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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.
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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