Bohr’s Correspondence Principle - Michigan State University · Bohr’s Correspondence Principle...
62
Bohr’s Correspondence Principle In limit that n →∞, quantum mechanics must agree with classical physics photon 2 i 2 f photon h n 1 n 1 eV 6 . 13 E f = - = In this limit, n i → n f , and then f photon → electron’s frequency of revolution in orbit. a Extension of Bohr theory to other “Hydrogen-like” atoms
Transcript of Bohr’s Correspondence Principle - Michigan State University · Bohr’s Correspondence Principle...
Bohr’s Correspondence Principle
In limit that n → ∞, quantum mechanics must agree with classical physics
photon2i
2f
photon hn1
n1
eV 6.13E f=
−=
In this limit, ni → nf , and thenf photon → electron’s frequency of revolution in orbit. a
Extension of Bohr theory to other “Hydrogen-like” atoms
Bohr’s Correspondence Principle
In limit that n → ∞, quantum mechanics must agree with classical physics
photon2i
2f
photon hn1
n1
eV 6.13E f=
−=
In this limit, ni → nf , and thenf photon → electron’s frequency of revolution in orbit. a
Extension of Bohr theory to other “Hydrogen-like” atomsHe+, Li++, Be+++, etc. (one electron “orbiting” nucleus of Q = +Ze)
Bohr’s Correspondence Principle
In limit that n → ∞, quantum mechanics must agree with classical physics
photon2i
2f
photon hn1
n1
eV 6.13E f=
−=
In this limit, ni → nf , and thenf photon → electron’s frequency of revolution in orbit. a
Extension of Bohr theory to other “Hydrogen-like” atomsHe+, Li++, Be+++, etc. (one electron “orbiting” nucleus of Q = +Ze)
22
2
422ee
2nn
eV 6.13Z-
2eZkm
n1
E =
−=
h
Bohr’s Correspondence Principle
In limit that n → ∞, quantum mechanics must agree with classical physics
photon2i
2f
photon hn1
n1
eV 6.13E f=
−=
In this limit, ni → nf , and thenf photon → electron’s frequency of revolution in orbit. a
Extension of Bohr theory to other “Hydrogen-like” atomsHe+, Li++, Be+++, etc. (one electron “orbiting” nucleus of Q = +Ze)
22
2
422ee
2nn
eV 6.13Z-
2eZkm
n1
E =
−=
h
Intrinsic linewidth ∆λ of emitted photons
Intrinsic linewidth ∆λ of emitted photons
Ei
Ef ground state
photon
excited state
Intrinsic linewidth ∆λ of emitted photons
Ei
Ef ground state
photon λ==−
hcEEE photonfi
excited state
Intrinsic linewidth ∆λ of emitted photons
Ei
Ef ground state
photon λ==−
hcEEE photonfi
Time-energy uncertainty principle:
excited state
Intrinsic linewidth ∆λ of emitted photons
Ei
Ef ground state
photon λ==−
hcEEE photonfi
Time-energy uncertainty principle:π
≥∆⋅∆4h
tE ii
excited state
Intrinsic linewidth ∆λ of emitted photons
Ei
Ef ground state
photon λ==−
hcEEE photonfi
Time-energy uncertainty principle:π
≥∆⋅∆4h
tE ii
The excited state of an atom is short lived (∆ti ~ 10-8 s ) before a photon is emitted.
excited state
Intrinsic linewidth ∆λ of emitted photons
Ei
Ef ground state
photon λ==−
hcEEE photonfi
Time-energy uncertainty principle:π
≥∆⋅∆4h
tE ii
The excited state of an atom is short lived (∆ti ~ 10-8 s ) before a photon is emitted. This causes an uncertainty in Ei (∆Ei) that induces an uncertainty in Ephoton, which in turn produces an uncertainty in λ.
excited state
Intrinsic linewidth ∆λ of emitted photons
Ei
Ef ground state
photon λ==−
hcEEE photonfi
Time-energy uncertainty principle:π
≥∆⋅∆4h
tE ii
The excited state of an atom is short lived (∆ti ~ 10-8 s ) before a photon is emitted. This causes an uncertainty in Ei (∆Ei) that induces an uncertainty in Ephoton, which in turn produces an uncertainty in λ.
excited state
For Ephoton ~ 2 eV (visible spectrum), ∆λ/λ ~ 10-8.
De Broglie electron waves and the Hydrogen atom
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λ
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie standing wavesin an electron orbit
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie standing wavesin an electron orbit
λ=π nr2
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie standing wavesin an electron orbit
λ=π nr2n = 1, 2, 3…
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie standing wavesin an electron orbit
λ=π nr2n = 1, 2, 3…
vmh
e=λ
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie standing wavesin an electron orbit
λ=π nr2n = 1, 2, 3…
vmh
e=λ
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie standing wavesin an electron orbit
λ=π nr2n = 1, 2, 3…
vmh
e=λ
De Broglie electron waves and the Hydrogen atom
nrvme h⋅=⋅⋅Meaning of: ?Analogy with standing waves on a vibrating string—get standing waves if have integer number of λ’s, in this case 3λ.
λ
3λInstead wrap string into circle,standing wave pattern is similar.
De Broglie standing wavesin an electron orbit
λ=π nr2n = 1, 2, 3…
vmh
e=λ
hnrvme =⋅⋅
Quantum Mechanics and the Hydrogen Atom
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
A revision of Bohr theory: n = 1 state actually has zero angular momentum!
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
A revision of Bohr theory: n = 1 state actually has zero angular momentum!
How is this possible? Won’t electron fall into proton?
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
A revision of Bohr theory: n = 1 state actually has zero angular momentum!
How is this possible? Won’t electron fall into proton?
Invoke Heisenberg Uncertainty Principle
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
A revision of Bohr theory: n = 1 state actually has zero angular momentum!
How is this possible? Won’t electron fall into proton?
Invoke Heisenberg Uncertainty PrincipleAs electron is localized near proton, the uncertainty of linear momentum will increase, causing its kinetic energy to rise.
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
A revision of Bohr theory: n = 1 state actually has zero angular momentum!
How is this possible? Won’t electron fall into proton?
Invoke Heisenberg Uncertainty PrincipleAs electron is localized near proton, the uncertainty of linear momentum will increase, causing its kinetic energy to rise.
π≥∆⋅∆
4h
pr r
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
A revision of Bohr theory: n = 1 state actually has zero angular momentum!
How is this possible? Won’t electron fall into proton?
Invoke Heisenberg Uncertainty PrincipleAs electron is localized near proton, the uncertainty of linear momentum will increase, causing its kinetic energy to rise.
Thus electron never “falls” into proton. Instead it forms a spherical “probability cloud” around nucleus.
π≥∆⋅∆
4h
pr r
Quantum Mechanics and the Hydrogen Atom
Schrodinger wave equation was solved for Hydrogen atom
A revision of Bohr theory: n = 1 state actually has zero angular momentum!
How is this possible? Won’t electron fall into proton?
Invoke Heisenberg Uncertainty PrincipleAs electron is localized near proton, the uncertainty of linear momentum will increase, causing its kinetic energy to rise.
“probability cloud”
nucleus
Thus electron never “falls” into proton. Instead it forms a spherical “probability cloud” around nucleus.
π≥∆⋅∆
4h
pr r
n = 1
Quantum Mechanics and the Hydrogen Atom (cont.)Quantum numbers
Quantum Mechanics and the Hydrogen Atom (cont.)Quantum numbers
Quantum Mechanics and the Hydrogen Atom (cont.)Quantum numbers
Need to include Spin Magnetic Quantum Number: ms = ± ½
Quantum Mechanics and the Hydrogen Atom (cont.)Quantum numbers
Need to include Spin Magnetic Quantum Number: ms = ± ½
ms = + ½
Quantum Mechanics and the Hydrogen Atom (cont.)Quantum numbers
Need to include Spin Magnetic Quantum Number: ms = ± ½
ms = + ½
ms = - ½
Pauli Exclusion Principle (1925) and the Periodic TableWolfgang Pauli (1900-1958)
Pauli Exclusion Principle (1925) and the Periodic TableWolfgang Pauli (1900-1958)
No two electrons in an atom can ever be in the same quantum state; that is, no two electrons in the same atom can have exactly the same value for the set of quantum numbers: n, l, ml, ms.
Pauli Exclusion Principle (1925) and the Periodic TableWolfgang Pauli (1900-1958)
No two electrons in an atom can ever be in the same quantum state; that is, no two electrons in the same atom can have exactly the same value for the set of quantum numbers: n, l, ml, ms.
+
+
+
Characteristic X-rays Mo: Z = 42
Characteristic X-rays Mo: Z = 42
n = 1K shell
n = 2L shell
State 1
Characteristic X-rays Mo: Z = 42
n = 1K shell
n = 2L shell
n = 1K shell
n = 2L shell
State 1High energy
electron
Characteristic X-rays Mo: Z = 42
n = 1K shell
n = 2L shell
n = 1K shell
n = 2L shell
State 1
Characteristic X-rays Mo: Z = 42
n = 1K shell
n = 2L shell
n = 1K shell
n = 2L shell
State 1 State 2
Characteristic X-rays Mo: Z = 42
n = 1K shell
n = 2L shell
n = 1K shell
n = 1K shell
n = 2L shell
n = 2L shell
State 1 State 2
Characteristic X-rays Mo: Z = 42
n = 1K shell
n = 2L shell
Eph
n = 1K shell
n = 1K shell
n = 2L shell
n = 2L shell
State 1 State 2 State 3
Characteristic X-rays Mo: Z = 42
n = 1K shell
n = 2L shell
Eph
n = 1K shell
n = 1K shell
n = 2L shell
n = 2L shell
In K shell for State 1, each electron partially shields the other. Thus effective nuclear charge ≡ Zeff = 42 – 1 = 41. In State 2, there is only one electron between L-shell electrons and nucleus, thus Zeff = 42 –1 = 41.
State 1 State 2 State 3
2eff2
2eff
K ZeV 6.13n
ZeV 6.13E ⋅−=⋅−=
2eff2
2eff
K ZeV 6.13n
ZeV 6.13E ⋅−=⋅−=
4Z
eV 6.13n
ZeV 6.13E
2eff
2
2eff
L ⋅−=⋅−=
2eff2
2eff
K ZeV 6.13n
ZeV 6.13E ⋅−=⋅−=
4Z
eV 6.13n
ZeV 6.13E
2eff
2
2eff
L ⋅−=⋅−=
keV 1.17EEE KLph =−=
2eff2
2eff
K ZeV 6.13n
ZeV 6.13E ⋅−=⋅−=
keV 17.1meV 1024.1
Ech 6
phK
⋅×=
⋅=λ
−
α
4Z
eV 6.13n
ZeV 6.13E
2eff
2
2eff
L ⋅−=⋅−=
keV 1.17EEE KLph =−=
2eff2
2eff
K ZeV 6.13n
ZeV 6.13E ⋅−=⋅−=
keV 17.1meV 1024.1
Ech 6
phK
⋅×=
⋅=λ
−
α
4Z
eV 6.13n
ZeV 6.13E
2eff
2
2eff
L ⋅−=⋅−=
pm 72K =λα
keV 1.17EEE KLph =−=
2eff2
2eff
K ZeV 6.13n
ZeV 6.13E ⋅−=⋅−=
keV 17.1meV 1024.1
Ech 6
phK
⋅×=
⋅=λ
−
α
4Z
eV 6.13n
ZeV 6.13E
2eff
2
2eff
L ⋅−=⋅−=
pm 72K =λα
keV 1.17EEE KLph =−=
