GREPhysics.NET: GR0177 #46

GR0177 #46

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Quantum Mechanics $\Rightarrow$ }de Broglie Wavelength

Since $E=p^2/(2m)$ and the de Broglie wavelength gives $\lambda=h/p$ , one can find the initial wavelength $\lambda=h/sqrt{2mE}$ . This yields an expression for $h=\lambda \sqrt{2mE}$

When the particle enters the well, its energy becomes $E^{'}=E-V$ , and thus $\lambda^{'}=h/sqrt{2m(E-V)}$ .

Plugging h in, one has $\lambda^{'}=\lambda \sqrt{2mE}/sqrt{2m(E-V)}$ , which is choice (E).

(This above solution is due to the lout Marko.)

Also, the current author's original approach leads to the right result, but it hand-waves the massiveness of the particle:

The initial kinetic energy of the free particle is its total energy $E=h v/\lambda \Rightarrow \lambda =h v/E$ . But, since one's dealing with de Broglie, one has $p=h/\lambda=mv$ . Thus, $\lambda^2 = h^2/(mE)$ .

In the potential, its energy becomes $E-V$ . Replace E above with that to get $\lambda^{'2}=h^2/(m(E-V))$ . But, from the same relation above, one has $h^2 = mE \lambda^2$ . Thus, $\lambda^{'2} = \lambda^2 E/(E-V) = \lambda^2 (1-V/E)^-1$ . This is choice (E).

Alternate Solutions

BerkeleyEric 2010-07-17 17:13:11	I think the quickest way to do this one is to look at $\lambda = \frac{h}{p}$ and $E=\frac{p^2}{2m}$ and realize that there must be some sort of power of 2 or 1/2. This eliminates all but (D) and (E). Then (if there is enough time), one can consider that the wavelength should increase so only (E) works. Reply to this comment
Jeremy 2007-09-20 17:40:02	The great thing about the Physics GRE is that eliminating the 4 wrong answers is just as good as picking the 1 right answer ;) This means you may be able to pick the right answer without solving (or even knowing how to solve) the problem. When recognized right away, this is often the fastest method of solution... Here's how to eliminate the four wrong answers on this problem: Think of extreme, special, or limiting cases, like when V > E. This is corresponds to a classically forbidden region, i.e. a region of exponential decay for the wave function. Therefore, $\lambda^'$ cannot be real, IT MUST BE COMPLEX. This eliminates everything but (E). These kinds of elimination arguments can often seem flimsy, but this one comes straight from the equations... $\frac{-\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}+V\psi=E\psi$ becomes $\frac{\partial^2\psi}{\partial x^2}+{k^'}^2\psi=0$ , with $k^'=\pm \sqrt{2m(E-V)}/\hbar$ . The solution has the form of a complex exponential ( $\psi \prop e^{ik^{'}x}$ ), so we see that $k^'$ is actually the wave vector ( $k^'=2\pi/\lambda^'$ ). V > E means $k^'$ , and therefore $\lambda^'$ , is imaginary. Of course, you don't really need to work with the differential equation, remembering the "classically forbidden region" part would be good enough for solving the problem. Reply to this comment

Comments

BerkeleyEric
2010-07-17 17:13:11

I think the quickest way to do this one is to look at $\lambda = \frac{h}{p}$ and $E=\frac{p^2}{2m}$ and realize that there must be some sort of power of 2 or 1/2. This eliminates all but (D) and (E).

Then (if there is enough time), one can consider that the wavelength should increase so only (E) works.

Reply to this comment

tau1777
2008-11-03 18:51:04

i had been having a tough time with this problem and i finally just got, i used that the initial one is $\lambda= h/\sqrt(2mE)$ . then going into a potential the energy increases to E+V. Then the new lambda is $\lambda= h/\sqrt(2mE+V)$ . Then I just thought of what I would have to multiply the first equation with to get the second one with the V.
After finally getting it I see that (E) is the only choice that lists $\lambda4 to the -1/2 power. this should have given the problem away at the very beginning i can't believe i never realized this before.
i guess the only issue with my solution is that I don't get that minus sign in (E) and i guess that's because the problem wants you to assume that the particle fell into some sort of potential well, instead of lets say a repulsive potential. vagueness of the ETS at its best.
hope this helps someone.

sonnb
2008-11-08 19:58:35

The free particle has an inital total energy of E, which is equal to its kinetic energy. When it enters the region of potential energy V, its total energy remains the same (by conservation of energy) so that its new kinetic energy is E-V.

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Poop Loops
2008-11-01 16:47:16

So how come I can't use $E^2 = (pc)^2 + (mc^2)^2$ and then ignore $mc^2$ ?

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gre..
2007-10-24 06:06:45

err.. E = $\ p^2/2m + V$ and $\ p = h$ $/\lambda$ (from $\lambda = h/p$ ) p -> E then $\lambda$ = $\ h/sqrt{2m(E-V)}$ now solve the prob, when V=0 $\lambda$ = $\ h/sqrt{2mE}$ and for V : not zero the new wavelength $\lambda'$ is $\lambda'$ = $\ h/sqrt{2m(E-V)}$ = ( $\ h/sqrt{2mE}$ )/( $\sqrt{1-V/E}$ ) = $\lambda/sqrt{1-V/E}$ (E).

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gre..
2007-10-24 06:02:40

err.. rnE = $\ p^2/2m + V$ and $\ p = h$ $/\lambda$ (from $\lambda = h/p$ )rnrnp -> Ernrnthenrn $\lambda$ = $\ h/sqrt{2m(E-V)}$ rnrnrnrnnow solve the prob,rnwhen V=0rn $\lambda$ = $\ h/sqrt{2mE}$ rnrnand for V : not zerornthe new wavelength $\lambda'$ isrn $\lambda'$ = $\ h/sqrt{2m(E-V)}$ = ( $\ h/sqrt{2mE}$ )/( $\sqrt{1-V/E}$ )rn = $\lambda/sqrt{1-V/E}$ rnrn(E).

Reply to this comment

Jeremy
2007-09-20 17:40:02

The great thing about the Physics GRE is that eliminating the 4 wrong answers is just as good as picking the 1 right answer ;) This means you may be able to pick the right answer without solving (or even knowing how to solve) the problem. When recognized right away, this is often the fastest method of solution... Here's how to eliminate the four wrong answers on this problem:

Think of extreme, special, or limiting cases, like when V > E. This is corresponds to a classically forbidden region, i.e. a region of exponential decay for the wave function. Therefore, $\lambda^'$ cannot be real, IT MUST BE COMPLEX. This eliminates everything but (E).

These kinds of elimination arguments can often seem flimsy, but this one comes straight from the equations...

$\frac{-\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2}+V\psi=E\psi$

becomes

$\frac{\partial^2\psi}{\partial x^2}+{k^'}^2\psi=0$ ,

with

$k^'=\pm \sqrt{2m(E-V)}/\hbar$ .

The solution has the form of a complex exponential ( $\psi \prop e^{ik^{'}x}$ ), so we see that $k^'$ is actually the wave vector ( $k^'=2\pi/\lambda^'$ ). V > E means $k^'$ , and therefore $\lambda^'$ , is imaginary. Of course, you don't really need to work with the differential equation, remembering the "classically forbidden region" part would be good enough for solving the problem.

Reply to this comment

Jeremy
2007-09-20 16:14:51

To clarify: The snake in the grass with $E=h v/\lambda$ is that here $v$ represents the phase velocity (wave velocity), which is not always the same as the group velocity (particle velocity). However, this is what is being assumed in the originally posted solution. In equations:

$E=h v_{phase}/\lambda$ and $p=m v_{group}$ ,

so the original solution implicitly - and erroneously - uses $v_{phase}=v_{group}$ . In general, the two are not equal because

$v_{phase} = \omega/k$ ,

whereas

$v_{group} = d\omega/dk$ .

For a plane wave solution in free space, the dispersion relation is: $\omega=\hbar k^2/(2m)$ , so in fact, $v_{phase}=\frac{1}{2}v_{group}$ . This explains the difference of a factor of 2 in the two solutions' equation for $\lambda$ ; one (correctly) has $h=\lambda \sqrt{2mE}$ , while the original version has $\lambda^2=h^2/(mE)$ . The error of equating the two velocities happens twice (once for $\lambda$ , and once for $\lambda^'$ ) and cancels in division in the algebra to solve the problem. This is why the first approach still gets the right answer.

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joshuacc1
2005-12-03 15:01:42

marko,

The question said a free particle with initial kinetic energy E. A free particle doesn't necessarily need to be a photon. Besides, a de broglie wave is calculated from a particle with a mass.

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Marko
2005-11-11 13:43:35

This derivation here gives the correct result (E) but is fundumentally flawed. E is not hv/\lambda. That for a photon with no mass. In this problem E=p^{2}/2m as always. Indeed \lambda=h/p and it follows from these two CORRECT statements that indeed the answer is E. Who ever did it the other was nothing more but a lucky son-of-a-bitch.

yosun
2005-11-11 13:58:27

-.- daughter of a bitch. btw marko, the bitch suggests that you put dollar-sign wrappers around your equations so that you get $p^2/2m$ instead of p^2/2m. your solution has been added above.

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