1. MS310 Quantum Physical Chemistry
Ch 9. The Hydrogen Atom
- Historical hydrogen atom model (Plum pudding model,
Rutherford model) vs. quantum mechanical model
- Formulate the Schrödinger equation for hydrogen atom
and solve it
Study the energy level and orbitals of hydrogen atom
MS310 Quantum Physical Chemistry

2. MS310 Quantum Physical Chemistry
9.1 Formulating the Schrödinger equation
Rutherford vs shell model
electrons are confined in spherical shells centered at nucleus
orbit the nucleus, accelerating motion and energy radiation
→ atom is ‘not’ stable!
This problem is solved by Quantum approach.
We consider the Coulomb potential x z y e– e+
H atom : 1 proton + 1e–
H-like atom : Z proton + 1e– ex) He+
Hamiltonian of hydrogen atom is
MS310 Quantum Physical Chemistry

3. MS310 Quantum Physical Chemistry
Use the center of mass(already discuss in chapter 7), we can divide it by 2 equations.
Focus on the internal motion(center of mass motion : translation)
Therefore, Schrödinger equation is written as
(in generally, we must use μ instead of me. However, in the hydrogen atom case, both are almost same and this book use me.)
MS310 Quantum Physical Chemistry

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9.2 Solving the Schrödinger equation for the Hydrogen atom
Use the separation of variable : ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)
We know the form of from chapter 7
Rewrite the Schrödinger equation using the angular momentum
Focus on the radial part(we already know the angular solution)
MS310 Quantum Physical Chemistry

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coefficient of second term : effective potential
First term : centripetal potential, related to 1/r2
Second term : coulomb potential, related to -1/r
Unless the l=0 case, centripetal potential is dominant → electron of p, d, f orbital(l>0) far from nucleus than electron of s orbital(l=0)
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9.3 Eigenvalues and eigenfunctions for the total energy
We can divide the radial part equation
i) V=0
Solution : spherical Bessel function
MS310 Quantum Physical Chemistry

7. The equation becomes to
ii) V≠0
Again introduce dimensionless quantities for convenience
The equation becomes to

8. MS310 Quantum Physical Chemistry
Energy of hydrogen atom :
Define the constant
: 0.529Å for hydrogen atom : Bohr radius
Therefore, energy is
For n>5 state, states are in the narrow range,
0 to -1x10-19 J.
Potential of H atom : very narrow for first few states,
but very wide for large n
As known from a particle in a box : energy spacing is
inverse of the square of box length
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9. MS310 Quantum Physical Chemistry
Energy : depends on only the principle quantum number n
However, wavefunction depends on the 3 quantum numbers, n, l, and ml
The relationship is given by
n : 1, 2, 3, 4, …
l : 0, 1, 2, 3, …, n-1
ml : 0, ±1, ±2, …, ±l
(existence of these quantum numbers are from boundary condition)
Radial function R(r) : product of exponential function with a polynomial, dimensionless variable r/a0
Functional form of radial function : depends on the quantum numbers n and l.
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First few radial functions Rnl(r) are given by
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Therefore, wavefunction of hydrogen atom, ψnlml is given by
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15. MS310 Quantum Physical Chemistry
Hydrogen wave function
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These functions are called to both eigenfunction and H atom ‘orbitals’
There are some property of orbital
1) Letter s, p, d, f are used to denote l = 0, 1, 2, 3
2) ψ100(r) : 1s orbital or wave function
3) all 3 wave functions with n=2, l=1 : 2p orbitals
4) wavefunction is real when ml=0, complex when otherwise
Wavefunction is normalized to generate the probability density.(postulate 3)
Energy of H atom : degenerated
- n=1 : no degeneracy
- n=2 : 4-fold degeneracy
- n=3 : 9-fold degeneracy
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Use the superposition principle : if y1 and y2 are solutions of DE, then c1y1+c2y2 is also solution of DE.
→ can make the complex functions to real
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Real form : more convenient to visualize the chemical bond
However, real form is not an eigenfunction of
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Experimental emission spectra is given by
Use the reduced mass instead of me, frequency of spectral line is given by
Sometimes, we use wave number instead of frequency
Rydberg constant : mee4/8ε02h3c : 2.180x10-18 J , cm-1
Reduced mass of H : 0.05% greater than me
Spectral line of H is given by
MS310 Quantum Physical Chemistry

23. MS310 Quantum Physical Chemistry
9.4 The hydrogen atom orbital
Bohr model : electron orbit around the nucleus and only certain orbits allowed
Probability : proportional to ψ*(r,θ,φ)ψ(r,θ,φ)dτ
Our focus is on the
1) wave function ψnlml(r,θ,φ)
2) probability of finding electron ψ2nlml(r,θ,φ)sinθdrdθdφ
3) define radial distribution function and
Ground state of H atom : ψ100(r)
Plot it : need 4 coordinate(x, y, z and P(x,y,z))
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a) 3D plot on x-y half plane
b) contour plot on x-y half plane
red : high probability, blue : low probability
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Plot of 1s, 2s, and 3s orbital
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Plot R(r) vs r : 1s, 2s, 2p, 3s, 3p, and 3d orbital
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Ground state : no radial nodes
Quantum number increase : # of nodes increase
When 2s and 3s, node has constant r : spherical ‘nodal surface’
Then, what about the node of another orbitals?
→ l>0 : not spherical symmetry : ‘angular shape’ of orbital
See contour of 2py, 3py, 3dxy and 3dz2 orbital
→ we can see the angular nodal surface
Nodal surface of 2py : y=0m no radial nodes
Generally, l nodal surfaces in angular part and n-l-1 radial nodal surfaces, n-1 total nodal surfaces
3py : additional nodal plane x=0 : radial node
3dxy : 2 nodal planes intersect the z axis
3dz2 : 2 ‘conical’ nodal surfaces, rotating the z axis
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Ex)9.3
Locate the nodal surfaces in
Sol) consider the radial and angular part separately.
Angular part : cos θ
Node : cos θ = 0, θ = π/2
It means the plane z=0 in Cartesian coordinate
Radial part :
Node depends on only (exponential term cannot be zero)
→ r=0 and r=6a0
r=0 : a point → no meaning, r=6a0 : a surface
Therefore, there are 1 angular and 1 radial node.
It is same as the general result(l angular nodes and n-l-1 radial nodes)
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9.5 The radial probability distribution function
See the ψ2n00(r,θ,φ) : n=1,2 and 3
Maxima is at r = 0
Consider the ψ2nlml(r,θ,φ) in general case(l>0)
→ centripetal barrier, nonzero angular momentum
→ wavefunction is not spherically symmetric.
: p and d orbitals
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Ex)9.4
a) Where is the maximum probability point?
b) Assume nucleus diameter of H is 2x10-15m. Then, probability of electron of 2s orbital is in the nucleus?
Sol)
a) The point : maximum value of ψ*(τ)ψ(τ)dτ
Only see the and differentiate it
However, r cannot be negative → consider the ρ=0
Therefore, maximum point is ρ=0 → r=0
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b) Result of a) : unphysical
Probability is given by
Assume the integrand is constant on the interval(rnucleus : small)
Approximately, because of small rnucleus
Finally we can obtain
Therefore, probability of finding the electron in the nucleus is essentially zero.
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Plot of a03R2(r) vs r/a0
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R(r) cannot describe the ‘real’ distribution!
→ radial probability depends on the ‘summation over all θ and φ’
For 1s orbital,
Generally, introduce the ‘radial distribution function’ P(r)
We can determine the most probable position of electron.
Understand the difference between radial distribution and probability density function is very important.
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9.6 The validity of the shell model of an atom
See the plot of radial distribution function
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Quantum distribution of 1s, 2s, and 3s
Classical shell model
Classical shell model is not valid any more.
However, we can see the most dense point of 1s orbital is very less intensity when 2s and 3s orbital!
→ classical shell model is useful although reducing a complex
function to a single number is unwise.
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Summary
Result of solving the Schrödinger equation for hydrogen atom is exactly equal to experimental data.
Shape of orbital is changed by the quantum number and probability of finding electron depends on the shape of orbital.
There are n-l-1 radial nodal surfaces and l angular nodal surfaces, # of total nodes is n-1 for nth state.
MS310 Quantum Physical Chemistry