1. For an electron in the ground state of a hydrogen atom:
a) calculate the expectation value of \(r^2\) . [This requires an integral over all 3D space. Don't forget to include the \( 4 \pi r^2 \) factor and to understand where it comes from.]
b) Calculate the expectation value of the potential energy.
c) Evaluate the expectation value of the potential energy in eV. What is its sign?
d) Think about the relationship between PE, KE and E for this state. Since you probably already know that the energy is -13.6 eV (1 Rydberg), can you guess what the KE must me?
2. Consider the wave-function of an electron in the ground state of a hydrogen atom, \( \psi _{100} = ( \pi a^3)^{-1/2} e^{-r/a} \) :
a) For y=z=0 plot this wave-function as a function of x along the x axis.
b) Plot this wave-function as a function of r. (What is the range of r?)
3. Consider the wave-function of an electron in a 1st-excited state of a hydrogen atom called \( \psi _{200} \), ( \( \psi _{200} = (8 \pi a^3)^{-1/2} (1-r/2a) e^{-r/2a} \))
a) For y=z=0 plot this wave-function as a function of x along the x axis.
b) Plot this wave-function as a function of \( r = +(x^2 + y^2 + z^2)^{1/2}\).
c) Where is this wave-function zero?
d) Use a (constant) contour plot to illustrate the nature of this wave-function.
4. Consider the wave-function of an electron in a 1st-excited state of a hydrogen atom \( \psi _{21x} = (8 \pi a^3)^{-1/2} (x/2a) e^{-r/2a} \) :
a) For y=z=0 plot this wave-function as a function of x along the x axis.
b) For x=z=0 plot this wave-function as a function of y along the y axis.
c) Where is this wave-function zero?
5. Use a (constant) contour plot to illustrate the nature of:
a) \( \psi _{21x} = (8 \pi a^3)^{-1/2} (x/2a) e^{-r/2a} \)
b) What axes did you use for this ?
6. Use a (constant) contour plots to illustrate the nature of:
a) \( \psi _{21y} = (8 \pi a^3)^{-1/2} (y/2a) e^{-r/2a} \)
b) \( ( \psi _{21y} + \psi _{21x})/2 \) . Along what direction (in the x-y plane) does the maximum of this wave function lie?
7. In the hydrogen atom, 1st-excited states can come in many forms. One can add together two, or more, 1st-excited states to make another one. (This is called hybridization.) We include coefficients on these "linear combinations" to maintain normalization. For example, consider the state: \( (\psi _{21x} - \psi _{200}) / \sqrt{2} \) .
a) For an electron in this state, calculate the expectation value of x, y and z.
b) Which one is not zero. (Why is that?)
c) what are the units of this expectation value? Is that what you expected?
8. Consider the state: \( (2/3)^{1/2} \psi _{21y} - (1/3)^{1/2} \psi _{200} \) .
a) For this state would the expectation value of x, y or z be non-zero?
b) Whichever one is non-zero, calculate it.
c) Comparing the non-zero expectation value from this problem to the one from the previous problem: which is larger? (extra credit: What do you think might be the origin of that difference?)
9. Consider the two states: \( (1/2)^{1/2} \psi _{21x} -(1/6)^{1/2} \psi _{21y} - (1/3)^{1/2} \psi _{200} \) and \( (2/3)^{1/2} \psi _{21y} - (1/3)^{1/2} \psi _{200} \) .
a) are these states othogonal to each other?
b) are these states normalized? [For these problems you can start from the assumption that {
\( \psi _{21x} \), \( \psi _{21y} \), \( \psi _{21z} \) and \( \psi
_{200} \)} are mutually orthogonal and normalized.]
c) (extra credit) Can you find a 3rd state constructed using 1st excited states that is orthogonal to each of these states (and is normalized)? Can you find a 3rd state constructed using only \( \psi _{200} \), \( \psi _{21x} \) and \( \psi _{21y} \) that is orthogonal to each of these states (and is normalized)?
10. a) Calculate the expectation value of the kinetic energy of an electron in the ground state of a hydrogen atom.
b) Evaluate this in eV. Is it positive or negative?
For #1, when we find the expectation value do we consider the complex conjugate of the wave function this time instead of taking the square of it?
ReplyDeleteAlso, when taking the integral using spherical coordinates, would we use the 4pi r^2 instead of r^2sin (theta). That way, we would take the integral from 0 to infinity with respect to dr? I'm not sure if I'm thinking in the right direction.
You are very much thinking in the right direction. The wave functions we are working with are still "real". As soon as that changes, we will start using the complex conjugate.
ReplyDeleteWhat you say makes sense, but just to be extra clear: r^2 sin(theta) and 4 pi r^2 are intimately related (and both arise in spherical coordinate integrals. 4 pi r^2 is what you get when you integrate r^2 sin(theta) over phi and theta. This only happens when the rest of the integrand is purely radial, i.e., not dependent on theta and phi. Does that makes sense?
PS. This is done in the video called HW1 video..., posted w while back to help with HW1.
This comment has been removed by the author.
DeleteQuestion 1
ReplyDeletepart a) is the answer 3*a^(2)
i think yes.
DeleteIt has units that make sense (length^2).
DeleteThis comment has been removed by the author.
ReplyDeleteFor I had a question about 1c, my answer for 1b ended up being (-e^2)/pi(epsilon)a, and I'm not sure if that is correct, but my problem is what do i do with the Epsilons units of coulombs? I can't seem to get rid of it.
ReplyDeleteWould those Coulombs cancel with the e^2 Coulombs? Check out HW 1, problem 6.
DeleteMy understanding was that e was the 2.71828 value, not the electron charge. If so, then how do some of the integrals from 1a and 1b work out?
DeleteIt depends on context. In the potential, U(x), it is the magnitude of the charge of an electron (or a proton). In the expression for "a" it is the strength of the potential and thus the same.
DeleteHowever, in all the wave-functions it is that 2.718 value, as you mentioned.
Looking at problem 6 from HW 1 may also be helpful.
For numbers 2-4, do we let r = sqrt(x^2 + y^2 + z^2) and then have a function as x as we let y = z = 0?
ReplyDeletesounds right to me.
DeleteShould all the wave functions in the assumption part of 9b be the same?
ReplyDeletenope. thanks. i fixed that.
DeleteThis comment has been removed by the author.
ReplyDeleteI think I am missing the point here, but when plotting psi as a function of r, how do we handle plotting with resepect to x,y, and z? Is psi a 4d function that we are trying to make 3d slices of?
ReplyDeleteEach psi is different. And they are each 3D functions. {x,y z} (cartesian) and {r, theta phi} spherical are 2 different coordinate systems. Either system can be used; the utility of a particular coordinate system depends on the nature (symmetry) of the function to be plotted.
ReplyDeleteYou are only asked to plot as a function of r when it is a function that has no dependence on theta or phi. In general, one cannot plot psi as a function of r. Only for a few special cases (e.g., only one out of 4 of our 1st excited states).
For 7b, I haven't come up with a good reason as to why < x > is nonzero, and not < y > and < z >. Surely, the name of the function Psi21x-Psi200 suggests it, and I was able to show that < y > and < z > are zero by integrating (integrating, integrating...), but I don't see a general reason/argument yet. Anyone have an idea on this?
ReplyDelete