Chapter 8 - Atomic Structure · Chapter 8 - Atomic Structure Hydrogen Atom Hydrogen Atom Combining...

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

Hydrogen Atom

The hydrogen atom is composed of a proton and

an electron with potential energy:

U(r) =1

4πε0

Now, our well is not square but follows an inverse law.

Hydrogen Atom

The hydrogen atom is composed of a proton and

an electron with potential energy:

U(r) =1

4πε0

Now, our well is not square but follows an inverse law.

Hydrogen Atom

All we need to do is solve Schrödinger’s equation

in spherical coordinates with this radial potential.

r2 sin θ

[sin θ

(r2∂ψ

(sin θ

1sin θ

∂2ψ

∂φ2

]− e2

4πε0rψ = Eψ

We’re not there yet, so let’s try a different way...

Hydrogen Atom

r2 sin θ

[sin θ

(r2∂ψ

(sin θ

1sin θ

∂2ψ

∂φ2

]− e2

4πε0rψ = Eψ

Hydrogen Atom

r2 sin θ

[sin θ

(r2∂ψ

(sin θ

1sin θ

∂2ψ

∂φ2

]− e2

4πε0rψ = Eψ

Hydrogen Atom

Instead, let’s think of the electron as orbiting the

proton classically.

Using centripetal acceleration, we get

F = ma1

4πε0

r2 = m(−v2

Hydrogen Atom

Instead, let’s think of the electron as orbiting the

proton classically.

Using centripetal acceleration, we get

F = ma1

4πε0

r2 = m(−v2

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, ....

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, ....

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!

Hydrogen Atom

r =4πε0~2

me2 n2 =ε0h2

πme2 n2

πme2 .

and a ≈ 52.92 pm.

Hydrogen Atom

r =4πε0~2

me2 n2 =ε0h2

πme2 n2

πme2 .

and a ≈ 52.92 pm.

Hydrogen Atom

To find the energy, we combine kinetic and

potential and then use the orbital equation.

14πε0

r2 = mv2

mv2 − 14πε0

r= − 1

8πε0

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!

Hydrogen Atom

14πε0

r2 = mv2

mv2 − 14πε0

r= − 1

8πε0

En = − me4

8ε0h21n2 = −E0

n2 for n = 1, 2, 3, ...

where E0 = 2.180× 10−18J.

Hydrogen Atom

14πε0

r2 = mv2

mv2 − 14πε0

r= − 1

8πε0

En = − me4

8ε0h21n2 = −E0

n2 for n = 1, 2, 3, ...

where E0 = 2.180× 10−18J.

Hydrogen Atom

The hydrogen atom can absorb a photon if its

energy matches an electron transition energy.

Eγ = hf = Ehigh − Elow

= −E0

n2high− 1

= −E0

n2high− 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.

Hydrogen Atom

The hydrogen atom can absorb a photon if its

energy matches an electron transition energy.

Eγ = hf = Ehigh − Elow

= −E0

n2high− 1

= −E0

n2high− 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.

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.

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.

Hydrogen Atom

But the electron may end up in another state

instead.

Hydrogen Atom

Remember this?

r2 sin θ

[sin θ

(r2∂ψ

(sin θ

1sin θ

∂2ψ

∂φ2

]− e2

4πε0rψ = Eψ

The ground state solution is:

ψ(x) =1

√πa3/2 e−r/a

Hydrogen Atom

Remember this?

r2 sin θ

[sin θ

(r2∂ψ

(sin θ

1sin θ

∂2ψ

∂φ2

]− e2

4πε0rψ = Eψ

The ground state solution is:

ψ(x) =1

√πa3/2 e−r/a

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!

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!

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).

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).

Hydrogen Atom

There are three quantum numbers for the

hydrogen atom.

Hydrogen Atom

There are three quantum numbers for the

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.

Chapter 8 - Atomic Structure · Chapter 8 - Atomic Structure Hydrogen Atom Hydrogen Atom Combining...

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