De Broglie Wavelength

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There was a physics passage in TBR that explained that an increase in the momentum of an electron = an increase in velocity, which would decrease the electron wavelength.

I thought that velocity = lambda * frequency, so I thought an increase in velocity would cause an increase in wavelength (direct relationship). Does this not apply in the case of an electron/quantum mechanics for some reason?

I did an internet search and found this relationship; lamda = h/mv, where velocity and wavelength are inversely proportional.

Could someone tell me why the first equation I listed does not work when considering wave particle duality(what this passage was talking about)?




EDIT:
Second part of this question. I don't know if anyone has the physics II TBR book, but I'm talking about the first passage on the De Broglie Wavelength on the first practice exam in the back of the book.

In the passage explanations, they say a couple of things that confuse me.

"By reducing the net resistance of the circuit, the current is increased. This means the potential drop across the 20 ohm resistor would go up."

Also, "Increasing the resistance of the circuit causes the current in the circuit to decrease, and this in turn causes the potential difference that the proton experiences to go down."

I thought that potential difference would remain constant with a change in both resistance and current, due to V = IR.

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So this is just completely different than EM radiation and the classic v=lamda*f relationship?

It's not completely different but it's also not completely the same. The differences are much beyond the scope of the MCAT - you should read the linked article if you are interested in the physical differences. But for the purposes of the MCAT, whenever de Broglie is invoked, think lambda = h/p.
 
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It's not completely different but it's also not completely the same. The differences are much beyond the scope of the MCAT - you should read the linked article if you are interested in the physical differences. But for the purposes of the MCAT, whenever de Broglie is invoked, think lambda = h/p.


I guess my poor performance on this passage was due to a failure to read closely enough. They had the de Broglie relationship in the passage, but I just used the EM radiation relationship for the questions
 
The de Broglie wavelength is a matter wave. The fundamental equation is lambda = h/mv, which gives you the relationship between velocity and wavelength. Some more advanced reading: de Broglie relation | Reading Feynman.
So this is just completely different than EM radiation and the classic v=lamda*f relationship?

Thanks for the quick response by the way
It's not completely different but it's also not completely the same. The differences are much beyond the scope of the MCAT - you should read the linked article if you are interested in the physical differences. But for the purposes of the MCAT, whenever de Broglie is invoked, think lambda = h/p.

Just to clarify, the relationship of v = λf applies to all waves, including matter waves. Same with the relationship of E = hf. The de Broglie wavelength formula of λ = h/(mv) makes it seem that λ and v are inversely related, but you can see the apparent relationship by combining the formulas together.

E = hf --> f = E/h;
v = λf --> v = λ*(E/h) --> λ = (hv)/E;

λ = h/(mv) -->
(hv/E) = (h/mv) -->
(v/E) = (1/mv) -->
(E/v) = mv -->
E = mv^2

The equation E = mv^2 is essentially the mass-energy equivalence for v approaching the speed of light, which applies for all waves. And so the calculations simply reveal that de Broglie matter waves are a type of wave that follows the usual principles.
 
Just to clarify, the relationship of v = λf applies to all waves, including matter waves. Same with the relationship of E = hf. The de Broglie wavelength formula of λ = h/(mv) makes it seem that λ and v are inversely related, but you can see the apparent relationship by combining the formulas together.

E = hf --> f = E/h;
v = λf --> v = λ*(E/h) --> λ = (hv)/E;

λ = h/(mv) -->
(hv/E) = (h/mv) -->
(v/E) = (1/mv) -->
(E/v) = mv -->
E = mv^2

The equation E = mv^2 is essentially the mass-energy equivalence for v approaching the speed of light, which applies for all waves. And so the calculations simply reveal that de Broglie matter waves are a type of wave that follows the usual principles.

The lambda equation you've derived is problematic because velocity factors not only into the explicit "v" term in the numerator but also in the "E" term in the denominator, as things that are moving are said to have energy. So you cannot easily parse apart the numerator vs. denominator velocity term contributions to lambda. Furthermore, increasing velocity does in fact decrease an object's wavelength. By the de Broglie relation, lambda = h/mv. h is a constant, m is essentially constant for non-relativistic speeds, and v is therefore the only variable.

Your calculations were in the introduction to the link I posted but from your derivation, you will see that the expression you get (E = m*v^2) is not easily reconcilable with what we see classically. Sure, E = m*v^2 for a photon, where v = c. But where else have you seen anything like that in classical mechanics? It might bring to mind the kinetic energy equation, but it's missing a factor of 1/2. The explanation for all this lies deep within quantum mechanics, which I don't feel qualified enough to explain in detail, but the article I posted does a good job.
 
The lambda equation you've derived is problematic because velocity factors not only into the explicit "v" term in the numerator but also in the "E" term in the denominator, as things that are moving are said to have energy. So you cannot easily parse apart the numerator vs. denominator velocity term contributions to lambda. Furthermore, increasing velocity does in fact decrease an object's wavelength. By the de Broglie relation, lambda = h/mv. h is a constant, m is essentially constant for non-relativistic speeds, and v is therefore the only variable.

Your calculations were in the introduction to the link I posted but from your derivation, you will see that the expression you get (E = m*v^2) is not easily reconcilable with what we see classically. Sure, E = m*v^2 for a photon, where v = c. But where else have you seen anything like that in classical mechanics? It might bring to mind the kinetic energy equation, but it's missing a factor of 1/2. The explanation for all this lies deep within quantum mechanics, which I don't feel qualified enough to explain in detail, but the article I posted does a good job.

Yeah I would refer to the link you provided for sure. I just think matter waves are a type of wave that follows all wave principles, so the equations like E = hf would still apply. The E = mv^2 could in a way represent the total energy of a wave but the details are likely out of scope.
 
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