DQW

how'd i do?

I think being triatomic has something to do with it - you don't get certain modes in diatomic molecules.

In diatomic (and linear triatomic) molecules, the moment of inertia about the internuclear axis is negligible.

The high specific heat capacity of Li is due to Dulong and Petit (in a crystalline solid the number of available modes is constant).

 

The high specific heat of water is essentially due to the fact that it is one of the lightest non-linear molecules. As far as I know, only ammonia (NH3), has a higher specific heat at RT - as is expected.

First approach :

 

Use the condition for equal slopes. This gives 2ax = 1 or x = 1/2a. From this, you have the point of intersection in terms of 'a', which you can plug into the equation for the parabola and solve for 'a'.

Alternative approach : For a line to be tangent to a curve, the line and curve must have exactly one point of intersection. The points of intersection between the two is clearly given by the quadratic equation got from eliminating y between the two equations. What is the condition for a quadratic equation to have exactly one real solution ?

 

PS : 'a' is not 1/2

What ideas do you have? Tell us what you think, and we'll suggest improvements or corrections, if required.

How do you solve this problem?

The parabola y=ax^2 + 6 is tangent to the line y=x.

Find a.

A necessary condition for a line to be a tangent at some point of a curve is that the line have the same slope as the curve does at that point.

How much centrifugal force is created by the Earth's rotation, if any? It is obviously not enough to counteract the Earth's gravity, and send us all flying into space, but is there still some acting on the bodies on the Earth?

At the equator, [imath]v^2/Rg \approx 0.3 [/imath]%.

Flux pinning in a Type II SC happens when the SC is in the Abrikosov phase. This is a phase that is intermediate to the Meissner phase and the normal phase. When you increase the temperature or field from the Meissner phase in certain types of SCs, the SC finds that it is energetically less expensive to partially allow some of the field through it (than to exclude the entire field or allow the entire field). So, rather than remain in the perfect Meissner phase or turn completely into a normal material, the SC nucleates tiny lines of normal material that carry flux through them. In a perfectly defect-free SC crystal, these lines, known as flux vortices, arrange themselves along a triangular lattice. However, in a real material (with impurities and defects), these vortices get pinned at the locations of these defects. This is what is known as flux pinning in Type II SCs, and is what helps stabilize a levitation demo.

http://www.chemistrycoach.com/quantum.htm

 

http://hyperphysics.phy-astr.gsu.edu/hbase/qunoh.html

 

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/quantum.html

 

http://scienceworld.wolfram.com/physics/QuantumNumbers.html

Or is there a difference between s/p/d/f and E=1,2,3, etc?

Yes there is a big difference. For hydrogenlike atoms/ions, the energy depends only on the principal quantum number 'n'. This relationship is what Tom wrote out (following the Bohr picture). For any value 'n' of the principal quantum number, there are n allowed values of the angular momentum quantum number 'l' (from 0 to n-1). It is these numbers that are designated s (l=0), p (l=1), d (l=2), f (l=4), ... by spectroscopists.

 

The principal quantum number does not determine angular momentum. And in multielectron atoms (in the absense of applied fields), the energy depends both on 'n' and 'l'.

Can quantum waves be used as a means to el communicato

One thing that comes close to an answer, in the spirit that I imagine this question was intended is the speculation that you can have communication between a pair of nanodevices on a chip, by placing them at the foci of a quantum corral.

http://www.scienceforums.net/forums/showthread.php?t=5230

 

Another answer that comes to mind is the possible use of photon polarizations to designate binary bits to achieve encrypted communication that overcomes the key distribution problem.

http://www.scienceforums.net/forums/showthread.php?t=13255

The safest place would be at the ends. The "unsafest" would be in the middle. This can be intuited from a symmetry argument, but can also be reasoned out pretty unmessily (just cumbersome to LaTeX it all out). Particularly, notice that contributions from separations greater than L/2 are smaller than those from separations smaller than L/2. This will demonstrate that x=L/2 provides the maximum probability of getting hit. A similar argument tells you that the ends are the safest. The trick is in adding amplitudes - that's all.

It's about this :

 

oh by the way' date=' how do you make an equation go to the next line using latex?[/quote']

Okay, I get what you're saying now. But the transmission of radiation by a transparent medium is not given by the Stefan-Boltzmann Law. That only talks about radiation from a heated surface. The energy transmitted through a transparent medium is a function of the transmission coefficient, not the emissivity.

 

Oh, wait, you're treating the inside surface of the glass window as the radiator ? I'm not sure I see how exactly that works, but I haven't really given it much thought.

I think it was supposed to be simple, and for the material in question, adding the heat loss mechanisms is a reasonable approximation.

I can't say I agree with this - it is completely unphysical. If the outside of the wall is assumed to be at T(out) - this is the assumption you make in justifying the conduction equation as written by Sarah - there will be NO radiation. And if the outside wall is assumed to be at T(in) - this is the assumption made in writing the radiation equation in that form - there will be no conduction.

 

As, I said before, conduction and radiation are not processes that happen in parallel here. "Adding them up" makes no sense to me (and this answer will be an overestimate; albeit by a fraction comparable to P(rad)/P(cond) as calculated by Sarah).

I'm not sure what mezarashi means by this interaction "from a distance". I am talking about specular reflection - which, if I'm not mistaken (and I could be), is not empirically different from diffuse reflection at the microscopic scale. Diffuse reflection, in my opinion, is merely a macroscopic effect arising from surface topographical features.

 

For a metal, the Fresnel reflectance is essentially a constant (angle independent), and reduces to Snell's Law. Also, for a metal, one can easily derive Snell's Law from a simple calculation based on the Maxwell Equations (and angle conservation comes directly from requiring that the tangential component of the E-field be continuous at an interface). However, this classical explanation provides little intuition for the mechanism at a microscopic level. As you refine your models and go from a classical model to QM to QED..., you have different explanations that approximate the real phenomenon to different extents.

 

For a simple semiclassical treatment, see Feynman's Lectures Vol 2 Ch. 32.

Also see http://en.wikipedia.org/wiki/Plasmon

The photon is actually absorbed and re-emitted. It is absorbed by one of the conduction electrons. This transfer of energy and momentum to the conduction electron creates a longitudinal compression wave within the conduction ("free") electrons. So, the impinging photon causes the plasma of free electrons to compress and the flex back (at the natural frequency of the free electron gas) spitting out a photon identical to the incident one.

No, sarah, that's not correct. The correct answer will be what you get from following some nerd's suggestion (post #5). Unforunately, this does give you a 4th order polynomial. If you're clever, you can use a Taylor expansion and throw away high order terms and make life easier for yourself.

 

Conduction through the pane and radiation off its surface are processes in series not in parallel. You can not add them up.

But why would heating cause *more* excitation *possibilities* ?. It seems to me (but I may be wrong) that it will just excite more electrons to an already possible excitation states, but not create new ones...

That's essentially correct, but not entirely. The most important effect of increasing temperature is that you now populate states that were unoccupied at lower temperatures - specifically some of the states above E(F) in the picture below.

 

 

In any case, this is more or less irrelevant to reflection off a metallic surface.

DQW, did you also affirmed the following sentence ?:

Yes, I did - in the context of reflection at least. There are processes like x-ray photoemission (of core electrons) that deal with individual atoms, but that's not what we're talking about here.

Wait, so a photon does not interact with each atom of a solid, by means of excitation and de-excitation ?. If this is the case, then using the flashlight I described in the post above, will you still get reflectance from the aluminum mirror ?.

Absolutely. And at frequencies below the plasma frequency, you will have every one of your incident frequencies be reflected, not just certain special frequencies. And this is not realy because you have a band of energies but because you are exciting collective oscillatory modes known as plasmons - kind of like bouncing a steel ball off a rubber floor.

The nexusmagazine article is pretty funny. One person who I suspect mightn't have appreciated the humor would be Pauli. He would have been furious that some bloke was giving Einstein the credit for predicting the neutrino. Grrr...

Rick Santorum is a #$&^%$, IMO, but that's besides the point.

 

Just a few days ago, when I was talking about this wth friends, they were speculating that Bush will perhaps throw the left some scraps (as a trade) in the form of loosening up little on stem cell research. I was highly sceptical. And I was right.

 

http://news.bbc.co.uk/2/hi/americas/4567253.stm

 

The more I read stuff like this and the fact that the Republican political machine is pushing Lynn Swann (yikes!!) to run for Goernor in PA, makes me believe more strongly in the Puppet Theory of Right Wing politics.

IIRC, there IS a way to detect if a fiber-optic cable has been intercepted and "Listened to", it`s something to do with QM, and that a photon once "Observed" will alter, and will thus be detectable by the receiver.

YT : As I explained in my previous pair of posts. You alter the photon polarization only if you do not know what the polarization is (and hence use an arbitrary filter). And with any given filter, the best you can do is eliminate one polarization - namely the one normal to the filter orientation. But if you are told that the polarization is along one of two normal directions, then you can eliminate one direction and hence determine the polarization direction. This is essentially what JohnWB's evesdropper "Eve" has achieved. So, it is possible for Eve to intercept the message and prepare an identical message. The only difficulty I can see, will be in making up for the time lag. In theory, this difficulty could be overcome by transmitting over air (conventionally) for some distance.