GREPhysics.NET: GR8677 #90

GR8677 #90

Problem

GREPhysics.NET Official Solution

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

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Comments

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 $\frac{1}{2}$ .

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FortranMan
2008-10-30 07:41:56

Try not to confuse this with the harmonic oscillator where the ground state is n=0.

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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.

jesford
2008-04-03 17:58:41

oops I guess 1 eV is less than 2, eh?

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Richard
2007-11-01 12:37:50

You wrote "load," but you should have written "node."

:)

realcomfy
2008-11-01 09:50:32

I also believe that it should state that the particle has $n=2$ nodes thus giving us $E_2 = k \left( 2 \right)^2 = 4k$

realcomfy
2008-11-01 09:51:31

Check here and correct me if I am wrong

http://en.wikipedia.org/wiki/Image:Particle_in_a_box_wavefunctions.png

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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.

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You wrote "load," but you should have written "node." :)

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