Upload
others
View
23
Download
0
Embed Size (px)
Citation preview
Chapter 8 - AtomicStructure
Hydrogen Atom
Chapter 8 - Atomic Structure
“Oh, the humanity!”
-Herbert Morrison, radioreporter of the Hindenburg
disaster
David J. StarlingPenn State Hazleton
PHYS 214
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
The hydrogen atom is composed of a proton and
an electron with potential energy:
U(r) =1
4πε0
e2
r.
Now, our well is not square but follows an inverse law.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
The hydrogen atom is composed of a proton and
an electron with potential energy:
U(r) =1
4πε0
e2
r.
Now, our well is not square but follows an inverse law.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
All we need to do is solve Schrödinger’s equation
in spherical coordinates with this radial potential.
−~2
2µ1
r2 sin θ
[sin θ
∂
∂r
(r2∂ψ
∂r
)+
∂
∂θ
(sin θ
∂ψ
∂θ
)+
1sin θ
∂2ψ
∂φ2
]− e2
4πε0rψ = Eψ
We’re not there yet, so let’s try a different way...
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
All we need to do is solve Schrödinger’s equation
in spherical coordinates with this radial potential.
−~2
2µ1
r2 sin θ
[sin θ
∂
∂r
(r2∂ψ
∂r
)+
∂
∂θ
(sin θ
∂ψ
∂θ
)+
1sin θ
∂2ψ
∂φ2
]− e2
4πε0rψ = Eψ
We’re not there yet, so let’s try a different way...
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
All we need to do is solve Schrödinger’s equation
in spherical coordinates with this radial potential.
−~2
2µ1
r2 sin θ
[sin θ
∂
∂r
(r2∂ψ
∂r
)+
∂
∂θ
(sin θ
∂ψ
∂θ
)+
1sin θ
∂2ψ
∂φ2
]− e2
4πε0rψ = Eψ
We’re not there yet, so let’s try a different way...
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Instead, let’s think of the electron as orbiting the
proton classically.
Using centripetal acceleration, we get
F = ma1
4πε0
e2
r2 = m(−v2
r
)
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Instead, let’s think of the electron as orbiting the
proton classically.
Using centripetal acceleration, we get
F = ma1
4πε0
e2
r2 = m(−v2
r
)
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Based on experimental evidence, he know that the
energy levels are quantized.
So let’s guess that the angularmomentum l is alsoquantized:
l = rmv = n~
v =n~rm
for n = 1, 2, 3, ....
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Based on experimental evidence, he know that the
energy levels are quantized.
So let’s guess that the angularmomentum l is alsoquantized:
l = rmv = n~
v =n~rm
for n = 1, 2, 3, ....
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Combining our quantized angular momentum
assumption with the orbital equation, we get
r =4πε0~2
me2 n2 =ε0h2
πme2 n2
We call the multiplier of n2 the Bohr Radius a.
r = an2 with a =ε0h2
πme2 .
and a ≈ 52.92 pm.
This result is surprisingly accurate!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Combining our quantized angular momentum
assumption with the orbital equation, we get
r =4πε0~2
me2 n2 =ε0h2
πme2 n2
We call the multiplier of n2 the Bohr Radius a.
r = an2 with a =ε0h2
πme2 .
and a ≈ 52.92 pm.
This result is surprisingly accurate!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Combining our quantized angular momentum
assumption with the orbital equation, we get
r =4πε0~2
me2 n2 =ε0h2
πme2 n2
We call the multiplier of n2 the Bohr Radius a.
r = an2 with a =ε0h2
πme2 .
and a ≈ 52.92 pm.
This result is surprisingly accurate!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
To find the energy, we combine kinetic and
potential and then use the orbital equation.
14πε0
e2
r2 = mv2
r
E =12
mv2 − 14πε0
e2
r= − 1
8πε0
e2
r2
Subbing in the Bohr radius formula for r, we get:
En = − me4
8ε0h21n2 = −E0
n2 for n = 1, 2, 3, ...
where E0 = 2.180× 10−18J.
This result agrees with quantum theory!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
To find the energy, we combine kinetic and
potential and then use the orbital equation.
14πε0
e2
r2 = mv2
r
E =12
mv2 − 14πε0
e2
r= − 1
8πε0
e2
r2
Subbing in the Bohr radius formula for r, we get:
En = − me4
8ε0h21n2 = −E0
n2 for n = 1, 2, 3, ...
where E0 = 2.180× 10−18J.
This result agrees with quantum theory!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
To find the energy, we combine kinetic and
potential and then use the orbital equation.
14πε0
e2
r2 = mv2
r
E =12
mv2 − 14πε0
e2
r= − 1
8πε0
e2
r2
Subbing in the Bohr radius formula for r, we get:
En = − me4
8ε0h21n2 = −E0
n2 for n = 1, 2, 3, ...
where E0 = 2.180× 10−18J.
This result agrees with quantum theory!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
The hydrogen atom can absorb a photon if its
energy matches an electron transition energy.
Eγ = hf = Ehigh − Elow
hcλ
= −E0
(1
n2high− 1
n2low
)1λ
= −E0
hc
(1
n2high− 1
n2low
)1λ
= R
(1
n2low− 1
n2high
)
with R = 1.097373× 107 m−1 (Rydberg Constant).
This is how we find the wavelength of photons emitted fromelectronic transitions.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
The hydrogen atom can absorb a photon if its
energy matches an electron transition energy.
Eγ = hf = Ehigh − Elow
hcλ
= −E0
(1
n2high− 1
n2low
)1λ
= −E0
hc
(1
n2high− 1
n2low
)1λ
= R
(1
n2low− 1
n2high
)
with R = 1.097373× 107 m−1 (Rydberg Constant).
This is how we find the wavelength of photons emitted fromelectronic transitions.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Each transition has a unique energy with a photon
of a different wavelength.
If the ending state is n = 1, then that series of wavelengthsis known as the Lyman series.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Each transition has a unique energy with a photon
of a different wavelength.
If the ending state is n = 1, then that series of wavelengthsis known as the Lyman series.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
But the electron may end up in another state
instead.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Remember this?
−~2
2µ1
r2 sin θ
[sin θ
∂
∂r
(r2∂ψ
∂r
)+
∂
∂θ
(sin θ
∂ψ
∂θ
)+
1sin θ
∂2ψ
∂φ2
]− e2
4πε0rψ = Eψ
The ground state solution is:
ψ(x) =1
√πa3/2 e−r/a
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Remember this?
−~2
2µ1
r2 sin θ
[sin θ
∂
∂r
(r2∂ψ
∂r
)+
∂
∂θ
(sin θ
∂ψ
∂θ
)+
1sin θ
∂2ψ
∂φ2
]− e2
4πε0rψ = Eψ
The ground state solution is:
ψ(x) =1
√πa3/2 e−r/a
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
The hydrogen atom’s electron, in the ground state,
can exist at all radii (except 0).
P(r) = 4πr2|ψ(x)|2 =4a3 r2e−2r/a
The peak of this function is at r = a, the Bohr radius!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
The hydrogen atom’s electron, in the ground state,
can exist at all radii (except 0).
P(r) = 4πr2|ψ(x)|2 =4a3 r2e−2r/a
The peak of this function is at r = a, the Bohr radius!
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Higher energy states of the hydrogen atom get
more complex due to the quantized angular
momentum.
These states all have the same energy (n = 2) but differentangular momenta (l = 0, 1).
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Higher energy states of the hydrogen atom get
more complex due to the quantized angular
momentum.
These states all have the same energy (n = 2) but differentangular momenta (l = 0, 1).
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
There are three quantum numbers for the
hydrogen atom.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
There are three quantum numbers for the
hydrogen atom.
Chapter 8 - AtomicStructure
Hydrogen Atom
Hydrogen Atom
Lecture Question 8.1Which of the following most closely resembles the Bohr
model of the hydrogen atom?
(a) A solid metal sphere with a net positive charge.
(b) A hollow metal sphere with a net negative charge.
(c) A tray full of mud with pebbles uniformly distributed
throughout.
(d) The Moon orbiting the Earth.
(e) Two balls, one large and one small, connected by a
spring.