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College Physics 150 Chapter 28 – Quantum Physics
•  Wave-­‐‑Particle Duality
•  MaAer Waves
•  The Electron Microscope
•  The Heisenberg Uncertainty Principle
•  Wave Functions for a Confined Particle
•  The Hydrogen Atom
•  The Pauli Exclusion Principle
•  Electron Energy Levels in a Solid
•  The Laser
•  Quantum Mechanical Tunneling
The Wave-­‐‑Particle Duality
Interference and diffraction experiments show that light behaves like a wave. The photoelectric effect, the Compton effect, and pair production demonstrate that light behaves like a particle. Consider a double slit experiment in which only one photon at a time leaves the light source. After a long time, the screen will show a typical interference paAern (c). Even though there is only one photon emiAed at a time, we cannot determine which slit it will pass through nor where it will land on the screen.
The intensity paAern on the screen is representative of the probability that a photon will land in a given location (higher intensity = higher probability). For an EM wave I∝E2, so E2∝ probability of a photon striking the screen at a given location. For an EM, wave E represents the wave function.
MaAer Waves
If a wave (EM radiation) can behave like a particle, might a particle act like a wave?
The answer is yes. If a beam of electrons with appropriate momentum is incident on a sample of material, a diffraction paAern will be evident.
On the right is a diffraction paAern made by x-­‐‑rays incident on a sample. On the left is a diffraction paAern made by an electron beam incident on the same sample. Like photons, the wavelength of a maAer wave is given by
h
λ= .
p
This is known as the de Broglie wavelength.
Example (text problem 28.8): What are the de Broglie wavelengths of electrons with the following values of kinetic energy? (a) 1.0 eV and (b) 1.0 keV.
(a) The momentum of the electron is
p = 2mK
= 2 ( 9.11×10 −31 kg) (1.0 eV) (1.60 ×10 −19 J/eV)
= 5.4 ×10 −25 kg m/s
and
h
6.626 ×10 −34 Js
−9
λ= =
=
1.23×10
m = 1.23 nm.
−25
p 5.4 ×10 kg m/s
Example (text problem 28.8): What are the de Broglie wavelengths of electrons with the following values of kinetic energy? (a) 1.0 eV and (b) 1.0 keV.
(b) The momentum of the electron is
p = 2mK
(
)(
)(
= 2 9.11×10 −31 kg 1.0 ×10 3 eV 1.60 ×10 −31 J/eV
)
= 1.7 ×10 −23 kg m/s
and
h
6.626 ×10 −34 Js
−11
λ= =
=
3
.
88
×
10
m = 38.8 pm.
− 23
p 1.7 ×10 kg m/s
Example (text problem 28.5): What is the de Broglie wavelength of an electron moving with a speed of 0.6c?
γ=
This is a relativistic electron with
1
2
1−
v
c2
= 1.25.
Its wavelength is
h
h
λ= =
p γmv
6.626 ×10 −34 Js
−12
=
=
3
.
23
×
10
m.
−31
8
(1.25) 9.11×10 kg 1.8 ×10 m/s
(
)(
)
A beam of electrons may be used in a double slit experiment instead of a light beam. If this is done, a typical interference paAern will be produced on the screen indicating electrons act like waves.
If a detector is placed to try to determine which of the two slits the electron goes through, the interference paAern disappears indicating the electron now behaves like a particle.
Electron Microscope
The resolution of a light microscope is limited by diffraction effects. The smallest structure that can be resolved is about half the wavelength of light used by the microscope. An electron beam can be produced with much smaller wavelengths than visible light, allowing for resolution of much smaller structures.
Example (text problem 28.15): An image of a biological sample is to have a resolution of 5 nm.
(a) What is the kinetic energy of a beam of electrons with a de Broglie wavelength of 5.0 nm?
p2
h2
K=
=
2m 2mλ2
= 9.64 × 10 − 21 J = 0.060 eV
(b) Through what potential difference should the electrons be accelerated to have this wavelength?
ΔK = ΔU
= −qΔV = eΔV
ΔK 0.060 eV
ΔV =
=
= 0.060 Volts
e
e
(c) Why not just use a light microscope with a wavelength of 5 nm to image the sample?
An EM wave with λ= 5 nm would be an x-­‐‑ray. The Uncertainty Principle
The uncertainty principle sets limits on how precise measurements of a particle’s momentum and position can be.
1
ΔxΔp x ≥ !
2
h
where! =
2π
The more precise a measurement of position, the more uncertain the measurement of momentum will be and the more precise a measurement of momentum, the more uncertain the measurement of the position will be. The energy-­‐‑time uncertainty principle is
ΔEΔt
1
≥ !.
2
Example (text problem 28.18): An electron passes through a slit of width 1.0×10-­‐‑8 m. What is the uncertainty in the electron’s momentum component in the direction perpendicular to the slit but in the plane containing the slit?
The uncertainty in the electron’s position is half the slit width Δx=0.5a (the electron must pass through the slit).
!
!
Δpx ≥
= = 1.1×10 −26 kg m/s
2Δx a
Example (text problem 28.19): At a baseball game, a radar gun measures the speed of a 144 gram baseball to be 137.32±0.10 km/hr.
(a) What is the minimum uncertainty of the position of the baseball?
Δpx = mΔvx and Δvx = 0.10 km/hr = 0.028 m/s.
1
ΔxΔp x = mΔxΔvx ≥ !
2
!
Δx =
= 1.3 ×10 −32 m
2mΔvx
(b) If the speed of a proton is measured to the same precision, what is the minimum uncertainty in its position?
!
Δx =
= 1.1×10 −6 m
2m p Δvx
Wave Functions for a Confined Particle
A particle confined to a region of space will have quantized energy levels.
Consider a particle in a box of width L that has impenetrable walls, that is, the particle can never leave the box.
Since the particle cannot be found outside of the box, its wave function must be zero at the walls. This is analogous to a standing wave on a string.
This particle can have
2L
λn =
n
h
nh
pn =
=
.
λn 2L
With n = 1, 2, 3,…
2
p
The kinetic energy of the particle is KE =
.
2m
Wave Functions for a Confined Particle
This particle can have
2L
λn =
n
h
nh
pn =
=
.
λn 2L
With n = 1, 2, 3,…
2
p
The kinetic energy of the particle is KE =
.
2m
E = K +U
And its total energy is
p2
n2h2
=
+0=
.
2
2m
8mL
The energy of the particle is quantized. The ground state (n = 1) energy is
h2
E1 =
8mL2
E = n 2 E1.
so that n
Example (text problem 28.29): A marble of mass 10 g is confined to a box 10 cm long and moves with a speed of 2 cm/s. (a) What is the marble’s quantum number n?
The total energy of the marble is 1 2
En = mv + 0 = 2.0 ×10 −6 J.
2
In generalEn
2
= n E1
Solving for n:n
=
h2
E1 =
.
2
8mL
En
8mEn L2
28
=
=
6
×
10
.
2
E1
h
Example (text problem 28.29): A marble of mass 10 g is confined to a box 10 cm long and moves with a speed of 2 cm/s. (b) Why do we not observe the quantization of the marble’s energy? The difference in energy between the energy levels n and n+1 is
2
En +1 − En = (n + 1) E1 − n 2 E1
= 2nE1 + E1
= (2n + 1)E1 = 6.6 ×10 −35 J.
The change in kinetic energy of the marble would be
1 2 1 2
mv f − mvi
2
2
1
= m v 2f − vi2
Assume vf ≈ vi.
2
1
= m(v f − vi )(v f + vi ) = mvi (v f − vi ).
2
ΔK =
(
)
To make a transition to the level n+1, the ball’s speed must change by
ΔK
(v f − vi ) = = 3.3 ×10 −31 m/s.
mvi
If a container has walls of finite height, a particle in the box will have quantized energy levels, but the number of bound states (E < 0 ) will be finite. In this situation the wave functions of the particle in the box extend past the walls of the container. This means there is a nonzero probability that the particle can “tunnel” its way through the walls and escape the box. The probability of finding a particle is proportional to the square of its wave function.
The Hydrogen Atom: Wave Functions and Quantum Numbers
In the quantum picture of the atom the electron does not orbit the nucleus. Quantum mechanics can be used to determine the allowed energy levels and wave functions for the electrons.
The wave function allows the determination of the probability of finding the electron in a given region of space. The allowed energy levels in the hydrogen atom are
mk 2 e 4
2
En = −
=
n
E1
2
2!
where E1 = -­‐‑13.6 eV.
n is the principle quantum number.
Even though the electron does not orbit the nucleus, it has angular momentum.
L = l l +1 ! Where l = 0, 1, 2,…n-­‐‑1
(
(
))
l is known as the orbital angular momentum quantum number.
For a given n and l, the angular momentum about the z-­‐‑axis (an arbitrary choice) is also quantized.
Lz = ml !
ml = -­‐‑l, -­‐‑l+1,…, -­‐‑1, 0, +1,…l-­‐‑1, l
ml is the orbital magnetic quantum number.
The spectrum of hydrogen can only be fully explained if the electron has an intrinsic spin. It is useful to compare this to the Earth spinning on its axis. This cannot be truly what is happening since the surface of the electron would be traveling faster than the speed of light.
S z = ms !
ms = ±½ for an electron
ms is the spin magnetic quantum number.
Electron cloud representations of the electron probability density for an H atom:
The Pauli Exclusion Principle
The Pauli Exclusion Principle says no two electrons in an atom can have the same set of quantum numbers. An electron’s state is fully described by four quantum numbers n, l ,ml, and ms.
In an atom:
A shell is the set of electron states with the same quantum number n.
A subshell is a unique combination of n and l. A subshell is labeled by its value of n and quantum number l by using spectroscopic notation.
Each subshell consists of one or more orbitals specified by the quantum numbers n, l, and ml. There are 2l+1 orbitals in each subshell.
The number of electron states in a subshell is 2(2l+1), and the number of states in a shell is 2n2.
The subshells are filled by electrons in order of increasing energy.
1s,2s,2 p,3s,3 p,4s,3d ,4 p,5s,4d ,5 p,6s,4 f ,5d ,6 p,7 s
Beware! There are exceptions to this rule.
The electron configuration for helium is:
1s
2
Specifies n
Specifies l
specifies the number of electrons in this orbital
Example (text problem 28.36): How many electron states of the H atom have the quantum numbers n = 3 and l = 1? Identify each state by listing its quantum numbers.
Here ml = -­‐‑1, 0, 1 and since 2 electrons can be placed in each orbital, there can be 6 electron states.
n
l
ml
ms
3
1
-½
3
1
-1
-1
+½
3
1
0
-½
3
1
0
+½
3
1
+1
-½
3
1
+1
+½
Example (text problem 28.46): (a) Find the magnitude of the angular momentum L for an electron with n = 2 and l = 1?
L = l (l +1)! = 1(1+1)! = 2!
(b) What are the allowed values of Lz?
The allowed values of ml are +1,0,-­‐‑1 so that Lz can be
+1!, 0!, −1!.
(c) What are the angles between the positive z-­‐‑axis and L so that the quantized components, Lz, have allowed values?
L = l (l +1)! = 1(1+1)! = 2!
The allowed values of ml are +1,0,-­‐‑1 so that Lz can be
+1!, 0!, −1!.
Lz
When l = 1, ml = -­‐‑1,0,+1
Lz
ml !
cosθ =
=
L
l (l + 1)!
+ 1!
1
cosθ1 =
=
⇒ θ1 = 45°
2!
2
0!
cosθ 2 =
= 0 ⇒ θ 2 = 90°
2!
− 1!
1
cosθ 3 =
=−
⇒ θ 3 = −45° = 135°
2!
2
1!
0!
− 1!
θ1
θ2
θ3
Electron Energy Levels in a Solid
An atom in isolation will only be able to emit photons of energy E that correspond to the difference in energies between the energy levels in the atom (a line spectrum). When atoms are not in isolation, the wave functions overlap which causes the energy levels to split. As a result, a solid (a large collection of atoms close together) will emit a continuous spectrum.
In a solid, because of the large number of atoms (N) present, each energy level becomes a band of N closely spaced energy levels. Solids also show band gaps where there are no allowed electron energy levels.
A material is a conductor if the highest energy electron state filled at T= 0 is in the middle of the band (the band is only partially filled).
If electrons fill their allowed states right to the top of the band, the material is either a semiconductor or an insulator.
Lasers
Laser is an acronym for Light Amplification by Stimulated Emission of Radiation.
An electron can go to a higher energy level by the absorption of a photon.
When an electron is in an excited state, it can go into a lower energy level by the spontaneous emission a photon.
An electron in an excited state can also go into a lower energy level by the stimulated emission of a photon.
A photon of energy ΔE can stimulate the emission of a photon (by interacting with the excited electron). The emiAed photon will have the same energy, phase, and momentum of the stimulating photon. Example (text problem 28.52): In a ruby laser, laser light of wavelength 694.3 nm is emiAed. The ruby crystal is 6.00 cm long, and the index of refraction of the ruby is 1.75. Think of the light in the ruby crystal as a standing wave along the length of the crystal. How many wavelengths fit in the crystal?
The wavelength of light in the crystal is
694.3 nm
λ= =
= 396.7 nm
n
1.75
λ0
number of wavelengths =
L
λ
= 1.51×10 5.
Tunneling
For a wide barrier, the probability per unit time of a particle tunneling through the barrier is
P∝e
−2κa
where a is the width of barrier and κ is a measure of the barrier height. 2m
(U 0 − E )
κ=
2
!