ELEMENTARY QUANTUM MECHANICS
Quantum Mechanics – 1
Page 1 of 9
ELEMENTARY QUANTUM MECHANICS 1. INTRODUCTION The triumph of modern physics is the triumph of quantum physics. Quantum physics (mechanics) → Grew out of failures of classical physics → Launched by experiments such as Photoelectric effect Compton scattering Blackbody radiation → Found some solutions in Planck’s hypothesis Wave-particle duality [De Broglie’s hypothesis] → Generated new ideas like Wave function Schröndinger equation Heisenberg’s uncertainty principle Quantum statistics 2. LIGHT AS A WAVE In classical elementary physics light is considered as an electromagnetic (EM) wave with a frequency. y Ey
Direction of Propagation
x
E0
Velocity = c
z
x Bz
B0
λ
An electromagnetic wave is a traveling wave with time-varying electric and magnetic fields that are perpendicular to each other and to the direction of propagation. Electric field Ey at position x at any time t,
E y ( x, t ) = E0 sin(kx − ωt ) Similar is the case for Bz.
⎛ ⎝
wavenumber = propagation constant ⎜ k =
2π ⎞ ⎟ λ ⎠
angular frequency (ω = 2πν ) , where ν = frequency
where λ is the wavelength 1 Intensity of the light wave = energy flowing per unit area per second, I = cε 0 E02 2 Velocity (phase velocity) of the wave,
c = νλ ,
absolute permittivity
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 2 of 9
Experiment 1: Young’s Double-Slit Experiment P
Constructive interference Destructive interference
Photographic plate showing Young’s fringes
Young’s fringes [dark & bright fringes] can only be explained by interference of light waves. Two waves from slits S1 and S2 when reach the photographic plate [P] in phase, they interfere constructively to give rise to bright fringe. Requirement for constructive interference: S1P – S2P = nλ When the two waves are out of phase by λ/2 they interfere destructively to give dark fringe. 1⎞ ⎛ S1P − S 2 P = ⎜ n + ⎟λ Requirement for destructive interference: 2⎠ ⎝
Experiment 2: X-ray Diffraction [Crystallography]
Atomic planes
Patterns (a) and (b) can only be explained by using wave concepts for light (c). The path difference of the two waves [fig. (c)] is 2d sin θ, where d is the separation of the atomic planes. For constructive interference this must be nλ. Otherwise, waves will interfere destructively to cancel each other. Thus, waves reflected from adjacent atomic planes interfere constitute a diffracted beam only when, 2d sin θ = nλ where n = 1, 2, 3, …
Bragg’s law → basis for identifying and studying various crystal structures (crystallography) Young’s double-slit and X-ray diffraction experiments clearly show that light exhibits wave-like properties. AUK (EEE, UIU)
Quantum Mechanics – 1
Page 3 of 9
3. FAILURE OF CLASSICAL PHYSICS Classical physics that considers light as wave fails to explain the results of some experiments. The results can only be explained by introducing new concepts that form the basis of quantum physics.
4. EXPERIMENT 1: THE PHOTOELECTRIC EFFECT The remarkable aspects of the photoelectric effect when it was first observed were:
The details of the photoelectric effect were in direct contradiction to the expectations of very well developed classical physics. The explanation marked one of the major steps toward quantum theory. Ref.: Hyperphysics (http://hyperphysics.phy-astr.gsu.edu)
1. The electrons were emitted immediately - no time lag! 2. Increasing the intensity of the light increased the number of photoelectrons, but not their maximum kinetic energy! 1 [Recall I = cε 0 E02 ] 2 3. Red light will not cause the ejection of electrons, no matter what the intensity! 4. A weak violet light will eject only a few electrons, but their maximum kinetic energies are greater than those for intense light of longer wavelengths!
Explanation of photoelectric effect Analysis of data from the photoelectric experiment showed that the energy of the ejected electrons was proportional to the frequency of the illuminating light. This showed that whatever was knocking the electrons out had an energy proportional to light frequency. The remarkable fact that the ejection energy was independent of the total energy of illumination (light intensity) showed that the interaction must be like that of a particle which gave all of its energy to the electron (on collision)! This fit in well with Planck's hypothesis that light in the blackbody radiation experiment could exist only in discrete bundles with energy E = hν, where h is the Planck’s constant (h = 6.626X10-34 Js)
Ref.: Hyperphysics (http://hyperphysics.phy-astr.gsu.edu)
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 4 of 9
Concept of Work Function In the photoelectric effect experiment, electrons ejected from a sodium metal surface were measured as an electric current. Finding the opposing voltage it took to stop all the electrons gave a measure of the maximum kinetic energy of the electrons in electron volts. 3
Maximum photoelectron kinetic energy KEm in eV
Slope = ΔE /Δν = 4.1X10-15 eV s = h
ΔE = 1.25 eV
2
1
Light below a frequency 14 of 4.39X10 Hz or wavelength longer than 683 nm would not eject electrons
Δν = 3X10
14
Hz
Data from Millikan, 1916 4
8
10
12X1014
Frequency, Hz ν0 The minimum energy required to eject an electron from the surface is called the photoelectric work function. The threshold for this element corresponds to a wavelength λ0 of 683 nm for sodium. The slope of KEm vs ν gives h, the Planck’s constant (= 4.1X10-15 eV s = 6.626X10-34 Js). Using the wavelength λ0 in the Planck relationship, E = hν0, (c = νλ0) gives a photon energy of 1.82 eV. 6
Einstein in 1905 successfully explained the photoelectric effect. Light consists of ‘energy packets’ or photons, each with energy hν. An electron in a metal is in a lower state of potential energy (PE) than in vacuum, by an amount Φ, the work function of the metal. This lower PE is a result of the Coulombic attraction interaction between the electron and the positive metal ions. Upon collision some of the photon energy goes toward overcoming the PE barrier. The energy left (hν – Φ) gives the electron its KE. Photoemission only occurs when hν is greater than Φ. At threshold, hν0 = Φ, where ν0 is the critical (cut-off) frequency, below which no photoemission occurs.
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 5 of 9
The Nobel Prize in Physics 1921 "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect" Albert Einstein (Germany and Switzerland) Kaiser-Wilhelm-Institut (now Max-Planck-Institut) für Physik Berlin, Germany, b. 1879 (in Ulm, Germany), d. 1955
Problem What is the energy of a blue photon that has a wavelength of 450 nm? Solution: We know the photon energy, E ph = hν = −34
hc
λ
−1
6.6 × 10 J s ⋅ 3 × 10 m s = 4.4 × 10 −19 J 450 × 10 −9 m Usually we express the photon energy in units of eV. [1 eV = 1.6 X 10-19 J] Therefore, Eblue =
Thus Eblue =
8
4.4 × 10 −19 J = 2.75 eV ↵ 1.6 × 10 −19 J eV
Problem In the photoelectric experiment, green light with wavelength of 522 nm is the longest wavelength radiation that can cause the photoemission from a clean Sodium surface.
a. What is the work function of Na? We know work function, Φ = hν 0 =
hc
λ0
Therefore work function of Sodium, Φ Na =
6.626 × 10 −34 J s ⋅ 3 × 10 8 m s −1 = 2.38 eV ↵ 522 × 10 −9 m ⋅ 1.6 × 10 −19 J eV
b. If UV (ultraviolet) radiation of wavelength 250 nm is incident to the Sodium surface, what would be the kinetic energy of the photoemitted electrons? The excess energy of the incident photon after overcoming the work function, i.e., E ph − Φ gives the photoemitted electron its K.E. The energy of the incident photon, EUV = Or, EUV
hc
λ
=
6.626 × 10 −34 Js ⋅ 3 × 10 8 ms −1 = 7.95 × 10 −19 J −9 250 × 10 m
7.95 × 10 −19 J = = 4.96 eV 1.6 × 10 −19 J eV
Therefore, K.E. = 4.96 eV – 2.38 eV = 2.58 eV ↵
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 6 of 9
5. EXPERIMENT 2: THE COMPTON SCATTERING
Arthur Holly Compton observed scattering of x-rays from electrons in a carbon target. When an X-ray with frequencyν, wavelength λ strikes an electron → it is deflected or scattered → electron recoils and moves away → scattered X-ray has a frequency ν′ which is less than ν Recoilingelectron X-ray photon
Electron
c
ν, λ
φ θ
y
Scattered photon
ν ,' λ '
Δλ = λ ′ − λ =
h (1 − cos θ ) me c
c
x
It is found that the recoiling electrons possess kinetic energy, K.E. = hν - hν′ Since the electron now also has a momentum pe, from law of conservation of momentum (momentum before collision = momentum after collision), we conclude that X-ray also has a momentum. h Momentum of the photon is related to its wavelength by p = .
λ
The Nobel Prize in Physics 1927 "for his discovery of the effect named after him" Arthur Holly Compton (USA) University of Chicago Chicago, IL, USA b. 1892 d. 1962
Problem Calculate the energy and momentum of an X-ray photon with a wavelength of 0.6 angstrom and the velocity of a corresponding electron that has the same momentum. Solution
6.6 × 10 −34 Js ⋅ 3 × 10 8 ms −1 hc = 2.06 × 10 4 eV = 10 19 − − J λe 0.6 × 10 m ⋅ 1.6 × 10 eV h 6.6 × 10 −34 Js Momentum of the X-ray photon, p = = = 1.1 × 10 − 23 kg m s −1 λ 0.6 × 10 −10 m Energy of the X-ray photon, E ph = hν =
Corresponding electron with the momentum p = me v will have a velocity, v=
p 1.1 × 10 −23 kg m s −1 = = 1.2 × 10 7 ms −1 ↵ −31 me 9.1 × 10 kg
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 7 of 9
6. EXPERIMENT 3: BLACK BODY RADIATION
All objects emit and absorb energy in the form of radiation (thermal radiation).
Intensity of the radiation depends on the radiation wavelength and temperature of the object.
At thermal equilibrium (object and surroundings at the same temperature), Energy absorbed = Energy emitted
When object temp. > surrounding temp. ⇒ There is a net emission of radiation energy.
Black body radiation = maximum amount of radiation energy that can be emitted by an object
In general ⇒ intensity of the radiated energy depends on the material’s surface
But radiation emitted from a cavity [box] with a small aperture ⇒ independent of the material of the cavity and corresponds very closely to black body radiation Iλ ec an id ar ri la rt ce p S
3000 K
Classical theory Planck's radiation law
2500 K
0
1
2
3
λ ( μm)
5
4
Schematic illustration of black body radiation and its characteristics (spectral irradiance Iλ is the emitted radiation intensity (power per unit area) per unit wavelength)
Classical physics explanation ⇒ acceleration, deceleration of the charges due to various thermal vibrations, oscillations, or motions of the atoms in the surface region of the cavity material result in electromagnetic waves of emission ⇒ these EM waves interfere with each other giving rise to many types of standing EM radiation waves (modes) with different wavelengths in the cavity mode
⇒ each mode contributes an energy kT to the emitted intensity ⇒ classical Rayleigh-Jeans law predicts 1 Spectral irradiance, Iλ ∝ 4 and Iλ ∝ T
λ
⇒ but the law predicts continued Iλ increase at lower Wavelength leading to ‘ultraviolet catastrophe’ ⇒ this is not in agreement with the experiment
modes developed in cavity
Classical physics failed to explain the black body radiation experimental results!
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 8 of 9
Max Planck’s explanation
Max Planck explained the black body radiation experimental results in 1900. ⇒ radiation within the cavity involves the emission and absorption of discrete (separate, isolated, distinct) amounts of light energy by the modes developed in the cavity ⇒ the quantity of the discrete energy quantum is determined by the frequency ν of the radiation and given by hν ⇒ modes emit and absorb an integer multiple of the discrete energy quantum, i.e. nhν ⇒ probability of a mode to have nhν energy is proportional to the Boltzmann factor e
⎛ nhν ⎞ ⎜− ⎟ ⎝ kT ⎠
2πhc 2 Planck’s black body ⎛ hkTν ⎞ radiation formula 5 λ ⎜⎜ e − 1⎟⎟ ⎝ ⎠ ⇒ the formula is in excellent agreement with all observed black body radiation characteristics ⇒ spectral irradiance is given by, I λ =
The Nobel Prize in Physics 1918 "in recognition of the services he rendered to the advancement of Physics by his discovery of energy quanta" Max Karl Ernst Ludwig Planck (Germany) Berlin University, Berlin, Germany b. 1858, d. 1947
7. ELECTRON AS WAVE (DE BROGLIE RELATIONSHIP)
So far we have seen light can behave as if it were a stream of particle-like entities called photons. Can electrons (which are particles) exhibit wave-like properties (interference, diffraction)? To probe Young’s double-slit and Crystallographic experiments are carried out with energetic electron beams.
The Young’s double-slit experiment showed Young’s fringes (high an low-intensity regions) The Crystallographic experiment produced diffraction rings.
Indeed electrons exhibit wave-like properties! The electron beams obey Braggs diffraction condition, 2d sin θ = nλ , knowing the interatomic distance d and measuring the angle of diffraction θ we can evaluate the wavelength associated with the wave-like behaviour of the electrons. The momentum of the electrons can be determined from the fact that the kinetic energy gained by the electrons p 2 / 2me is equal to the energy eV from the accelerating voltage V in the electron tube.
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 9 of 9
It is found that an electron traveling with a momentum p behaves like a wave of wavelength λ h given by λ = p Earlier from the Compton scattering we found that the photon with wavelength λ behaves h like a particle of momentum p given by p =
λ
h h Thus λ = or p = relates wave-like and particle-like properties to and from each p λ other (wave-particle duality).
This is first hypothesized by de Broglie in 1924 → known as de Broglie relations
The Nobel Prize in Physics 1929 "for his discovery of the wave nature of electrons" Prince Louis-Victor Pierre Raymond de Broglie (France) Sorbonne University, Institut Henri Poincaré, Paris, France b. 1892, d. 1987
AUK (EEE, UIU)
Page 1 of 9
ELEMENTARY QUANTUM MECHANICS 1. INTRODUCTION The triumph of modern physics is the triumph of quantum physics. Quantum physics (mechanics) → Grew out of failures of classical physics → Launched by experiments such as Photoelectric effect Compton scattering Blackbody radiation → Found some solutions in Planck’s hypothesis Wave-particle duality [De Broglie’s hypothesis] → Generated new ideas like Wave function Schröndinger equation Heisenberg’s uncertainty principle Quantum statistics 2. LIGHT AS A WAVE In classical elementary physics light is considered as an electromagnetic (EM) wave with a frequency. y Ey
Direction of Propagation
x
E0
Velocity = c
z
x Bz
B0
λ
An electromagnetic wave is a traveling wave with time-varying electric and magnetic fields that are perpendicular to each other and to the direction of propagation. Electric field Ey at position x at any time t,
E y ( x, t ) = E0 sin(kx − ωt ) Similar is the case for Bz.
⎛ ⎝
wavenumber = propagation constant ⎜ k =
2π ⎞ ⎟ λ ⎠
angular frequency (ω = 2πν ) , where ν = frequency
where λ is the wavelength 1 Intensity of the light wave = energy flowing per unit area per second, I = cε 0 E02 2 Velocity (phase velocity) of the wave,
c = νλ ,
absolute permittivity
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 2 of 9
Experiment 1: Young’s Double-Slit Experiment P
Constructive interference Destructive interference
Photographic plate showing Young’s fringes
Young’s fringes [dark & bright fringes] can only be explained by interference of light waves. Two waves from slits S1 and S2 when reach the photographic plate [P] in phase, they interfere constructively to give rise to bright fringe. Requirement for constructive interference: S1P – S2P = nλ When the two waves are out of phase by λ/2 they interfere destructively to give dark fringe. 1⎞ ⎛ S1P − S 2 P = ⎜ n + ⎟λ Requirement for destructive interference: 2⎠ ⎝
Experiment 2: X-ray Diffraction [Crystallography]
Atomic planes
Patterns (a) and (b) can only be explained by using wave concepts for light (c). The path difference of the two waves [fig. (c)] is 2d sin θ, where d is the separation of the atomic planes. For constructive interference this must be nλ. Otherwise, waves will interfere destructively to cancel each other. Thus, waves reflected from adjacent atomic planes interfere constitute a diffracted beam only when, 2d sin θ = nλ where n = 1, 2, 3, …
Bragg’s law → basis for identifying and studying various crystal structures (crystallography) Young’s double-slit and X-ray diffraction experiments clearly show that light exhibits wave-like properties. AUK (EEE, UIU)
Quantum Mechanics – 1
Page 3 of 9
3. FAILURE OF CLASSICAL PHYSICS Classical physics that considers light as wave fails to explain the results of some experiments. The results can only be explained by introducing new concepts that form the basis of quantum physics.
4. EXPERIMENT 1: THE PHOTOELECTRIC EFFECT The remarkable aspects of the photoelectric effect when it was first observed were:
The details of the photoelectric effect were in direct contradiction to the expectations of very well developed classical physics. The explanation marked one of the major steps toward quantum theory. Ref.: Hyperphysics (http://hyperphysics.phy-astr.gsu.edu)
1. The electrons were emitted immediately - no time lag! 2. Increasing the intensity of the light increased the number of photoelectrons, but not their maximum kinetic energy! 1 [Recall I = cε 0 E02 ] 2 3. Red light will not cause the ejection of electrons, no matter what the intensity! 4. A weak violet light will eject only a few electrons, but their maximum kinetic energies are greater than those for intense light of longer wavelengths!
Explanation of photoelectric effect Analysis of data from the photoelectric experiment showed that the energy of the ejected electrons was proportional to the frequency of the illuminating light. This showed that whatever was knocking the electrons out had an energy proportional to light frequency. The remarkable fact that the ejection energy was independent of the total energy of illumination (light intensity) showed that the interaction must be like that of a particle which gave all of its energy to the electron (on collision)! This fit in well with Planck's hypothesis that light in the blackbody radiation experiment could exist only in discrete bundles with energy E = hν, where h is the Planck’s constant (h = 6.626X10-34 Js)
Ref.: Hyperphysics (http://hyperphysics.phy-astr.gsu.edu)
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 4 of 9
Concept of Work Function In the photoelectric effect experiment, electrons ejected from a sodium metal surface were measured as an electric current. Finding the opposing voltage it took to stop all the electrons gave a measure of the maximum kinetic energy of the electrons in electron volts. 3
Maximum photoelectron kinetic energy KEm in eV
Slope = ΔE /Δν = 4.1X10-15 eV s = h
ΔE = 1.25 eV
2
1
Light below a frequency 14 of 4.39X10 Hz or wavelength longer than 683 nm would not eject electrons
Δν = 3X10
14
Hz
Data from Millikan, 1916 4
8
10
12X1014
Frequency, Hz ν0 The minimum energy required to eject an electron from the surface is called the photoelectric work function. The threshold for this element corresponds to a wavelength λ0 of 683 nm for sodium. The slope of KEm vs ν gives h, the Planck’s constant (= 4.1X10-15 eV s = 6.626X10-34 Js). Using the wavelength λ0 in the Planck relationship, E = hν0, (c = νλ0) gives a photon energy of 1.82 eV. 6
Einstein in 1905 successfully explained the photoelectric effect. Light consists of ‘energy packets’ or photons, each with energy hν. An electron in a metal is in a lower state of potential energy (PE) than in vacuum, by an amount Φ, the work function of the metal. This lower PE is a result of the Coulombic attraction interaction between the electron and the positive metal ions. Upon collision some of the photon energy goes toward overcoming the PE barrier. The energy left (hν – Φ) gives the electron its KE. Photoemission only occurs when hν is greater than Φ. At threshold, hν0 = Φ, where ν0 is the critical (cut-off) frequency, below which no photoemission occurs.
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 5 of 9
The Nobel Prize in Physics 1921 "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect" Albert Einstein (Germany and Switzerland) Kaiser-Wilhelm-Institut (now Max-Planck-Institut) für Physik Berlin, Germany, b. 1879 (in Ulm, Germany), d. 1955
Problem What is the energy of a blue photon that has a wavelength of 450 nm? Solution: We know the photon energy, E ph = hν = −34
hc
λ
−1
6.6 × 10 J s ⋅ 3 × 10 m s = 4.4 × 10 −19 J 450 × 10 −9 m Usually we express the photon energy in units of eV. [1 eV = 1.6 X 10-19 J] Therefore, Eblue =
Thus Eblue =
8
4.4 × 10 −19 J = 2.75 eV ↵ 1.6 × 10 −19 J eV
Problem In the photoelectric experiment, green light with wavelength of 522 nm is the longest wavelength radiation that can cause the photoemission from a clean Sodium surface.
a. What is the work function of Na? We know work function, Φ = hν 0 =
hc
λ0
Therefore work function of Sodium, Φ Na =
6.626 × 10 −34 J s ⋅ 3 × 10 8 m s −1 = 2.38 eV ↵ 522 × 10 −9 m ⋅ 1.6 × 10 −19 J eV
b. If UV (ultraviolet) radiation of wavelength 250 nm is incident to the Sodium surface, what would be the kinetic energy of the photoemitted electrons? The excess energy of the incident photon after overcoming the work function, i.e., E ph − Φ gives the photoemitted electron its K.E. The energy of the incident photon, EUV = Or, EUV
hc
λ
=
6.626 × 10 −34 Js ⋅ 3 × 10 8 ms −1 = 7.95 × 10 −19 J −9 250 × 10 m
7.95 × 10 −19 J = = 4.96 eV 1.6 × 10 −19 J eV
Therefore, K.E. = 4.96 eV – 2.38 eV = 2.58 eV ↵
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 6 of 9
5. EXPERIMENT 2: THE COMPTON SCATTERING
Arthur Holly Compton observed scattering of x-rays from electrons in a carbon target. When an X-ray with frequencyν, wavelength λ strikes an electron → it is deflected or scattered → electron recoils and moves away → scattered X-ray has a frequency ν′ which is less than ν Recoilingelectron X-ray photon
Electron
c
ν, λ
φ θ
y
Scattered photon
ν ,' λ '
Δλ = λ ′ − λ =
h (1 − cos θ ) me c
c
x
It is found that the recoiling electrons possess kinetic energy, K.E. = hν - hν′ Since the electron now also has a momentum pe, from law of conservation of momentum (momentum before collision = momentum after collision), we conclude that X-ray also has a momentum. h Momentum of the photon is related to its wavelength by p = .
λ
The Nobel Prize in Physics 1927 "for his discovery of the effect named after him" Arthur Holly Compton (USA) University of Chicago Chicago, IL, USA b. 1892 d. 1962
Problem Calculate the energy and momentum of an X-ray photon with a wavelength of 0.6 angstrom and the velocity of a corresponding electron that has the same momentum. Solution
6.6 × 10 −34 Js ⋅ 3 × 10 8 ms −1 hc = 2.06 × 10 4 eV = 10 19 − − J λe 0.6 × 10 m ⋅ 1.6 × 10 eV h 6.6 × 10 −34 Js Momentum of the X-ray photon, p = = = 1.1 × 10 − 23 kg m s −1 λ 0.6 × 10 −10 m Energy of the X-ray photon, E ph = hν =
Corresponding electron with the momentum p = me v will have a velocity, v=
p 1.1 × 10 −23 kg m s −1 = = 1.2 × 10 7 ms −1 ↵ −31 me 9.1 × 10 kg
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 7 of 9
6. EXPERIMENT 3: BLACK BODY RADIATION
All objects emit and absorb energy in the form of radiation (thermal radiation).
Intensity of the radiation depends on the radiation wavelength and temperature of the object.
At thermal equilibrium (object and surroundings at the same temperature), Energy absorbed = Energy emitted
When object temp. > surrounding temp. ⇒ There is a net emission of radiation energy.
Black body radiation = maximum amount of radiation energy that can be emitted by an object
In general ⇒ intensity of the radiated energy depends on the material’s surface
But radiation emitted from a cavity [box] with a small aperture ⇒ independent of the material of the cavity and corresponds very closely to black body radiation Iλ ec an id ar ri la rt ce p S
3000 K
Classical theory Planck's radiation law
2500 K
0
1
2
3
λ ( μm)
5
4
Schematic illustration of black body radiation and its characteristics (spectral irradiance Iλ is the emitted radiation intensity (power per unit area) per unit wavelength)
Classical physics explanation ⇒ acceleration, deceleration of the charges due to various thermal vibrations, oscillations, or motions of the atoms in the surface region of the cavity material result in electromagnetic waves of emission ⇒ these EM waves interfere with each other giving rise to many types of standing EM radiation waves (modes) with different wavelengths in the cavity mode
⇒ each mode contributes an energy kT to the emitted intensity ⇒ classical Rayleigh-Jeans law predicts 1 Spectral irradiance, Iλ ∝ 4 and Iλ ∝ T
λ
⇒ but the law predicts continued Iλ increase at lower Wavelength leading to ‘ultraviolet catastrophe’ ⇒ this is not in agreement with the experiment
modes developed in cavity
Classical physics failed to explain the black body radiation experimental results!
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 8 of 9
Max Planck’s explanation
Max Planck explained the black body radiation experimental results in 1900. ⇒ radiation within the cavity involves the emission and absorption of discrete (separate, isolated, distinct) amounts of light energy by the modes developed in the cavity ⇒ the quantity of the discrete energy quantum is determined by the frequency ν of the radiation and given by hν ⇒ modes emit and absorb an integer multiple of the discrete energy quantum, i.e. nhν ⇒ probability of a mode to have nhν energy is proportional to the Boltzmann factor e
⎛ nhν ⎞ ⎜− ⎟ ⎝ kT ⎠
2πhc 2 Planck’s black body ⎛ hkTν ⎞ radiation formula 5 λ ⎜⎜ e − 1⎟⎟ ⎝ ⎠ ⇒ the formula is in excellent agreement with all observed black body radiation characteristics ⇒ spectral irradiance is given by, I λ =
The Nobel Prize in Physics 1918 "in recognition of the services he rendered to the advancement of Physics by his discovery of energy quanta" Max Karl Ernst Ludwig Planck (Germany) Berlin University, Berlin, Germany b. 1858, d. 1947
7. ELECTRON AS WAVE (DE BROGLIE RELATIONSHIP)
So far we have seen light can behave as if it were a stream of particle-like entities called photons. Can electrons (which are particles) exhibit wave-like properties (interference, diffraction)? To probe Young’s double-slit and Crystallographic experiments are carried out with energetic electron beams.
The Young’s double-slit experiment showed Young’s fringes (high an low-intensity regions) The Crystallographic experiment produced diffraction rings.
Indeed electrons exhibit wave-like properties! The electron beams obey Braggs diffraction condition, 2d sin θ = nλ , knowing the interatomic distance d and measuring the angle of diffraction θ we can evaluate the wavelength associated with the wave-like behaviour of the electrons. The momentum of the electrons can be determined from the fact that the kinetic energy gained by the electrons p 2 / 2me is equal to the energy eV from the accelerating voltage V in the electron tube.
AUK (EEE, UIU)
Quantum Mechanics – 1
Page 9 of 9
It is found that an electron traveling with a momentum p behaves like a wave of wavelength λ h given by λ = p Earlier from the Compton scattering we found that the photon with wavelength λ behaves h like a particle of momentum p given by p =
λ
h h Thus λ = or p = relates wave-like and particle-like properties to and from each p λ other (wave-particle duality).
This is first hypothesized by de Broglie in 1924 → known as de Broglie relations
The Nobel Prize in Physics 1929 "for his discovery of the wave nature of electrons" Prince Louis-Victor Pierre Raymond de Broglie (France) Sorbonne University, Institut Henri Poincaré, Paris, France b. 1892, d. 1987
AUK (EEE, UIU)
