Wave Particle Duality Quantum Physics Lesson 3 Today’s Objectives Explain what is meant by...

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Wave Particle DualityWave Particle DualityQuantum Physics Lesson 3Quantum Physics Lesson 3
Today’s ObjectivesToday’s Objectives Explain what is meant by waveparticle duality.Explain what is meant by waveparticle duality.
Describe the main points of de Broglie’s Describe the main points of de Broglie’s hypothesis that matter particles also have a hypothesis that matter particles also have a wavelike nature.wavelike nature.
State and use the equation State and use the equation λλ = h/p = h/mv = h/p = h/mv
Describe evidence for de Broglie’s hypothesis.Describe evidence for de Broglie’s hypothesis.
Wave particle dualityWave particle duality We have seen……………..We have seen……………..
Photons : Quanta (particles) of light
Electrons: Being diffracted. A property of waves
Prince Louis de BrogliePrince Louis de Broglie1892198718921987
Electrons should not Electrons should not be considered simply be considered simply as particles, but that as particles, but that frequency must be frequency must be assigned to them also.assigned to them also.
(1929, Nobel Prize (1929, Nobel Prize Speech)Speech)
De Broglie (1924)De Broglie (1924) Suggested that particles such as electrons Suggested that particles such as electrons
might show wave properties.might show wave properties. He summised that the He summised that the de Broglie de Broglie
wavelengthwavelength, , λλ was given by: was given by:
m = massv = velocity of the particle
mvh
ph
Note that:• This is a matter wave equation not electromagnetic wave• The de Broglie wavelength can be altered by changing the velocity of the particle.
In words...In words...
momentumconstant sPlanck'h wavelengtBroglie de
velocitymassconstant sPlanck'h wavelengtBroglie de
The diffraction tubeThe diffraction tube
Summary of ExperimentSummary of Experiment Beam of electrons directed at a thin Beam of electrons directed at a thin
metal foil.metal foil. Rows of atoms cause the electron Rows of atoms cause the electron
beam to be diffracted in certain beam to be diffracted in certain directions only.directions only.
We observe rings due to electrons We observe rings due to electrons being diffracted by the same amount being diffracted by the same amount from grains of different orientations, at from grains of different orientations, at the same angle to the incident beam.the same angle to the incident beam.
What we should seeWhat we should see
Electron diffractionElectron diffraction 1927: Davisson & Gerner confirmed 1927: Davisson & Gerner confirmed
this prediction with experiments this prediction with experiments using electron beams.using electron beams.
They actually used a nickel target They actually used a nickel target instead of a carbon one (we used)instead of a carbon one (we used)
The wavelength they measured The wavelength they measured agreed with de Broglieagreed with de Broglie
There is a relationship between the There is a relationship between the accelerating voltage V and the k.e. of accelerating voltage V and the k.e. of the particlesthe particles
Diffraction effects have been shown forDiffraction effects have been shown for
Hydrogen atomsHydrogen atomsHelium atomsHelium atoms
Neutrons Neutrons
Neutron diffraction is an excellent way Neutron diffraction is an excellent way of studying crystal structures.of studying crystal structures.
mvh
ph
What is the wavelength of a human being, assuming he/she weighs 70 kg, and is running at 25 m/s?
De Broglie WavelengthDe Broglie Wavelength In 1932, De In 1932, De
Broglie Broglie discovered that discovered that all particles with all particles with momentum have momentum have an associated an associated wavelength.wavelength.
Practice QuestionsPractice Questions 1.Find the wavelength of an electron of mass 1.Find the wavelength of an electron of mass
9.00 × 109.00 × 103131 kg moving at 3.00 × 10 kg moving at 3.00 × 1077 m s m s11
2. Find the wavelength of a cricket ball of 2. Find the wavelength of a cricket ball of mass 0.15 kg moving at 30 m smass 0.15 kg moving at 30 m s11..
3. It is also desirable to be able to calculate 3. It is also desirable to be able to calculate the wavelength associated with an electron the wavelength associated with an electron when the accelerating voltage is known. when the accelerating voltage is known. There are 3 steps in the calculation. There are 3 steps in the calculation. Calculate the wavelength of an electron Calculate the wavelength of an electron accelerated through a potential difference of accelerated through a potential difference of 10 kV.10 kV.
Step 1: Kinetic energy Step 1: Kinetic energy EEKK = = eVeV = 1.6 × 10 = 1.6 × 101919 × 10000 = 1.6 × 10 × 10000 = 1.6 × 101515 J J
Step 2: Step 2: EEKK = ½ mv = ½ mv22 = ½m (mv) = ½m (mv) 22 = p = p22 / 2m, / 2m,
so momentumso momentumpp = √2m = √2mEEkk = √2 × 9.1 × 10 = √2 × 9.1 × 103131 × 1.6 × 10 × 1.6 × 101515
= 5.4 × 10= 5.4 × 102323 kg m s kg m s11
Step 3: Wavelength Step 3: Wavelength λλ = h / p = 6.63 × 10 = h / p = 6.63 × 103434 / 5.4 × 10 / 5.4 × 102323 = 1.2 × 10 = 1.2 × 10
1111 m m = 0.012 nm.= 0.012 nm.
Slits
Laser
Screen 1
L1
Slit spacing, dWavelength, Distance to screen, LFringe spacing, x
Screen 2L2
d1d2
EndEnd