Download PHYS150-Ch27

College Physics 150 Chapter 27 – Quantum Physics and the Photon
•  Blackbody Radiation
•  The Photoelectric Effect
•  X-­‐‑ray Production
•  Compton ScaGering
•  Early Models of the Atom
•  The Bohr Model for the Hydrogen Atom
•  Pair Production/Annihilation
Everything is solved!
Everything is solved!.....just a few things to clean up!
Photoelectric Effect
Emission Spectra
It was thought that classical mechanic and electricity and magnetism could explain everything. Just a few more questions to clear up.
Ultraviolet Catastrophe
Quantum Mechanics!
Niels Bohr
Developed quantum theory to lay down the foundation for the understanding of the atom.
Max Planck
The founder of quantum mechanics and quantum physics.
His constant relates energy to the frequency of light
Lois De Broglie
Discovered the wave nature of electrons and therefore all maGer.
Quantum Mechanics!
Werner Heisenberg
Developed wave mechanics in quantum theory. Well known for his development of the uncertainty principle. Albert Einstein
Solves the photoelectric effect and develops special and general relativity (next chapter)
Erwin Schrödinger
Developed wave-­‐‑matrix mechanics for quantum theory. Also, known for the Schrödinger cat paradox
Quantization
A quantity is quantized if its possible values are limited to a discrete set.
An example from classical physics is the allowed frequencies of standing waves on a stretched string. Only integer multiples of the fundamental frequency produce standing waves. Blackbody Radiation
A blackbody emits a continuous spectrum of radiation. The spectrum is determined only by the temperature of the blackbody.
E0 = hf
The quantum of EM radiation is the photon.
where h = 6.626×10-­‐‑34 J s is called Planck’s constant.
To correctly explain the shape of the blackbody spectrum Planck proposed that the energy absorbed or emiGed by oscillating charges came in discrete bundles called quanta. The energy of the quanta are
The Photoelectric Effect Under certain circumstances EM radiation incident on a metal will eject electrons from the metal. This is the photoelectric effect. Experiments show:
1.  Brighter light causes more electrons to be ejected, but not with more kinetic energy.
2.  The maximum KE of ejected electrons depends on the frequency of the incident light.
3.  The frequency of the incident light must exceed a certain threshold, otherwise no
electrons are ejected.
4.  Electrons are ejected with no observed time delay regardless of the intensity of the
incident light.
The wave theory of light says EM waves carry energy. The energy is absorbed by electrons in the metal target which are then ejected when they accumulate enough energy to escape. However the wave theory is unable to completely explain the photoelectric effect. Einstein proposed a particle theory of light.
Particle or Wave?
Experiments show:
1.  Brighter light causes more electrons to be ejected, but not with more kinetic energy.
Wave theory predicts a more intense beam of light, having more energy, should cause more electrons to be emiGed and they should have more kinetic energy.
Particle theory predicts a more intense beam of light to have more photons so more electrons should be emiGed, but since the energy of a photon does not change with beam intensity, the kinetic energy of the ejected electrons should not change.
The particle theory is consistent with observation 1.
Experiments show:
2. The maximum KE of ejected electrons depends on the frequency of the incident light.
Wave theory cannot explain the frequency dependence of the maximum kinetic energy. Particle theory predicts the maximum kinetic energy of the ejected electrons to show a dependence on the frequency of the incident light. Each electron in the metal absorbs a whole photon: some of the energy is used to eject the electron and the rest goes into the KE of the electron.
The particle theory is consistent with observation 2.
The maximum KE of an ejected electron is
KEmax = hf − φ
where φ is called the work function and is the energy needed to break the bond between the electron and the metal. Experiments show:
3. The frequency of the incident light must exceed a certain threshold, otherwise no
electrons are ejected.
Wave theory can offer no explanation.
Particle theory predicts a threshold frequency is needed. Only the incident photons with f > fthreshold will have enough energy to free the electron from the metal. The particle theory is consistent with observation 3.
The electron is ejected from the metal when the energy supplied by the photon exactly equals the work function. This defines the threshold frequency.
hf threshold − φ = 0
f threshold =
φ
h
Here it is often convenient to use h = 4.136×10-­‐‑15 eV s.
Experiments show:
4. Electrons are ejected with no observed time delay regardless of the intensity of the
incident light.
Wave theory predicts that if the intensity of the light is low, then it will take some time before an electron absorbs enough energy to be ejected from the metal.
Particle theory predicts a low intensity light beam will just have a low number of photons, but as long as f > fthreshold an electron that absorbs a whole photon will be ejected; no time delay should be observed.
The particle theory is consistent with observation 4.
The particle theory of light is needed to explain the photoelectric effect (and Compton scaGering and pair production). A wave theory of light is needed to explain interference paGerns. Both are correct!
Wave-­‐‑Particle Duality
Example (text problem 27.1): A 200 W infrared laser emits photons with a wavelength of 2.0×10-­‐‑6 m while a 200 W ultraviolet laser emits photons with a wavelength of
7.0×10-­‐‑8 m.
(a) Which has greater energy, a single infrared photon or a single ultraviolet photon?
E = hf =
hc
The UV photon has the greater energy; its wavelength is smallest.
λ
(b) What is the energy of a single infrared photon and the energy of a single ultraviolet photon?
EUV =
EIR =
hc
λUV
hc
λIR
= 2.8 ×10 −18 J = 18 eV
= 9.9 × 10 − 20 J = 0.62 eV
(c) How many photons of each kind are emiGed per second?
energy emitted/sec
number of photons emitted per second =
energy/photon
For both lasers the energy emiGed per second is 200 J. The UV laser emits 7.0×1019 photons/sec and the IR laser emits 2.0 ×1021 photons per second.
Example (text problem 27.4): The photoelectric threshold frequency of silver is 1.04×1015 Hz. What is the minimum energy required to remove an electron from silver?
KEmax = hf threshold − φ = 0
φ = hf threshold
(
)(
= 6.626 ×10 −34 Js 1.04 ×1015 Hz
= 6.89 ×10 −19 J = 4.30 eV
)
Example (text problem 27.11): Two different monochromatic light sources, one yellow (580 nm) and one violet (425 nm), are used in a photoelectric effect experiment. The metal surface has a photoelectric threshold frequency of
6.20×1014 Hz. (a) Are both sources able to eject photoelectrons from the metal? Explain.
c
The frequency of each source is 14
f yellow =
f violet =
λyellow
c
λviolet
= 5.17 ×10 Hz
= 7.06 ×1014 Hz.
Only the violet light is above the threshold frequency.
(b) How much energy is required to eject an electron from the metal?
KEmax = hf threshold − φ = 0
φ = hf threshold
(
)(
= 6.626 ×10 −34 Js 6.20 ×1014 Hz
= 4.11 ×10 −19 J = 2.56 eV
)
X-­‐‑ray Production
When high energy electrons impact a target x-­‐‑ray photons can be emiGed as the electrons are slowed. This process is called bremsstrahlung (German for breaking radiation).
There is a continuous spectrum of radiation emiGed up to a cutoff frequency. The cutoff frequency is defined by the maximum kinetic energy of the electron K = hfmax. The spikes in the spectrum are called characteristic x-­‐‑
rays. These peaks depend on the target material.
Example (text problem 27.17): What is the cutoff frequency for an x-­‐‑ray tube operating at 46 kV?
An electron is accelerated through a potential difference ΔU, use energy conservation to determine the final KE of the electron if it starts from rest. The final KE will determine the cutoff frequency.
ΔK = −ΔU
K f = eΔV
hf max = eΔV
f max
eΔV
46 ×10 3 eV
19
=
=
=
1
.
1
×
10
Hz
−15
h
4.136 ×10 eV s
Compton ScaGering
Before Collision
y
After Collision
Photon (E1, p1)
θ
x
Photon (E0, p0)
Free electron at rest
φ
Free electron (K, p)
After Collision
Before Collision
y
Photon (E1, p1)
θ
x
Photon (E0, p0)
φ
Free electron at rest
Conserve momentum and energy during the collision:
pi = p f
x : p0 = p1 cos θ + p cos φ
y : 0 = p1 sin θ − p sin φ
Free electron (K, p)
Ei = E f
E0 + me c 2 = E1 + K + me c 2
∴ K = E0 − E1 = c( p0 − p1 )
E=pc for a photon
Manipulating the previous expressions gives Δλ is the Compton shift.
h
(1 − cosθ )
λ1 − λ0 =
me c
Δλ = λc (1 − cosθ ).
h
λc =
= 2.426 pm
me c
The Compton wavelength
Example (text problem 27.27): A photon is incident on an electron at rest. The scaGered photon has a wavelength of 2.81 pm and moves at an angle of 29.5° with respect to the direction of the incident photon.
(a) What is the wavelength of the incident photon?
Δλ = λc (1 − cosθ )
= (2.43 pm )(1 − cos 29.5°)
The Compton shift is
= 0.315 pm.
The incident wavelength is
λ0 = λ1 − Δλ
= 2.81 pm − 0.314 pm = 2.50 pm.
(b) What is the final kinetic energy of the electron?
The final kinetic energy of the electron is equal to the change in the photon’s energy.
K = c( p0 − p1 )
⎛ h h ⎞
⎛ 1 1 ⎞
= c⎜⎜ − ⎟⎟ = hc ⎜⎜ − ⎟⎟
⎝ λ0 λ1 ⎠
⎝ λ0 λ1 ⎠
= 8.77 × 10 −15 J = 55 keV
Spectroscopy and Early Models of the Atom
A hot, solid object will emit a continuous spectrum. A hot gas will show an emission or line spectrum (dark background with bright lines). Each element has its own unique set of spectral lines.
Examples of emission spectra:
An absorption spectrum (bright background with dark lines) is seen if a hot source is viewed though a gas.
Before the structure of the atom was known, an empirical result was derived for the wavelengths of the spectral lines of hydrogen in the visible portion of the spectrum (the Balmer series).
1
⎛ 1 1 ⎞
= R⎜ − 2 ⎟
λ
⎝ 4 n ⎠
Where R = 1.097×107, m-­‐‑1 is the Rydberg constant and n ≥3.
The Bohr Model of the Hydrogen Atom
The Thomson model of the atom had a volume of positive charge with the negatively charged electrons embedded within the volume.
ScaGering experiments by Rutherford led to the conclusion that an atom had a very small nucleus of positive charge (10-­‐‑5 times the size of the atom containing nearly all of the mass) that was surrounded by the electrons.
Problems to be solved!
It was thought that the electrons in their orbits should radiate (they are accelerated) causing the electron’s orbit to decay, implying that atoms are not stable. This is obviously false. Any model of the atom must also explain the line spectra of the elements.
The Bohr Model of the Hydrogen Atom
The Bohr model assumes:
The electron is allowed to be in only one of a discrete set of states called stationary states. The electron orbits have quantized radii, energy, and angular momentum. Newtonian physics applies to an electron in a stationary state.
The electron can transition between one stationary state and another provided it can absorb/emit a photon of energy equal to the energy difference between the states. ΔE = hf. The stationary states have quantized angular momentum in the amount h
Ln = n
= n!;
2π
n is an integer.
The allowed radii are n 2! 2
2
rn =
=
n
a0
2
me ke
where a0 = 52.9 pm is the Bohr radius.
2 4
m
k
e
E1
The energy levels are given byE = − e
= 2
n
2 2
2n !
n
where E1 = -­‐‑13.6 eV is the energy of the ground state, the lowest possible energy of the electron. When n > 1 the electron is in an excited state. The quantity n is an integer and is the principal quantum number.
Energy level diagram for hydrogen
The energy of a photon emiGed (absorbed) by an electron during a transition is
⎛ 1
1 ⎞⎟
⎜
E=
= Ei − E f = − E1 2 − 2
⎜ n
⎟
λ
n
f
i
⎝
⎠
1
E1 ⎛⎜ 1
1 ⎞⎟
=−
− 2
2
⎜
λ
hc ⎝ n f ni ⎟⎠
hc
where
−
E1
= 1.097 ×10 7 m −1 = R
hc
is the Rydberg constant.
When nf=2, the above result reduces to the Balmer formula.
The Bohr model correctly predicted the wavelengths of the spectral lines of hydrogen in the visible. There are several problems with the Bohr model. Bohr’s model, while successful at predicting the spectrum of hydrogen, fails at predicting the spectra of most other elements. Only hydrogenic atoms (atoms that only have one electron; Li2+ for example) can have their spectra computed using the Bohr model.
n 2 a0
n2!2
The allowed radii are r =
=
.
n
2
me kZe
Z
2 2 4
2
m
k
Z
e
Z
E1H
The energy levels areE = − e
=
n
2 2
2n !
n2
where Z is the atomic number of the atom and E1H = -­‐‑13.6 eV .
Example (text problem 27.34): Find the Bohr radius of doubly ionized lithium (Li2+).
2
n
a
0
What is r1? r =
n
Z
12 a0 1
r1 =
= a0 = 17.6 pm
3
3
The inner most energy level is closer to the nucleus than in an H atom.
Example (text problem 27.43): A hydrogen atom has an electron in the n = 5 level. (a) If the electron returns to the ground state, what is the minimum number of photons that can be emiGed?
One photon; the electron may transition from the n = 5 level to the n = 1 level. (b) What is the maximum number that might be emiGed?
Four photons; the electron may cascade from n = 5 to 4 to 3 to 2 to n = 1.
Pair Annihilation and Pair Production
A photon can interact with an atomic nucleus and change itself into an electron-­‐‑
positron pair (or some particle-­‐‑antiparticle pair.) A positron is an antielectron. The nucleus is needed to ensure that momentum is conserved.
The energy of the photon must be at least 2mec2. If E > 2mec2 , then the additional energy goes into the kinetic energy of the electron-­‐‑positron pair. This is pair production.
e + e → 2 photons.
The inverse process is −
+
The process of pair production created protons, neutrons, and electrons in the earliest moments after the Big Bang.
To have enough energy, the photons must be “hot” enough. Electrons need T~1010 K and for protons/neutrons T~1013 K. The early Universe must have been much hoGer than it is today.
Pair production creates equal amounts of maGer and antimaGer. Where in the Universe is all of the antimaGer?
Example (text problem 27.57): A muon and an antimuon, each with mass 207 times greater than an electron, were at rest when they annihilated and produced two photons of equal energy. What is the wavelength of each of the photons?
For an electron-­‐‑positron pair2m
2
c
= 1.022 MeV.
e
2mµ c 2
For a muon-­‐‑antimuon pair
(
)
= 207 2me c 2 = 212 MeV.
The created photons each have 106 MeV of energy. Their wavelengths are
hc 1240 eV nm
λ= =
E 106 ×10 6 eV
= 1.17 ×10 −5 nm = 1.17 ×10 −14 m.