FAQ:
o So, was the quiz worth anything?
-The point of the quiz was to:
“…help you be ready for Tuesday's class, in which we will solve the H-atom problem. In fact just being aware of these questions will help you be prepared”
The real value of the quiz would be if it helped you be ready for and appreciate today’s class, especially the subtle, but important role of the quantized length scale that emerges at the very end of the calculation. It is easy to miss that after the fatigue of a long calculation.
o Can't we leave it at,"a length scale constructed from e^2, hbar, and m is hbar^2/me^2"?
-Yes
o Is there any way we can make up the quiz?
-If you post the answer to the quiz here you will get full credit for the quiz. If you also did it correctly in class you get extra credit.
o "people have been punished directly for showing up late (understandable), and indirectly for finishing homework on time…" -Jon
-Perhaps it might seem that way to you now, but from my point of view no one is being punished. Punishment is not really an issue here at all. This is about trying to create opportunities for you to learn and remember interesting and important physics.
The idea was that if people spent 1/2 hour thinking about length scales and attractive potentials they would get more out of the class.
Subscribe to:
Post Comments (Atom)
In an earlier post, Zack mentioned that we should read several pages from Liboff's book, and it actually helps to do that! The answer to the quiz is on page 447, equation 10.119 but Liboff uses the reduced mass and calls the length scale the Bohr radius. But ignoring that and just using class notation, the length scale would be (hbar^2)/m*e^2.
ReplyDeleteI got there late, and I don't think the K.E. and P.E. question was on the quiz, but with the little reading that I did, I think that problem 10.43 is relevant to finding those values, but I'm not sure.
It makes me feel a bit better knowing I'm not the only one having a hard time following the class. Zack, I think part of the reason the lack of hw structure is such a problem is the fact that the course does not closely follow any textbook, and some of the things we talk about are not necessarily in a textbook. I'm not saying that's a bad thing (and you definitely gave us reading for the hydrogen atom), but it does make it a lot harder to know what we should study, and what we should understand. As another student put it, it's hard if you don't speak Zack.
ReplyDeleteSo here's my check of the length scale:
ReplyDelete(h-bar)^2
----------
m e^2
=
J^2 s^2
--------
kg J m
=
kg m^2 s^2
------------
s^2 kg m
= m
Me too. V=-e^2/r
ReplyDeletehbar^2 has units of J^2 s^2
m(kg). e^2 has units of C^2
J^2 s^2/kgC^2 has units of meters.
So, a characteristic length 'a' would be:
a=hbar^2/me^2
Ok...here is my new solution to the quiz with my corrected understanding of what it was asking. (sorry about that :) )
ReplyDeleteThe value a_0 is the characteristic length scale in units of meters. We will pretend that we don't know what the Bohr Radius is yet, however, this is a_0 which can be derived from the Uncertainty Principle and looking at KE and PE as we did for problem 6 in HW1.
Anyway...if we just look at the units and the only thing we are given is the electron charge square e^2 [Joule*m], hbar [J*sec] and m [kg] there is a way to arrange these values in order to create a new constant in units of meters.
A Joule has units of kg*m^2/s^2. First e^2 ends up with a m^3 plus other stuff. We just want one meter, so if we take hbar^2 we will have m^4, so we can stick the hbar^2 in the numerator and devide by e^2. Unfortunately, when we do this there will also be a leftover [kg], so to get rid of it, just devide by the mass. The seconds cancel perfectly which leaves just mass...thus the characteristic lenth scale can be written as a_0 = hbar^2/(me^2) or in units: [((kg*m^2/s^2)^2)/((kg*m^2/s^2)*m)] = m.
Tadda! :)
Also, Energy is related to this characteristic length scale. This was a fundamental discovery in QM!
Yeah, pretty simple.
ReplyDeleteAnalogous to the attractive delta-function potential where
V(x) = -alpha*delta(x)
and
a = (hbar^2)/(m*alpha)
alpha is the strength of the delta function potential.
In the hydrogen atom,
V(r) = -(e^2)/r
so (e^2) is the strength of the 1/r (hydrogen atom) potential. So, by analogy,
a = (hbar^2)/(m * e^2)
This will work out for the units as well, units of length of course, resulting in meters [after converting joules to kg*(m/s)^2]
This comment has been removed by the author.
ReplyDeleteHere is my quiz. I hope this is visible.
ReplyDeletehttp://i137.photobucket.com/albums/q218/deluzian/quizqm.jpg
This is Ariel, for some reason my email is under my mom's name. sorry for the confusion.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletee^2 (Energy * length)
ReplyDeletehbar (Energy * time)
m (mass)
Energy (mass * length^2/time^2)
Most length scales go as hbar^2/m
putting e^2 on the bottom makes the units work, so
a = hbar^2/me^2
(Carlin, by the way)
e^2 = energy*length (J*m)
ReplyDeleteh_bar = energy*time (J*s) or (kg*m^2*s^-1)
m = mass (kg)
In the length scale form of h_bar^2/m or (J*m^2/kg)
Dividing by e^2 leaves length
(kg^2*m^4*s^4) / (kg^2*m^3*s^4) = m
Oh, this is Jennifer.
ReplyDeleteMy solution:
ReplyDeletehttp://people.ucsc.edu/~akunapul/Quiz2.pdf
@Adi Yay Zombie Feynman!
ReplyDelete