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  GR9277 #53
Problem
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\prob{53}
A particle of mass m is confined to an infinitely deep square-well potential:
<br />
 V(x)&=&\infty,\;\;x\leq0,x\geq a \\<br />
 V(x)&=&0,\;\;0\lt x \lt a<br />

The normalized eigenfunctions, labeled by the quantum number n, are $\psi_n=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}$

A measurement of energy E will always satisfy which of the following relationships?


  1. $E\leq\frac{\pi^2\hbar^2}{8ma^2}$
  2. $E\leq\frac{\pi^2\hbar^2}{2ma^2}$
  3. $E=\frac{\pi^2\hbar^2}{8ma^2}$
  4. $E=\frac{n^2\pi^2\hbar^2}{8ma^2}$
  5. $E=\frac{\pi^2\hbar^2}{2ma^2}$

Quantum Mechanics}Energy

If one forgets the energy of an infinite well, one can quickly derive it from the time-independent Schrodinger's Equation -\frac{\hbar}{2m}\psi^{''}+V\psi = E\psi. However, since V=0 inside, one has -\frac{\hbar}{2m}\psi^{''} = E\psi.

Plug in the ground-state 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).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
panos85
2007-10-30 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.Alternate Solution - Unverified
Comments
francesco
2017-09-14 20:26:58
if you look at the eigenfunctions (in particular, at the argument of the sine) you can deduce that k = \\frac{n\\pi}{a}, plug it in E = \\frac{\\hbar^2 k^2}{2m} and find choice BNEC
syreen
2013-09-18 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<=
NEC
Kabuto Yakushi
2010-09-04 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 = \frac{\pi^2\hbar^2}{2ma^2}

with a note that using \hbar^2 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.
NEC
Fizzics
2009-10-30 11:44:19
Well I tried to post that last thing as a typo. Typo Alert!
Fizzics
2009-10-30 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.
NEC
jw111
2008-11-04 11:45:30
Since E=\frac{p^2}{2m}
and  p=H*k where H = h bar

so E=\frac{H^2k^2}{2m}

with the solution given by question,
k=\frac{n\pi}{a}

thus
E=\frac{n^2\pi^2H^2}{2ma^2}
NEC
Monk
2008-10-13 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 :PNEC
adenos
2007-11-02 12:49:53
B should read E >= ....
evanb
2008-06-23 19:07:46
yes.
evanb
2008-06-23 19:11:02
Dammit, the preview de-TYPO-ified my entry.

Yes, (B) should be >= and not <=

and it should be the answer, not (E)
NEC
panos85
2007-10-30 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
2008-06-16 19:52:47
Unfortunately, choice D has an n^2 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=\frac{\vec{p}^2}{2m}, then the 8 in the denominator of D does not make sense.
hoyas08
2008-06-16 19:55:05
Unfortunately, choice D has an n^2 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=\frac{\vec{p}^2}{2m}, then the 8 in the denominator of D doesn't make sense, eliminating that answer.
Imperate
2008-10-16 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
2009-08-06 19:24:55
How does that eliminate D? D is dependent on n.
ramparts
2009-08-06 19:24:55
How does that eliminate D? D is dependent on n.
Alternate Solution - Unverified
antithesis
2007-10-05 18:17:17
Aside from typos mentioned, the question itself has a typo, B should have equal or greater thanNEC
Mexicana
2007-10-04 17:46:32
It is much easier to derive this by getting the zero point energy via Heisenberg Uncertainty Principle \delta x \delta p = h/2\pi and then square this expression. Now just substitute for \delta x= a where a is the width of the square well and then the usual expression for momentum (\delta p)^2 = 2m E .
Tommy Koulax
2007-10-29 09:53:59
doesn't that lead to E=\frac{(h bar)^2}{2*m*a^2} which is not an option provided?
NEC
georgi
2007-08-26 20:20:13
as everyone is posting the answer should be choice b, not e as writtenNEC
welshmj
2007-07-12 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. NEC
JB
2005-12-07 02:32:15
Isn't your answer choice (B)
jcain6
2005-12-08 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
2006-03-27 15:43:35
I also think it is (B).
sirius
2008-06-25 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
2008-06-25 13:09:59
nevermind, i misread which had the >=, i agree it is (B)
Typo Alert!
jcain6
2005-11-22 19:06:10
Your missing a (hbar)^2 in your numerator.
yosun
2005-11-23 02:00:51
jcain6: thanks for the typo-alert; it's been corrected.
Fixed Typos!

Post A Comment!
You are replying to:
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 = \frac{\pi^2\hbar^2}{2ma^2} with a note that using \hbar^2 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.

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