GR9277 #53


Problem


\prob{53}
A particle of mass m is confined to an infinitely deep squarewell potential:
The normalized eigenfunctions, labeled by the quantum number n, are
A measurement of energy E will always satisfy which of the following relationships?






Quantum Mechanics}Energy
If one forgets the energy of an infinite well, one can quickly derive it from the timeindependent Schrodinger's Equation . However, since inside, one has .
Plug in the groundstate wave function , where . Chunk out the second derivative to get . Plug in k to get .
Note that can be deduced from boundary conditions, i.e., the wave function vanishes at both ends ( and ). The second boundary condition forces the n's to be integers. Since one can't have a trivial wave function, , and thus . One finds that , since , as in choice (E).


Alternate Solutions 
panos85 20071030 11:31:21  The energy depends on n so it cannot be a constant. This eliminates C, D, E. Furthermore it cannot be bounded from above as it increases with n. (You can easily remember that.) So A is out. Choice B remains.  

Comments 
francesco 20170914 20:26:58  if you look at the eigenfunctions (in particular, at the argument of the sine) you can deduce that , plug it in and find choice B   syreen 20130918 13:47:54  The Schrodinger EQ should have hbar^2, not just hbar.
Also, as others have noted, the typed question has a typo. B. should have E>=, NOT E<=   Kabuto Yakushi 20100904 08:58:08  As long as you remember (which it is probably a good idea to memorize for this test) the energy in a potential well to be
E =
with a note that using results in a two not an eight on the bottom, choice (B) is obvious. (C),(D), and (E) are to precise to satisfy the "always" criteria.   Fizzics 20091030 11:44:19  Well I tried to post that last thing as a typo.   Fizzics 20091030 11:43:33  The Answer should be "B" not "E". And the answer shown in the problem as "B" is wrong. Should be E is GREATER THAN or EQUAL TO.
Typos.   jw111 20081104 11:45:30  Since
and where H = h bar
so
with the solution given by question,
thus
  Monk 20081013 21:46:07  (hbar)^2 is missing in the original presentation of the TISE, it just shows (hbar)...though most of you should know that anyhow :P   adenos 20071102 12:49:53  B should read E >= ....
evanb 20080623 19:07:46 
yes.

evanb 20080623 19:11:02 
Dammit, the preview deTYPOified my entry.
Yes, (B) should be >= and not <=
and it should be the answer, not (E)

  panos85 20071030 11:31:21  The energy depends on n so it cannot be a constant. This eliminates C, D, E. Furthermore it cannot be bounded from above as it increases with n. (You can easily remember that.) So A is out. Choice B remains.
hoyas08 20080616 19:52:47 
Unfortunately, choice D has an term in it, so you can't eliminate it by reasoning that the energy must depend on n. However, if you can remember that the energy of a particle in free space (V=0) is E=, then the 8 in the denominator of D does not make sense.

hoyas08 20080616 19:55:05 
Unfortunately, choice D has an term in it, so you can't eliminate it by reasoning that the energy must depend on . However, if you can remember that the energy of a particle in free space (V=0) is E=, then the 8 in the denominator of D doesn't make sense, eliminating that answer.

Imperate 20081016 10:18:38 
This is not true for D, D has E=n^2... which is the correct form.rnrnOne can throw away A becayse there is no upper bound to energy.rnOne can throw away C and E because they [b] are [/b] constants, which is unacceptable for the energy.rnrnThis leaves B and D, and D can only be eliminated by working out the correct coeff, as others have posted.rnrnI think this question is a bit harsh though because from elementary QM infinite square wells always have n^2 dependence so you see D whilst doing a high speed test like the GRE then you are going to pick D, well I did anyway....rnrnI

ramparts 20090806 19:24:55 
How does that eliminate D? D is dependent on n.

ramparts 20090806 19:24:55 
How does that eliminate D? D is dependent on n.

  antithesis 20071005 18:17:17  Aside from typos mentioned, the question itself has a typo, B should have equal or greater than   Mexicana 20071004 17:46:32  It is much easier to derive this by getting the zero point energy via Heisenberg Uncertainty Principle and then square this expression. Now just substitute for where is the width of the square well and then the usual expression for momentum .
Tommy Koulax 20071029 09:53:59 
doesn't that lead to E= which is not an option provided?

  georgi 20070826 20:20:13  as everyone is posting the answer should be choice b, not e as written   welshmj 20070712 15:51:47  Choice B is the correct answer not E, but choice B on this page has a typo. It should be greater than or equal to, not less than or equal to.   JB 20051207 02:32:15  Isn't your answer choice (B)
jcain6 20051208 19:20:24 
It should be choice (B) because the question requires E to ALWAYS satisfy the relationship. Choice (E) only works when the potential is 0 between o and a.

Andresito 20060327 15:43:35 
I also think it is (B).

sirius 20080625 13:05:18 
the potential is 0 from 0 to a. it says so in the problem. this is a particle in an infinite square well, and answer (E) is the only answer that makes sense.

sirius 20080625 13:09:59 
nevermind, i misread which had the >=, i agree it is (B)

  jcain6 20051122 19:06:10  Your missing a (hbar)^2 in your numerator.
yosun 20051123 02:00:51 
jcain6: thanks for the typoalert; it's been corrected.

 




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

