Download Bohr`s Correspondence Principle

Bohr’s Correspondence Principle
In limit that n → ∞,
quantum mechanics must agree with classical physics
E photon
 1

1
= 13.6 eV  2 − 2  = hf photon
 nf n i 
In this limit, ni → nf , and then
f 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
E photon
 1

1
= 13.6 eV  2 − 2  = hf photon
 nf n i 
In this limit, ni → nf , and then
f photon → electron’s frequency of revolution in orbit. a
Extension of Bohr theory to other “Hydrogen-like” atoms
He+, 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
E photon
 1

1
= 13.6 eV  2 − 2  = hf photon
 nf n i 
In this limit, ni → nf , and then
f photon → electron’s frequency of revolution in orbit. a
Extension of Bohr theory to other “Hydrogen-like” atoms
He+, Li++, Be+++, etc. (one electron “orbiting” nucleus of Q = +Ze)
2 2 4
 1  me ke Z e 
2 13.6 eV
E n = − 2 
=-Z
2
2


n
 n  2 h

Bohr’s Correspondence Principle
In limit that n → ∞,
quantum mechanics must agree with classical physics
E photon
 1

1
= 13.6 eV  2 − 2  = hf photon
 nf n i 
In this limit, ni → nf , and then
f photon → electron’s frequency of revolution in orbit. a
Extension of Bohr theory to other “Hydrogen-like” atoms
He+, Li++, Be+++, etc. (one electron “orbiting” nucleus of Q = +Ze)
2 2 4
 1  me ke Z e 
2 13.6 eV
E n = − 2 
=-Z
2
2


n
 n  2 h

Intrinsic linewidth ∆λ of emitted photons
Intrinsic linewidth ∆λ of emitted photons
Ei
excited state
photon
Ef
ground state
Intrinsic linewidth ∆λ of emitted photons
Ei
excited state
photon
Ef
ground state
Ei − Ef = E photon
hc
=
λ
Intrinsic linewidth ∆λ of emitted photons
Ei
excited state
photon
Ef
ground state
Time-energy uncertainty principle:
Ei − Ef = E photon
hc
=
λ
Intrinsic linewidth ∆λ of emitted photons
Ei
excited state
photon
Ef
Ei − Ef = E photon
hc
=
λ
ground state
Time-energy uncertainty principle:
h
∆E i ⋅ ∆t i ≥
4π
Intrinsic linewidth ∆λ of emitted photons
Ei
excited state
photon
Ef
Ei − Ef = E photon
hc
=
λ
ground state
Time-energy uncertainty principle:
h
∆E i ⋅ ∆t i ≥
4π
The excited state of an atom is short lived (∆ti ~ 10-8 s ) before a
photon is emitted.
Intrinsic linewidth ∆λ of emitted photons
Ei
excited state
photon
Ef
Ei − Ef = E photon
hc
=
λ
ground state
Time-energy uncertainty principle:
h
∆E i ⋅ ∆t i ≥
4π
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.
Intrinsic linewidth ∆λ of emitted photons
Ei
excited state
photon
Ef
Ei − Ef = E photon
hc
=
λ
ground state
Time-energy uncertainty principle:
h
∆E i ⋅ ∆t i ≥
4π
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.
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
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
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
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie standing waves
in an electron orbit
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie standing waves
in an electron orbit
2 πr = nλ
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie standing waves
in an electron orbit
2 πr = nλ
n = 1, 2, 3…
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie standing waves
in an electron orbit
h
λ=
me v
2 πr = nλ
n = 1, 2, 3…
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie standing waves
in an electron orbit
h
λ=
me v
2 πr = nλ
n = 1, 2, 3…
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie standing waves
in an electron orbit
h
λ=
me v
2 πr = nλ
n = 1, 2, 3…
De Broglie electron waves and the Hydrogen atom
Meaning of: m e ⋅ v ⋅ r = n ⋅ h ?
Analogy with standing waves on a vibrating string—get standing
waves if have integer number of λ’s, in this case 3λ.
λ
3λ
Instead wrap string into circle,
standing wave pattern is similar.
De Broglie standing waves
in an electron orbit
h
λ=
me v
2 πr = nλ
n = 1, 2, 3…
me ⋅ v ⋅ r = nh
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 Principle
As 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 Principle
As electron is localized near proton, the
uncertainty of linear momentum will
increase, causing its kinetic energy to
rise.
h
∆r ⋅ ∆p r ≥
4π
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
As electron is localized near proton, the
uncertainty of linear momentum will
increase, causing its kinetic energy to
rise.
h
∆r ⋅ ∆p r ≥
4π
Thus electron never “falls” into proton.
Instead it forms a spherical
“probability cloud” around nucleus.
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
As electron is localized near proton, the
uncertainty of linear momentum will
increase, causing its kinetic energy to
rise.
h
∆r ⋅ ∆p r ≥
4π
Thus electron never “falls” into proton.
Instead it forms a spherical
“probability cloud” around nucleus.
“probability cloud”
n=1
nucleus
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 Table
Wolfgang Pauli (1900-1958)
Pauli Exclusion Principle (1925) and the Periodic Table
Wolfgang 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,
m l, m s.
Pauli Exclusion Principle (1925) and the Periodic Table
Wolfgang 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,
m l, m s.
+
+
+
Characteristic X-rays Mo: Z = 42
Characteristic X-rays Mo: Z = 42
State 1
n=1
K shell
n=2
L shell
Characteristic X-rays Mo: Z = 42
State 1
High energy
electron
n=1
K shell
n=1
K shell
n=2
L shell
n=2
L shell
Characteristic X-rays Mo: Z = 42
State 1
n=1
K shell
n=1
K shell
n=2
L shell
n=2
L shell
Characteristic X-rays Mo: Z = 42
State 1
State 2
n=1
K shell
n=1
K shell
n=2
L shell
n=2
L shell
Characteristic X-rays Mo: Z = 42
State 1
State 2
n=1
K shell
n=1
K shell
n=1
K shell
n=2
L shell
n=2
L shell
n=2
L shell
Characteristic X-rays Mo: Z = 42
State 1
State 2
State 3
n=1
K shell
n=1
K shell
n=1
K shell
n=2
L shell
n=2
L shell
n=2
L shell
Eph
Characteristic X-rays Mo: Z = 42
State 1
State 2
State 3
n=1
K shell
n=1
K shell
n=1
K shell
n=2
L shell
n=2
L shell
n=2
L shell
Eph
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.
Z2eff
E K = −13.6 eV ⋅ 2 = −13.6 eV ⋅ Z2eff
n
Z2eff
E K = −13.6 eV ⋅ 2 = −13.6 eV ⋅ Z2eff
n
E L = −13.6 eV ⋅
Z2eff
n2
Z2eff
= −13.6 eV ⋅
4
Z2eff
E K = −13.6 eV ⋅ 2 = −13.6 eV ⋅ Z2eff
n
EL =
Z2eff
−13.6 eV ⋅ 2
n
= −13.6 eV ⋅
Z2eff
4
E ph = E L − E K = 17.1 keV
Z2eff
E K = −13.6 eV ⋅ 2 = −13.6 eV ⋅ Z2eff
n
EL =
Z2eff
−13.6 eV ⋅ 2
λ Kα
h ⋅ c 1.24 × 10 −6 eV ⋅ m
=
=
E ph
17.1 keV
n
= −13.6 eV ⋅
Z2eff
4
E ph = E L − E K = 17.1 keV
Z2eff
E K = −13.6 eV ⋅ 2 = −13.6 eV ⋅ Z2eff
n
EL =
Z2eff
−13.6 eV ⋅ 2
λ Kα
h ⋅ c 1.24 × 10 −6 eV ⋅ m
=
=
E ph
17.1 keV
n
λ K α = 72 pm
= −13.6 eV ⋅
Z2eff
4
E ph = E L − E K = 17.1 keV
Z2eff
E K = −13.6 eV ⋅ 2 = −13.6 eV ⋅ Z2eff
n
EL =
Z2eff
−13.6 eV ⋅ 2
λ Kα
h ⋅ c 1.24 × 10 −6 eV ⋅ m
=
=
E ph
17.1 keV
n
λ K α = 72 pm
= −13.6 eV ⋅
Z2eff
4
E ph = E L − E K = 17.1 keV