GR8677 #47


Problem



Quantum Mechanics}FranckHertz Experiment
Like the typical experiment in QM, the FrankHertz experiment has to do with shooting a bunch of particles through some oven. Its significance is in early observations of energy levels. Also, its key conclusion is that electrons are scattered elastically.
But, if one doesn't know the above, one can always use MOEthe Method of Elimination.
(A) Elastic collision requires conservation of kinetic energy. A bit restraining, but keep the choice.
(B) Never scattered elastically? Never is too strong a word to be favored by ETS.
(C) Seems reasonableish. Keep it.
(D) This is, again, too restraining. It makes sense that depending on the impact angle and momentum, different amount of energy would be lost.
(E) Discrete energy lost is mentioned in (C), but again, this choice is much too restraining, stating that ``there is no energy range..."too strong of a phrase to be favored by ETS.


Alternate Solutions 
There are no Alternate Solutions for this problem. Be the first to post one! 

Comments 
Almno10 20101111 21:26:36  The reason E is wrong is this. The experiment did not probe all energy ranges in existence. Even the LHC could not prove that "there does not exist an energy range such that 'x y and z' happens."   blue_down_quark 20080827 00:40:28  I don't understand why E is wrong. In FrankHertz experiment an electron either loses energy discretely or it doesn't . So simply because D is stated too strongly it cannot be omitted. Sometimes a strongly stated phrase can be the correct answer. So why can we disregard E so easily?
blue_down_quark 20080827 00:42:19 
I meant E in the third sentence, not D

Poop Loops 20080928 16:40:33 
It's all about scale. For example if you are operating on the eV scale, and an electron can only gain or lose energy in 1eV steps, that's very obviously discrete.
If, however, you are working on a 10,000,000 eV scale and an electron still only loses or gains 1eV, you can't tell. So it's a handwavy approximation, but apparently ETS thinks that's good enough.
Basically the higher you get on your energy scale, the more continuous energy levels of an electron seem.

Jeremy P 20081023 20:33:55 
rule out E:
The question concerns the results of the FranckHertz experiment. While one might observe discrete energy loss for some range, it would be impossible to prove by experiment that the loss is discrete for every range.
C and E seem to say the same thing about discrete vs continuous, but rule out E for it is too absolute (as Yosun suggests).
ETS might want to test that you know that scattering can be inelastic and elastic and so E seems redundant.

neon37 20081102 23:41:29 
Well the important thing to remember for this problem is exactly what the FrankHertz experiment proved. As madfish states it, it confirmed the discrete nature of the energy levels, which is what choice C says. The experiment did not have to do with the range of energy as E states (although I'm not quite sure what E means exactly). I guess it helps when you have done the experiment :P.

neon37 20081103 00:37:02 
Well the important thing to remember for this problem is exactly what the FrankHertz experiment proved. As madfish states it, it confirmed the discrete nature of the energy levels, which is what choice C says. The experiment did not have to do with the range of energy as E states (although I'm not quite sure what E means exactly). I guess it helps when you have done the experiment :P.

Almno10 20101111 21:25:29 
The reason E is wrong is this. The experiment did not probe all energy ranges in existence. Even the LHC could not prove that "there does not exist an energy range such that 'x y and z' happens."

resinoth 20150917 21:06:00 
Electrons can lose energy in ways besides scattering. Acceleration of a free electron can produce bremsstrahlung, which is continuous.\r\n\r\nThis is why E is false.

aprilrussell 20180530 03:51:45 
run 3

  madfish 20071102 13:45:34  The Hertz experiment confirmed the discrete energy levels predicted by quantum theory. (think discrete emission of spectral lines of light, what unit can the frequency of such light be measured in? ah yes...Hertz)
physicsisgod 20081105 20:23:37 
Just to be obnoxious, the Hertz of frequency fame and the Hertz of the FranckHertz experiment are not the same person. The SI unit of frequency is named after Heinrich Hertz, who did lots of work with the properties of electromagnetic waves. Gustav Ludwig Hertz, who was Heinrich's nephew, decided it would be more fun to shoot electrons into a gas, for which he won the Nobel Prize.
As a physicist, one should be wary of specious correlations....

flyboy621 20101114 20:47:49 
(physicsisgod) +1

justin_l 20121108 11:16:34 
it's a mnemonic...

  jax 20051202 11:42:24  I was thinking after reading this solution that A was correct even though I chose C. Sometimes it isn't clear which answer is correct (that's when I turn to the answer key at the back of the exam). If somebody could clarify why C is better than A it would be much appreciated!
(just posting to get this question back on the list of recent postings so maybe somebody will notice it)
timsaucer 20061030 16:53:06 
Electrons undergo elastic and inelastic collisions during this experiment. When the energy of the incident electrons is cannot excite the Mercury atoms (due to their discrete energy levels), the electrons undergo elastic collisions. When the electrons have an energy equal to the difference in two states of the Mercury atoms, they undergo inelastic collisions. This eliminates choices A & B. Lets say an electron excites an electron from the n=2 state to the n=3 state. The difference in energy lost by the electron during that inelastic collision is different from if it excites an electron from the n=1 to the n=3 state. You can refer to the Bohr model of the atom to get a rough (but sufficient for this problem) idea of the energy differences. This affirms C is correct and that D cannot be correct. E is not correct, because once you have reached energies on the order of the ionization energy, the energy levels "blur" together to give you a continuous (close enough) energy loss.

sawtooth 20071102 09:04:21 
I agree. This is a quite complete answer. The whole point is to observe the results of the inelastic collisions in the electron current. If the electrons gain enough energy to excite the gas, they lose this kinetic energy (inelastic collision) and a suitable negative potential prevents them reaching the anode and contributing to the current. If the accelarating potential is strong enough, the electrons can excite the gas in multiple instances. Quite a cool experiment, you can see the gas glowing in discreet areas. So. it can;t be A, it is Definitely C.

  mendelrat 20051110 21:33:24  Why is C the more correct answer than A? Your last comment in the introduction makes this seem confusing.  

Post A Comment! 

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .

type this... 
to get... 
$\int_0^\infty$ 

$\partial$ 

$\Rightarrow$ 

$\ddot{x},\dot{x}$ 

$\sqrt{z}$ 

$\langle my \rangle$ 

$\left( abacadabra \right)_{me}$ 

$\vec{E}$ 

$\frac{a}{b}$ 





The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... 

