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Transcript
Phys 102 – Lecture 24
Phys 102 Lecture 24
The classical and Bohr atom
1
th
State of late 19
Century Physics
• Two great theories
Two great theories
“Classical
Classical physics
physics”
Newton’s laws of mechanics & gravity
Maxwell’s theory of electricity & magnetism, including EM waves
• But... some unsettling problems
Stability of atom & atomic spectra
Photoelectric effect
...and others
d th
• New theory required
Quantum mechanics
Phys. 102, Lecture 24, Slide 2
The “classical” atom
Negatively charged electron orbits around positively charged nucleus
–
O biti
Orbiting e
h
has centripetal acceleration:
ti t l
l ti
e2 mv 2
FE = k 2 =
r
r
Total energy of electron:
r
+
ke2
so,
= mv 2
r
FE
Etot
–
1 2
e2
1 ke2
=−
= K + U = mv − k
2
r
2 r
Etot
v
0
r
Hydrogen atom
Recall Lect. 4
Phys. 102, Lecture 24, Slide 3
Stability of classical atom
Prediction – orbiting e– is an oscillating charge & should emit EM waves in every direction
EM waves in every direction
+
utot =
–
1
1 2
2
ε0 Erms
+
Brms
2
2 μ0
EM waves carry energy, so e– should lose energy & fall into nucleus! Classical atom is NOT stable!
Lifetime of classical atom = 10–11 s
Reality – Atoms are stable
Phys. 102, Lecture 24, Slide 4
Atomic spectrum
Prediction – e– should emit light at whatever frequency f it orbits nucleus
d sin θ = mλ
Discharge tube
Recall Lecture 22
Diffraction grating
Screen
Hydrogen
Reality – Only certain
Only certain
frequencies of light are emitted & are different for different elements
Helium
Nitrogen
Oxygen
Phys. 102, Lecture 24, Slide 5
Frequency
Quantum mechanics
Quantum mechanics explains stability of atom & atomic spectra (and many other phenomena )
spectra (and many other phenomena...)
QM is one of most successful and accurate scientific theories
Predicts measurements to <10
Predicts
measurements to <10–88 (ten parts per billion!) (ten parts per billion!)
Wave‐particle duality – matter behaves as a wave
Particles can be in many places at the same time
Particles
can be in many places at the same time
Processes are probabilistic not deterministic
Measurement changes behavior
Certain quantities (ex: energy) are quantized
Phys. 102, Lecture 24, Slide 6
Matter waves
Matter behaves as a wave with de Broglie wavelength Wavelength of matter wave
matter wave
h
λ=
p
Pl k’
Planck’s constant
t t
h = 6.626 ×10−34 J ⋅ s
Momentum M
of particle
Ex: a fastball (m = 0.5 kg, v
Ex: a fastball (m
0.5 kg, v = 100 mph 100 mph ≈ 45 m/s)
45 m/s)
λ fastball
h
=
p
Ex: an electron (m = 9.1×10–31 kg, v = 6×105 m/s)
h
λelectron =
p
Phys. 102, Lecture 24, Slide 7
X‐ray vs. electron diffraction
X‐ray
X
ray diffraction
diffraction
e– diffraction
DEMO
Identical pattern emerges if de Broglie wavelength of e– equals the X‐ray wavelength!
th X
l th!
Phys. 102, Lecture 24, Slide 8
Electron diffraction
Beam of monoenergetic e– passes through double slit
Interference pattern = probability
d sin θ = mλ
λ=
h
p
– –
–
– ––
Merli – 1974
Tonomura – 1989 Phys. 102, Lecture 24, Slide 9
ACT: Double slit interference
Consider the interference pattern from a beam of mono‐
energetic electrons A passing through a double slit.
energetic electrons A
passing through a double slit
Now a beam of electrons B with 4x th
the energy of A
f A enters the slits. t th lit
What happens to the spacing θ
between interference maxima?
A.
B
B.
C.
D.
E.
θ
θB = 4 θA
θB = 2 θ
= 2 θA
θB = θA
θB = θA / 2
θB = θA / 4
Phys. 102, Lecture 24, Slide 10
The Bohr model
e– behave as waves & only orbits that fit an integer number of wavelengths are allowed
g
Orbit circumference
2 = nλλ = n
2πr
+
h
p
n = 1,
1 22,3
3…
de Broglie wavelength
de Broglie wavelength
Angular momentum is quantized
Angular momentum is quantized
Ln = n
≡
h
2π
“h bar”
“Quantum number”
Phys. 102, Lecture 24, Slide 11
ACT: Bohr model
What is the quantum number n of this hydrogen atom?
+
A.
B.
C.
D.
n = 1
n = 3
n = 6
n = 12
Phys. 102, Lecture 24, Slide 12
Energy and orbit quantization
From classical atom:
Angular momentum is quantized
ke2
= mv 2
r
Ln ≡ pr = mvr = n
+
n = 1
E
n = 2
=2
n = 3
Smallest orbit is at a0, so e– cannot fall into nucleus
r
n = 4
n = 3
n = 2
Radius of orbit is quantized
n2 2
2
≡
n
a0
rn =
2
mke
Energy is quantized
Smallest orbit has energy –R
g
“ground state”
“Bohr radius”
n = 1
mk 2 e 4
R
En = − 2 2 ≡ − 2
n
2n
Rydberg
constant
Phys. 102, Lecture 24, Slide 13
ACT: CheckPoint 3.2
Suppose the charge of the nucleus is doubled (+2e), but the e– charge remains the same (
the e
remains the same (–e).
e). How does r
How does r for the for the
ground state (n = 1) orbit compare to that in hydrogen?
n2 2
For hydrogen: rn =
mke 2
+2e
A. 1/2 as large
B. 1/4 as large
+
–
C. the same
Phys. 102, Lecture 24, Slide 14
ACT: CheckPoint 3.3
There is a particle in nature called a muon, which has the same charge as the electron but is 207 times heavier A muon
same charge as the electron but is 207 times heavier. A muon
can form a hydrogen‐like atom by binding to a proton.
How does the radius of the ground state (n = 1) orbit for this hydrogen‐like atom compare to that in hydrogen?
A. 207× larger
B. The same
C. 207× smaller
Phys. 102, Lecture 24, Slide 15
ACT: Bohr model momentum
According to the Bohr model, how is the electron linear g
,
momentum p = mv quantized?
A. p is not quantized B. p is quantized as 1/n
C. p is quantized as n
Phys. 102, Lecture 24, Slide 16
Atomic units
At atomic scales, Joules, meters, kg, etc. are not convenient units “Electron Volt” – energy gained by charge +1e when accelerated by 1 Volt: U = qV
1e = 1.6×10–19 C, so 1 eV = 1.6×10–19 J
Planck constant: h = 6.626 × 10–34 J∙s
Speed of light: c = 3 × 108 m/s
Electron mass: m = 9.1 × 10–31 kg
hc ≈ 2 ×1025 J ⋅ m = 1240 eV ⋅ nm
mc 2 = 8.2 ×10−13 J = 511, 000 eV
ke2
Since U =
, ke2 has units of eV∙nm like hc
r
ke2
ke2
1
= 2π
≈
c
hc 137
ke2 ≈ 1.44 eV ⋅ nm
“Fine structure constant” (dimensionless)
Phys. 102, Lecture 24, Slide 17
Rydberg constant & Bohr radius
Energy of electron in H‐like atom (1 e–, nuclear charge +Ze):
2
2
2 4
Z
mk Z e
En = − 2 R = −
n
2n 2 2
2
Z
511, 000 eV Z 2
2 ⎛ ke ⎞
= − 2 mc ⎜
⎟ =−
2n
2 ⋅137 2 n 2
⎝ c ⎠
2
2
Z2
= −13.6 eV 2
n
Radius of electron orbit:
n 2 c ⎛ c ⎞ n 2 1240 eV ⋅ nm
n2 2
n2
=
137
rn =
=
a0 =
2 ⎜
2 ⎟
2
Z mc ⎝ ke ⎠
Z 2π ⋅ 511, 000 eV
Z
mkZe
n2
= 0.0529nm
0 0529nm
Z
Phys. 102, Lecture 24, Slide 18
ACT: Hydrogen‐like atoms
Consider an atom with a nuclear charge of +2e with a single electron orbiting, in its ground state (n = 1), i.e. He+. How much energy is required to ionize the atom totally?
A. 13
A
13.6 eV
6 V
B. 2 × 13.6 eV
C 4 ×
C.
4 × 13.6 eV
13 6 eV
Phys. 102, Lecture 24, Slide 19
Heisenberg Uncertainty Principle
e– beam passing through slit of width a will diffract
pf
Δy = a
–
Δpy
p0
p0
e– passed somewhere inside slit.
Uncertainty in y
Momentum was along x, but e– diffracts along y.
Uncertainty in y‐momentum
If slit narrows, diffraction pattern spreads out
Uncertainty in y
Uncertainty in y
Uncertainty in y‐momentum
Δy ⋅ Δp y ≥
2
Phys. 102, Lecture 23, Slide 20
ACT: CheckPoint 4
The Bohr model cannot be correct! To be consistent with the Heisenberg Uncertainty Principle which of the
the Heisenberg Uncertainty Principle, which of the following properties cannot be quantized? 1.
2.
3.
4.
En = −
R
n2
Energy is quantized
Angular momentum is quantized Ln = n
Radius is quantized rn = n 2 a0
Linear momentum & velocity are quantized pn =
na0
A. All of the above
B. #1 & 2
C. #3 & 4
Phys. 102, Lecture 24, Slide 21
Summary of today’s lecture
• Classical model of atom
Predicts unstable atom & cannot explain atomic spectra
• Quantum mechanics
Matter behaves as waves
Heisenberg Uncertainty Principle
• Bohr model
Only orbits that fit n electron wavelengths are allowed
Explains the stability of the atom
Energy quantization correct for single e– atoms (H, He+, Li++)
However, it is fundamentally incorrect
Need complete quantum theory (Lect. 26)
Phys. 102, Lecture 23, Slide 22