quantum physics
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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. - PowerPoint PPT PresentationTRANSCRIPT
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