GR9277 #90



Alternate Solutions 
cailh 20061031 07:15:01  A simpy way. Rotational energy of dimoleculars due to far infard ray emission. Lamda approx 1*E3 m.
Other quits:
vibration Lamada approx 1*E6 m  

Comments 
enterprise 20180331 21:08:51  You can quickly notice that the MeV range is out. 10^9 is too small. 10 ev ~ Hydrogen ionization energy which is UV. The only sensible choice is the milli ev range.   jondiced 20101109 17:22:12  Ground state of hydrogen's electron levels is 13.6 eV, so the rotational levels must be lower than that. This eliminates C, D, and E.
neon37 20101112 08:02:26 
why lower?

postal1248 20131002 16:52:11 
As nyuko says below, electronic levels are of greater energy than rotational levels.

  apps00 20101006 22:32:17  It may be useful to remember that
where is mass of a molecula and is the electron's mass. , and
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  BerkeleyEric 20100627 23:11:52  The thermal energy at room temp. is 1/40 eV, and these rotational energy levels are definitely accessible at that temp., so C, D, and E are eliminated. The value in A is really, really small, so we're left with B.
syreen 20130918 16:16:32 
I like this! I would add (if this works)
Each degree of freedom contributes 1/2 kT to the energy, so spacing~1/2 kT. kT at room temp~1/40 eV.
1/2(1/40)=1/8*10^2~~10^3.

  nyuko 20091030 22:16:26  I think it is worthwhile to know this:
Electronic levels: ~1 eV
Vibrational levels: ~0.1eV
Rotational levels: ~0.001eV
One can find this in Taylor's Modern Physics, summary for chapter 12   FortranMan 20081011 17:00:22  ...I just used fNkT/2, with N =1 and T = 300K, though you did have to know what k is in eV, which the reference sheet doesn't list. Is this a nono answer or it is okay since most of thermodynamics consists of approximations of quantum mechanical behaviors?
Poop Loops 20081025 20:07:23 
I'd say it's probably better, since you're thinking of a spinning rod. I find it kind of dubious to add the angular momenta of each particle and say that's the total angular momentum for the system. You're no longer thinking of one particle, but two together attached together...
It's like the angular momentum of Earth about its own axis and the Sun about its own axis *doesn't* give the total angular momentum of the EarthSun system.

john77 20090321 07:59:46 
This is the easiest way . And you don't have to know k in eV since your are given k in joules and the electron charge .

  cailh 20061031 07:15:01  A simpy way. Rotational energy of dimoleculars due to far infard ray emission. Lamda approx 1*E3 m.
Other quits:
vibration Lamada approx 1*E6 m
madfish 20071101 21:28:34 
This solutions yields order magnitude of 1 for E=h*c/lambda...doesnt seem to work

Jeremy 20071103 15:03:55 
I thought about it from a spectroscopy perspective as well. Recall the ever useful formula: . In words, the recipe is: take 1240, divide by a wavelength in nanometers, and get an energy in electronvolts. Couple this with the fact that the energy level spacing has infrared energies, and you quickly arrive at the correct answer.
When I first worked this problem, I wasn't too sure what wavelength range infrared corresponded to, but I knew that red is about 600 nm, or ~2 eV. Thus, the energies in answer choices (C), (D), and (E) are far too large. implies a wavelength of roughly , whereas implies a wavelength one million times larger, i.e. . Ergo, I chose (B).

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  jonestr 20051112 00:31:59  a handy value to remember is hc=1240eVnm which can make speed the computation here if you just account for the 4 pi^2
Andresito 20060329 21:37:26 
I also recommend using quantities in eVs.
You could almost approximate every quantity and work only with orders of magnitude as the answers are given in orders of magnitude.

 




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