The Hydrogen Atom Redux The first three wavefunctions (n=1,2,3) of a - - PowerPoint PPT Presentation

The Hydrogen Atom Redux

Physics 273 – The Hydrogen Atom Redux

The first three wavefunctions (n=1,2,3) of a 1D particle in a box of length L.

2

0.2 0.4 0.6 0.8 1.0

• 1.0
• 0.5

0.5 1.0

ψ(x/L) x/L

n=1 n=2 n=3

Physics 273 – The Hydrogen Atom Redux

Example: Let’s consider a particle-in-a-box in the n=2

• state. What are the probabilities that the particle is

located between: a) 0 < x < L/3 b) L/3 < x < 2L/3 c) 2L/3 < x < L Notice that 0.402+.196+0.402=1! Good: we’ve normalized the wavefunction properly.

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Ψ(x) = r 2 L sin ⇣nπx L ⌘ P1 = Z L/3 2 L sin2 ✓2πx L ◆ dx ≈ 0.402 P2 = Z 2L/3

L/3

2 L sin2 ✓2πx L ◆ dx ≈ 0.196 P2 = Z L

2L/3

2 L sin2 ✓2πx L ◆ dx ≈ 0.402

Physics 273 – The Hydrogen Atom Redux 4

0.2 0.4 0.6 0.8 1.0

• 1.0
• 0.5

0.5 1.0

ψ(x/L) x/L

0.2 0.4 0.6 0.8 1.0 0.5 1.0 1.5 2.0

x/L [ψ(x/L)]2

The n=2 wavefunction of a particle in a box The n=2 wavefunction squared of a particle in a box Total area=1

Physics 273 – The Hydrogen Atom Redux

Ok, let’s start tackling the hydrogen atom. The first problem we encounter is the fact that it is a 3- dimensional system. Since the potential has a radial symmetry, it will be best to move into spherical coordinates.

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V (r) = − 1 4⇡✏0 e2 r

Coulomb Potential:

As before, we can move into an inertial frame by substituting the reduced mass for the electron mass.

Physics 273 – The Hydrogen Atom Redux

The potential is independent of time, so we use the 3D time-independent Schrodinger equation: We need to do a coordinate transformation. It turns out that this yields the following: Yuck.

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~2 2mr2Ψ(~ x) + V (~ x)Ψ(~ x) = EΨ(~ x) − ~2 2µ  1 r2 @ @r ✓ r2 @Ψ @r ◆ + 1 r2 sin ✓ @ @✓ ✓ sin ✓@Ψ @✓ ◆ + 1 r2 sin2 ✓ @2Ψ @2

• −

1 4⇡✏0 e2 r Ψ = EΨ

Physics 273 – The Hydrogen Atom Redux

What we want to do is get this into the form [...] + [...]=0 This essentially turns the problem into solve two separate differential equations. So, let’s rewrite ψ by separating variables: and plug it into Schrodinger’s equation...

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Ψ(~ x) = Ψ(r, ✓, ) = R(r)Y (✓, )

depends only on r depends only on θ,Φ

Physics 273 – The Hydrogen Atom Redux

We get... Now, multiply (on the left) by to get This is exactly the form we want: [...] + [...] = 0

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− ~2 2µ  Y r2 ∂ ∂r ✓ r2 ∂R ∂r ◆ + R r2 sin θ ∂ ∂θ ✓ sin θ∂Y ∂θ ◆ + R r2 sin2 θ ∂2Y ∂φ2

• + V (r)RY = ERY

−2µ ~2 r2 Y R  1 R ∂ ∂r ✓ r2 ∂R ∂r ◆ − 2µr2 ~2 (V − E)

• +

 1 Y sin θ ∂ ∂θ ✓ sin θ∂Y ∂θ ◆ + 1 Y sin2 θ ∂2Y ∂φ2

• = 0

Physics 273 – The Hydrogen Atom Redux

So we have separated the “r” part from the “θ,Φ” part. Since varying one “part” cannot affect varying the other “part,” both parts must be independent and constant: Notice that the angular part is also independent of V. It turns out that the solution to this differential equation is known as the spherical harmonics. We can ignore it for now. It’s only important that it’s independent of r.

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 1 R @ @r ✓ r2 @R @r ◆ − 2µr2 ~2 (V − E)

• = `(` + 1)

 1 Y sin ✓ @ @✓ ✓ sin ✓@Y @✓ ◆ + 1 Y sin2 ✓ @2Y @2

• = −`(` + 1)

constants

Physics 273 – The Hydrogen Atom Redux

Let’s first consider the 𝓂=0 case. This is the easiest to solve. The solution will be of the form: Let’s plug it in. From the chain rule, we have So we have:

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1 R @ @r ✓ r2 @R @r ◆ + 2µr2 ~2 ✓ 1 4⇡✏0 e2 r + E ◆ = 0 R(r) = Ce−r/a0 1 R ∂ ∂r ✓ r2 ∂R ∂r ◆ = 1 R  2r∂R ∂r + r2 ∂2R ∂r2

• ∂R

∂r = − 1 a0 R , ∂2R ∂r2 = 1 a2 R → 1 R ∂ ∂r ✓ r2 ∂R ∂r ◆ = −2r a0 + r2 a2

Physics 273 – The Hydrogen Atom Redux

Rearranging terms, we get For this to hold for all values of r, the expression (...) must be equal to 0! Thus we can solve for a0 and E. This is the Bohr radius! Similarly, if (...)=0, then the energy is

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E = ✓ ~2 µa0 − e2 4⇡✏0 ◆ 1 r − ~2 2µa2 ✓ ~2 µa0 − e2 4⇡✏0 ◆ = 0 → a0 = 4⇡✏0~2 µe2 E = − ~2 2µa2 = − µe4 32⇡2✏2

0~2 ≈ −13.6 eV

Physics 273 – The Hydrogen Atom Redux

Finally, the last bit is to get the normalization. For the ground-state equation, the angular solution to Schrodinger’s equation is a constant. Therefore: Therefore, the ground state wavefunction of the hydrogen atom is

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Ψ(r, θ, φ) = Ce−r/a0 Z Ψ?Ψd3r = Z ∞ Ψ(r)?Ψ(r)4πr2dr = 1 Z ∞ 4πC2r2e−2r/a0dr = 1 → C = s 1 πa3 Ψ(r, θ, φ) = s 1 πa3 e−r/a0

Physics 273 – The Hydrogen Atom Redux

Example: Calculate the most probable value of the radius in the ground state. The probability for an electron between r and r+dr is: Find the maximum value of P(r): The only local maxima occurs when r=a0, that is, the Bohr radius is the most probable location of the electron.

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P(r)dr = Ψ?Ψ · 4πr2dr = 1 πa3 e−2r/a0 · 4πr2dr

dP(r) dr = 0 = 4 a3  2re−2r/a0 − 2 a0 r2e−2r/a0

• = 8

a3 re−2r/a0  1 − r a0