. What is the kinetic energy of emitted electrons if light of wavelength 450 nm shines on the metal?

A

l = 600 nmhc W Kl

0

hc hc Kl l

0

1240 1240450 nm 600 nm

hc hcKl l

; K = 2.76 eV – 2.07 eV

K = 0.690 eV Or K = 1.10 x 10-19 J

Stopping Potential

A

Cathode

AnodeIncident light

Potentiometer

+ -V

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

Kmax = eVo

0E hf W eV Photoelectric

equation:

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

0h WV fe e

Slope of a Straight Line (Review)

The general equation for a straight line is:

y = mx + b

The x-intercept xo

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

ySlope mx

xo x

y

The slope of a line:

yx

Slope

Finding Planck’s Constant, h

Using the apparatus on the previous slide, we determine the stopping potential for a number of incident light frequencies, then plot a graph.

Note that the x-intercept fo is the threshold

frequency.

0h WV fe e

hSlopee

fo

Stopping potential

Frequency

V

Finding h constant

yx

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 fe e

-154.13 x 10 V/HzhSlopee

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

Experimental 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

Equation:

0 0; eV E W W hf

W = (6.63 x 10-34 Js)(1.09 x 1015 Hz) =7.20 x 10-19 J -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 J1.6 x 10 J

V Stopping potential:

Vo = 0.800 V

A

Cathode

AnodeIncident

light

+ -V

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

Total Energy, E 2 2 20( )E m c p c

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

E = moc2

E = pc

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

hcE pcl

hp

l Wavelength of a photon:

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

de Broglie Wavelengt

h:

hmv

l

Finding Momentum from K.E.

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

K = ½mv2 ; p = mv

mK = ½m2v2 = ½p2

Multiply first Equation by m:

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.)

-e- 90 eV

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

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

-19-171.6 x 10 J90 eV 1.44 x 10 J

1 eVK

p = 5.12 x 10-24 kg m/s

h hp mv

l

-34

-24

6.23 x 10 J5.12 x 10 kg m/s

hp

l l = 0.122 nm

Summary

Planck’s Equation: E = hf (h = 6.626 x 10-34 J

s)

Apparently, light consists of tiny bundles of energy called photons, each having a well-defined quantum 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 l is in nm, the energy in eV is found from: 1240E

l

Wavelength in nm; Energy in eV

Cathode

AnodeIncident light

Ammeter+- A

AC

212

hcE W mvl

0

hcWl

Threshold wavelength

lo

Summary (Cont.)

A

Cathode

AnodeIncident light

Potentiometer

+ -V

Kmax = eVo

0h WV fe e

hSlopee

Planck’s Experiment:

fo

Stopping potential

Frequency

V

yx

Slope

Summary (Cont.)

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

E = moc2

E = pc

Quantum physics works for waves or particles:

hp

l Wavelength of a photon:

de Broglie Wavelengt

h:

hmv

l

CONCLUSION: Chapter 38BQuantum Physics