28 Quantum Physics

Physics 150 Quantum Physics Chapter 28 Ma8er waves In the previous chapter we showed that: Light is both an EM wave and a par4cle. Par4cle theory •  Photoelectric effect •  Compton sca8ering •  Pair producGon Wave theory •  Interference pa8erns (Young’s double slit experiment) If a wave (EM radiaGon) can behave like a parGcle, might a parGcle act like a wave? Electron can be both an EM wave and a par4cle. Louis de Broglie doctoral thesis 1924 (only 30 pages) – Nobel prize Physics 150, Prof. M. Nikolic 2 Ma8er waves If an electron is a wave, than it has its own wavelength: h
λ=
p
p is the momentum of an electron h is a Planck’s constant De Broglie wavelength A beam of electrons may be used in a double slit experiment instead of a light beam. You will see interference pa8ern! If you try to track which of the two slits the electron goes through, the interference pa8ern disappears. Physics 150, Prof. M. Nikolic 3 Exercise: De Broglie equaGon What is the raGo of the wavelength of a 0.1 keV photon to wavelength of a 0.1 keV electron? Assume that electron is nonrelaGvisGc. What is given: Ke = 0.1 keV = 100 eV Ephoton = 0.1 keV = 100 eV Wavelength of the photon (chapter 27): hc
E=
λp
λp
4.13×10
(
=
hc
λp =
E
−15
eV s) ⋅ (3×108 m/s)
100 eV
= 1.24 ×10 −8 m.
Wavelength of the electron: h
λe =
p
We need to find the momentum of the electron. Physics 150, Prof. M. Nikolic 4 Exercise: De Broglie equaGon What is the raGo of the wavelength of a 0.1 keV photon to wavelength of a 0.1 keV electron? Assume that electron is nonrelaGvisGc. What is given: Ke = 0.1 keV = 100 eV Ephoton = 0.1 keV = 100 eV If electron is considered a nonrelaGvisGc (speed<<speed of light) than Newtonian mechanics (Phys 140) can be applied: and 1
K e = me v 2 p = m v
e
2
1 ! p$
K e = me # &
2 " me %
2
p = 2me K e
p = 2 ⋅ 9.1×10 −31kg ⋅100 ⋅1.6 ×10 −19 J = 5.4 ×10 −24 kg m/s
Wavelength of the electron is then: 6.626 ×10 −34 Js
−10
λe =
=
1.23×10
m
−24
5.4 ×10 kg m/s
Physics 150, Prof. M. Nikolic And the raGo is λ p 1.24 ×10 −8 m
=
= 100.8
−10
λe 1.23×10 m
5 ApplicaGon – Electron microscope Uses the interference pa8ern from the beam of electrons to produce images of small objects •  The resoluGon of a light microscope is limited by diffracGon 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 resoluGon of much smaller structures. Physics 150, Prof. M. Nikolic 6 Exercise: Electron microscope A scanning electron microscope is used to look at cell structure with 10-­‐nm resoluGon. A beam of electrons from the hot filament is accelerated with a voltage of 12 kV and then focused to a small spot on the specimen. What is the wavelength of the beam of electrons? Electrons convert their potenGal energy to the kineGc energy. K e = ΔU = −qΔV = eΔV
Ager going through potenGal difference of 12 kV, kineGc energy of electrons is 12 keV. Then, similarly to the previous problem p = 2me K e
h
λe =
p
λe =
Physics 150, Prof. M. Nikolic h
λe =
2me K e
6.626 ×10 −34 Js
2 ⋅ 9.1×10 −31kg ⋅12000 ⋅1.6 ×10 −19 J
= 1.12 ×10 −11 m
7 The uncertainty principle Ok, so electron can be both parGcle and wave. But this means that electron interferes with itself?! ⓵
✹
⓶
Electron has to go through BOTH slits! When we detect which hole the electron went through, the interference pa8ern disappears ! Quantum mechanics is not determinisGc, it’s probabilis4c. We can only say that there’s a probability distribu4on for the electrons/photons, we can’t predict where any parGcular electron/photon will end up. Physics 150, Prof. M. Nikolic 8 Uncertainty principle In 1927, German physicist Werner Heisenberg formulated the Heisenberg uncertainty principle 1
ΔxΔp x ≥ !
2
where h
!=
2π
•  The more precise a measurement of posiGon è the more uncertain the measurement of momentum will be. •  The more precise a measurement of momentum è the more uncertain the measurement of the posiGon will be. Note, in the double slit experiment the uncertainty in the electron’s posiGon is half the width of the slit. Physics 150, Prof. M. Nikolic 9 Uncertainty principle Similar principle can be applied to energy and Gme 1
ΔEΔt ≥ !
2
If a system is in a quantum state for a Gme interval Δt è the uncertainty in the energy of that state is related to the Gme of that state (Δt) Physics 150, Prof. M. Nikolic 10 Exercise: Uncertainty principle At a baseball game, a radar gun measures the speed of a 144 gram baseball to be 137.32±0.10 km/hr. What is the minimum uncertainty of the posiGon of the baseball? What is given: m = 0.144 kg v+Δv = 137.32±0.10 km/hr 1
ΔxΔpx ≥ !
2
Δx =
!
2Δpx
" km %" 1000m %" 1h %
m
Convert km/hr to m/s: Δvx = $ 0.1 '$
'$
' = 0.028
#
h &# 1km &# 3600s &
s
Uncertainty in measuring the momentum is: Δpx = mΔvx
Δpx = 0.144kg ⋅ 0.0028m / s = 0.004kg m/s
6.626 ×10 −34 Js
Δx =
= 1.3×10 −32 m
(4 ⋅ 3.14 ⋅ 0.004 kg m/s)
Physics 150, Prof. M. Nikolic 11 Wave funcGons for a confined parGcle A parGcle confined to a region of space will have quanGzed energy levels. Consider a parGcle in a box of width L that has impenetrable walls è the parGcle can never leave the box (similar to standing waves) Wave funcGons n = 1 n = 2 n = 4 Physics 150, Prof. M. Nikolic De Broglie said: Electron waves in hydrogen were standing waves In a standing wave only certain wavelengths are allowed (Chapter 11) 2L
λn =
n
12 ParGcle in a box De Broglie wavelength of a parGcle is related to its momentum: KineGc energy of the parGcle: p2
K=
2m
h nh
pn =
=
λn 2L
PotenGal energy (we choose): U = 0 Total energy of the parGcle: p2
E = K +U =
+0
2m
Minimum possible energy è ground state (n=1) Physics 150, Prof. M. Nikolic n2h2
En =
8mL2
h2
E1 =
8mL2
2
En = n E1
13 Exercise: ParGcle in a box A neutron is located within a uranium nucleus (radius r = 7.5 x 10-­‐15 m). The neutron goes from the second excited state to the ground state, emiong a photon in the process. What is the wavelength of the photon? What is given: ni = 3 nf = 1 r = 7.5 x 10-­‐15 m n2h2
En =
8mL2
Neutron has similar mass as proton: m = 1.67 x 10-­‐27 kg If the radius of the nucleus is r = 7.5 x 10-­‐15 m, then width of the box that confines the neutron is L = 2r = 15 x 10-­‐15 m. h2
(6.626 ×10 −34 Js)2
−13
Energy of the ground state: E1 =
=
=
1.46
×10
J
2
−27
−15
2
8mL 8⋅1.67 ×10 kg ⋅ (15 ×10 m)
Energy of the second excited state: E3 = n 2 E1 = 9E1 = 1.314 ×10 −12 J
Physics 150, Prof. M. Nikolic 14 Exercise: ParGcle in a box A neutron is located within a uranium nucleus (radius r = 7.5 x 10-­‐15 m). The neutron goes from the second excited state to the ground state, emiong a photon in the process. What is the wavelength of the photon? What is given: ni = 3 nf = 1 r = 7.5 x 10-­‐15 m The transiGon energy: E = E3 − E1 = 1.314 ×10 −12 J −1.46 ×10 −13 J = 1.168 ×10 −12 J
Wavelength of the emi8ed photon: hc
E=
λ
Physics 150, Prof. M. Nikolic hc
λ=
E
6.626 ×10 −34 Js ⋅ 3×108 m / s
−13
λ=
=
1.7
×10
m
−12
1.168 ×10 J
15 ParGcle in a finite box If a container has walls of finite height, a parGcle in the box will have quanGzed energy levels, but the number of bound states (E < U0 ) will be finite. Experiments showed that the parGcle can extend past the walls of the container and “tunnel” its way out of the box. The probability of finding a parGcle is proporGonal to the square of its wave funcGon. When E > U0 è free parGcle è conGnuum of wavelengths and energies possible. Physics 150, Prof. M. Nikolic 16 The hydrogen atom In the quantum picture of the atom the electron does not really orbit the nucleus. e-
Square of wave funcGon è probability of finding electron in a given region of space p+
Quantum mechanics can be used to determine the allowed energy levels and wave funcGons for the electrons. Physics 150, Prof. M. Nikolic 17 Quantum numbers 1. Principal quantum number (n) è defined with allowed energy levels of atom (Bohr model) mk 2 e 4
2
En = −
=
n
E1
2
2!
where E1 = -­‐13.6 eV. 2. Orbital angular momentum quantum number (l) è defined with allowed angular momentum of the electron L = ! l (l +1)
Physics 150, Prof. M. Nikolic where l = 0, 1, 2,…n-­‐1 18 Quantum numbers 3. Orbital magne4c quantum number (ml) è defined with allowed angular momentum about the z-­‐axis Lz = ml !
where ml = -­‐l, -­‐l+1,…, -­‐1, 0, +1,…l-­‐1, l 2. Spin magne4c quantum number (ms) è due to the electron spin S z = ms !
Physics 150, Prof. M. Nikolic ms = ±½ for an electron 19 The Pauli exclusion principle Each electron’s state is fully described by four quantum numbers n, l ,ml, and ms. The Pauli Exclusion Principle says no two electrons in an atom can have the same set of quantum numbers. Physics 150, Prof. M. Nikolic 20 Shells and subshells A shell is the set of electron states with the same quantum number n. Each shell (n) is composed of one or more subshells (l) such that l<n 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. Physics 150, Prof. M. Nikolic 21 Understanding the periodic table of elements 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
4s orbital comes before 3d orbital! Example: Chlorine 17Cl Z = 17 → 17 protons → 17 electrons 1s
Specifies n Physics 150, Prof. M. Nikolic 2
n =1 → 1s2 (2 electrons) n = 2 → 2s2 2p6 (8 electrons) n = 3 → 3s2 3p5 (7 electrons) specifies the number of electrons in this orbital Specifies l 22 Exercise: Electron configuraGon (a) Find the magnitude of the angular momentum L for an electron with n = 2 and l = 1? Angular momentum is given with L = ! l (l +1)
L = ! 1(1+1) = 2!
(b) What are the allowed values of Lz? Angular momentum about z axis is given with Lz = ml !
The allowed values of ml are ml<l è +1,0,-­‐1 so that Lz can be Lz = +1!
Lz = 0!
Lz = −1!
Physics 150, Prof. M. Nikolic 23 Electron energy levels in solids A material is a conductor if the highest energy electron state filled at T= 0 is in the middle of the band. Physics 150, Prof. M. Nikolic If electrons fill their allowed states right to the top of the band, the material is either a semiconductor or an insulator. 24 Lasers Laser is an acronym for Light AmplificaGon by SGmulated Emission of RadiaGon. Physics 150, Prof. M. Nikolic 25 Lasers A photon of energy ΔE can sGmulate the emission of a photon (by interacGng with the excited electron). The emi8ed photon will have the same energy, phase, and momentum of the sGmulaGng photon. Typically the excited states of electrons have lifeGmes of about 10-­‐8 seconds. To make a laser, the material must have metastable states with lifeGmes of about 10-­‐3 seconds. This allows for a popula4on inversion in which more electrons are in a higher energy state rather than in a lower energy state. Physics 150, Prof. M. Nikolic 26