Chemistry 262
Professor S. Alex Kandel
Spring 2014
Due 2/28/2014
Problem Set 5
1. McQuarrie, problem 12-7. (Ask yourself: “do these equations I’ve derived make
sense?”)
2. McQuarrie, problem 12-10.
3. McQuarrie, problem 12-12.
4. McQuarrie, problem 12-21.
5. The wavefunctions for hydrogen’s 1s and 2pz orbitals are:
ψ1s = √
ψ2pz =
1
3/2
πa0
e−r/a0
1
re−r/2a0 cos θ
√
5/2
4 2πa0
Try to match the forms of these functions to your mental images of the corresponding orbitals. Here, a0 is the Bohr radius,
a0 ≈ 52.9 pm
Or in Mathematica,
psi1s = 1/Sqrt[Pi * a0^3] * Exp[-r/a0]
psi2p = 1/(4 * Sqrt[2 * Pi] * a0^(5/2)) * r * Exp[-r/(2 a0)]
* Cos[theta]
Mathematica can do multiple integrals; the following command integrates a
function of x, y, and z over all space:
Integrate[(1/Sqrt[Pi * a0^3] * Exp[-Sqrt[x^2 + y^2 + z^2]/a0])^2,
{x,-Infinity,Infinity},{y,-Infinity,Infinity},{z,-Infinity,Infinity}]
Use Mathematica (you can download it from the OIT web site, or you can find
it on campus cluster computers) for the following problems. Print out your
Mathematica notebook and hand it in with your problem set. If your notebook
contains a lot of other commands (mistakes, experiments, fun), either edit them
out, or highlight the commands that worked to get your answers.
1
(a) For any wavefunction ψ, the square modulus |ψ|2 is proportional to the
probability of finding the particle (for the H atom, the electron) at a given
location. The total probability of finding the electron, summed over all
space, must be equal to 1:
Z
dτ |ψ|2 = 1
R
Here, dτ is shorthand for a three-dimensional integral that covers all
space, so:
Z
dτ =
Z ∞
−∞
dx
Z ∞
−∞
dy
Z ∞
dz =
−∞
Z ∞
2
r dr
0
Z π
sin θdθ
0
Z 2π
dφ
0
Show that the integral of |ψ|2 over all space equals 1 for the 1s and 2pz
orbitals.
(b) The average radial distance (this is also called the expectation value) for
an electron in the atom is given by the integral:
hri =
Z
dτ r · |ψ|2
Note that this r is in addition to the r2 that is automatically in the integral.
Calculate hri for both orbitals.
(c) The average squared distance (that is, the average of the squares of the
distance) is
Z
hr2 i =
dτ r2 · |ψ|2
Calculate the root-mean-square distance for both orbitals:
rrms =
q
hr2 i
(d) For a 1s orbital, what is the probability that the electron is within a0 of
the nucleus? 2a0 ? (To calculate this, replace the ∞ limit on r with a0 or
2a0 .)
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