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GR8677 #90
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Quantum Mechanics$\Rightarrow$}Particle in a Box

The particle is in an infinite well (or box, if you will) of length 2a. (It's stuck forever bouncing around between the two walls.)

The number of nodes in the wave function determines the energy level. In this case, there is one load, thus this is $E_2$. The lowest state would be $E_1$

$E_n=k n^2$ eV for particle in a box. Given that $E_2 = 2 = 4 k$, one determines $k=1/2$ eV. Thus, $E_1 = k = 1/2$ eV. The answer is thus (C).

Alternate Solutions
 casseverhart132019-08-20 12:22:40 Helpful problem. Lucky me I found your site by accident. towing serviceReply to this comment BillNyeTheRussianSpy2018-05-22 03:06:49 I think, notationally, it is easier to have $E_{n}=n^2 E_{1}$. But it still comes out the same.\r\n\r\n$E_{2}=2eV=4E_{1}$ implies $E_{1}=1/2eV$Reply to this comment BillNyeTheRussianSpy2018-05-22 03:06:49 I think, notationally, it is easier to have $E_{n}=n^2 E_{1}$. But it still comes out the same.\r\n\r\n$E_{2}=2eV=4E_{1}$ implies $E_{1}=1/2eV$Reply to this comment
casseverhart13
2019-08-20 12:22:40
Helpful problem. Lucky me I found your site by accident. towing service
BillNyeTheRussianSpy
2018-05-22 03:06:49
I think, notationally, it is easier to have $E_{n}=n^2 E_{1}$. But it still comes out the same.\r\n\r\n$E_{2}=2eV=4E_{1}$ implies $E_{1}=1/2eV$
BillNyeTheRussianSpy
2018-05-22 03:06:49
I think, notationally, it is easier to have $E_{n}=n^2 E_{1}$. But it still comes out the same.\r\n\r\n$E_{2}=2eV=4E_{1}$ implies $E_{1}=1/2eV$
Moush
2010-09-19 06:20:35
Yosun, in the 2nd paragraph you have "load" instead of "node". Richard mentioned it but didn't classify it as a typo so you probably didn't see it.

I just noticed that the preview button resets the classification to NEC.

Thanks for this site!
neon37
2008-11-06 14:49:13
I solved this slightly differently,

We know,
$E=\frac{p^2}{2m}$
and De Broglie wavelength,
$\lambda=\frac{h}{p}$
Plug this into the E and we find out,
$E \infty \frac{1}{\lambda^2}$
So wavelength is doubled going from this state to ground state so the Energy has to be a quarter so $\frac{1}{2}$.
 wittensdog2009-07-27 13:10:26 That works in this case, but only because it makes sense to talk about a specific wavelength. The infinite square well is weird in the sense that because there are definite energy states in a region where there is no potential energy, you have E = p^2/2m for a set of definite energies, you end up with states that have a definite magnitude of momentum. So then these definite momentum magnitude states correspond to a precise wavelength. Normally there would be a spread in p, but the square well has this nice feature. I didn't even really think about this until a semester after taking my QM course, when I read it in the appendix of my statmech book. There was actually kind of an interesting derivation of the infinite square well properties, just making reference to the idea of de Broglie waves being fit inside the well. But if you were to try to apply the method to another potential, I imagine it would not be so successful. It makes a lot of sense thinking about the fourier transform. Inside the well you have pure sine waves, and the fourier transform of a pure sine wave with some frequency w just has two delta spikes at +/- w, each normalized with a factor of 1/2 to make the total probability equal to 1. Of course there would be mixing in from other wavelengths if you were to consider the entire range of x values, outside of -a and a, because then you have a wave pulse.
 wittensdog2009-07-27 13:11:30 But I guess my point was, I always thought that was a really interesting property of the square well.
FortranMan
2008-10-30 07:41:56
Try not to confuse this with the harmonic oscillator where the ground state is n=0.
jesford
2008-04-03 17:57:00
Even if you can't remember the infinite-square-well energies, the problem asks for the LOWEST possible energy... so it can't be more than the 2 eV energy given for the diagram. This rules out choices D and E. Also, recall that the particle can't have zero energy, so the only options left are B and C.
 jesford2008-04-03 17:58:41 oops I guess 1 eV is less than 2, eh?
Richard
2007-11-01 12:37:50
You wrote "load," but you should have written "node."

:)

 realcomfy2008-11-01 09:50:32 I also believe that it should state that the particle has $n=2$ nodes thus giving us $E_2 = k \left( 2 \right)^2 = 4k$
 realcomfy2008-11-01 09:51:31 Check here and correct me if I am wrong http://en.wikipedia.org/wiki/Image:Particle_in_a_box_wavefunctions.png
 gt20092009-06-25 17:17:22 2 antinodes, not nodes
 wittensdog2009-10-08 11:37:54 This isn't very scientific, but instead of fretting over nodes vs. anti-nodes, I always just remember that the state number is the number of "bumps."
 IRFAN 2011-08-18 02:25:34 Actually you ppl confuse node i thimk it might be loops
grae313
2007-10-07 19:01:27
Alternately, knowing $E = kn^2$ we can write

$\ 2 = 4k$

$\ X = 1k$

Divide the first equation by the second and solve for x.

You are replying to:
Helpful problem. Lucky me I found your site by accident. towing service

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$