AAMC 3 #10 Voltage question

[Jwinsler7]

The tube contains a heated filament cathode (C), which emits electrons. A power supply (LV) regulates the filament temperature, the electrical current in the tube, and the number of X-rays produced at anode (A). A power supply (HV) regulates electron acceleration.

Q. In order to increase the maximum kinetic energy of electrons colliding with the anode, the scientist made which of the following changes?

A) The voltage of HV was increased.
B) The voltage of HV was decreased.
C) The voltage of LV was increased.
D) The voltage of LV was decreased.

I always thought kinetic energy was directly proportional to temperature. Since LV is the one controlling temperature, I assumed increasing LV voltage would increase the max. kinetic energy of electrons.

But the answer is A. How does increasing the acceleration of electrons increase their maximum kinetic energy?

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[shaboobly]

The tube contains a heated filament cathode (C), which emits electrons. A power supply (LV) regulates the filament temperature, the electrical current in the tube, and the number of X-rays produced at anode (A). A power supply (HV) regulates electron acceleration.

Q. In order to increase the maximum kinetic energy of electrons colliding with the anode, the scientist made which of the following changes?

A) The voltage of HV was increased.
B) The voltage of HV was decreased.
C) The voltage of LV was increased.
D) The voltage of LV was decreased.

I always thought kinetic energy was directly proportional to temperature. Since LV is the one controlling temperature, I assumed increasing LV voltage would increase the max. kinetic energy of electrons.

But the answer is A. How does increasing the acceleration of electrons increase their maximum kinetic energy?

when a particle is accelerating the speed is increasing linearly with time. kinetic energy is directly proportional to the square of the velocity, so if the electron has a higher velocity, it will have a higher kinetic energy.

 

[Jwinsler7]

Yeah, I see the relationship now. But wouldn't choice C work as well? Increasing the filament temperature, hence increasing the KE of electrons?

 

[shaboobly]

Yeah, I see the relationship now. But wouldn't choice C work as well? Increasing the filament temperature, hence increasing the KE of electrons?

maybe, but the directions of each section is to choose the best answer and if the acceleration of the e increases it will have the most profound effect on the kinetic energy. the energy supplied by the filament would have to be in a heat transfer of one of the ways: conduction, convection, or radiation which is not as profound as just increasing the the particles velocity because KE = 1/2 mv^2 = 3/2 kT where k is boltzman's constant. the temperature of the filament is not the same as the temperature of the electrons.

 

[Jwinsler7]

Makes sense. Thanks for your quick reply!

 

[shaboobly]

Makes sense. Thanks for your quick reply!

np dude. im always on here in the mornings.. wake up and review and help people with questions 🙂

btw i remember that question. i had to think about it for like 30 sec then i was like ohhh yea and i got it right 😎

 

[sciencebooks]

when a particle is accelerating the speed is increasing linearly with time. kinetic energy is directly proportional to the square of the velocity, so if the electron has a higher velocity, it will have a higher kinetic energy.

I went with the KE=W=qV equation for this one because the proportionality was there and then, like you said, chose A because it was the most direct effect -- the question asked "electrons colliding" and the root stated that HV "regulates electron acceleration."

 

[shaboobly]

I went with the KE=W=qV equation for this one because the proportionality was there and then, like you said, chose A because it was the most direct effect -- the question asked "electrons colliding" and the root stated that HV "regulates electron acceleration."

how can you use that work energy theorem? it's not just KE, but change in and according to heisenberg you cant know the velocity of an electron so how can you determine the change in kinetic energy to figure the work? seems to me you got kinda lucky with this one.

 

[stevenz]

how can you use that work energy theorem? it's not just KE, but change in and according to heisenberg you cant know the velocity of an electron so how can you determine the change in kinetic energy to figure the work? seems to me you got kinda lucky with this one.

Don't use HUP to guide your understanding of situations like these. Unless stated, passages on the MCAT will almost certainly pertain only to classical mechanics.

For your information, though, to say that a particle does not have well defined velocity is not to say that a particle does not have well defined energy. In QM, energy is not defined via the equation 1/2 mv^2, but by the Hamiltonian operator, of which the kinetic energy portion is proportional to the momentum operator (recall, KE = p^2/2m; this holds for quantum mechanical systems except that p is the momentum operator and not your ordinary classical momentum, mv).

In the case that we are talking about, the electrons are travelling as a beam, which, in QM, is a state of precise velocity (and, by HUP, undefined position). In other words, the wavefunction of the electron is spread over all space (in the dimension the beam is pointed in), but it forms a beam of precise momentum and therefore precise kinetic energy (again, in the direction the beam is pointed in). Thus an electron beam as described has a well defined kinetic energy and, in fact, the situation for energy reduces to the classical case (except for some exotic cases of tunneling, etc.)

Source: I'm a physics major

 

[sciencebooks]

how can you use that work energy theorem? it's not just KE, but change in and according to heisenberg you cant know the velocity of an electron so how can you determine the change in kinetic energy to figure the work? seems to me you got kinda lucky with this one.

So how did you go from

kinetic energy is directly proportional to the square of the velocity, so if the electron has a higher velocity, it will have a higher kinetic energy.

to higher velocity = higher voltage?

 

[shaboobly]

So how did you go from



to higher velocity = higher voltage?

one of the voltages supplies the electrons with acceleration, and velocity is directly proportional to acceleration so increasing the voltage of that battery would cause the electron ro accelerate more and have a higher velocity thus kinetic energy. the voltage is part of the passage, not something you have to know.

 

[shaboobly]

Don't use HUP to guide your understanding of situations like these. Unless stated, passages on the MCAT will almost certainly pertain only to classical mechanics.

For your information, though, to say that a particle does not have well defined velocity is not to say that a particle does not have well defined energy. In QM, energy is not defined via the equation 1/2 mv^2, but by the Hamiltonian operator, of which the kinetic energy portion is proportional to the momentum operator (recall, KE = p^2/2m; this holds for quantum mechanical systems except that p is the momentum operator and not your ordinary classical momentum, mv).

In the case that we are talking about, the electrons are travelling as a beam, which, in QM, is a state of precise velocity (and, by HUP, undefined position). In other words, the wavefunction of the electron is spread over all space (in the dimension the beam is pointed in), but it forms a beam of precise momentum and therefore precise kinetic energy (again, in the direction the beam is pointed in). Thus an electron beam as described has a well defined kinetic energy and, in fact, the situation for energy reduces to the classical case (except for some exotic cases of tunneling, etc.)

Source: I'm a physics major

i'm a chem major, it's irrelevant. what im saying is you dont know the velocity before the electron is accelerating so you cant find the change in kinetic energy to find the work.

source: my brain

 

[sciencebooks]

one of the voltages supplies the electrons with acceleration, and velocity is directly proportional to acceleration so increasing the voltage of that battery would cause the electron ro accelerate more and have a higher velocity thus kinetic energy. the voltage is part of the passage, not something you have to know.

Oh, I see what you're saying. Straightforward.

 

[stevenz]

You CAN use the work energy theorem in this case. Work is defined as the change in kinetic energy and there is certainly a change in kinetic energy of the electron. You cannot calculate it without knowing its initial velocity but you can certainly reason that the work done on an electron is proportional to the potential it is under, and that increasing the potential would therefore increase the velocity of an electron.

Now, I STILL don't see why you would bring up Heisenberg. The electron in this case has a (theoretically) precisely defined velocity that we could measure to (again, theoretically) arbitrary precision. I was just trying to clear up some misconceptions about quantum mechanics. If you think that HUP applies in this situation whatsoever, then it appears that I was not incorrect.

 

[sciencebooks]

You CAN use the work energy theorem in this case. Work is defined as the change in kinetic energy and there is certainly a change in kinetic energy of the electron. You cannot calculate it without knowing its initial velocity but you can certainly reason that the work done on an electron is proportional to the potential it is under, and that increasing the potential would therefore increase the velocity of an electron.

Now, I STILL don't see why you would bring up Heisenberg. The electron in this case has a (theoretically) precisely defined velocity that we could measure to (again, theoretically) arbitrary precision. I was just trying to clear up some misconceptions about quantum mechanics. If you think that HUP applies in this situation whatsoever, then it appears that I was not incorrect.

That's what I was thinking too?

 

[Eastfolding]

A power supply (LV) regulates the filament temperature, the electrical current in the tube, and the number of X-rays produced at anode (A).

The filament temperature, electrical current, and number of X-rays are all related. What the power supply does is increase the temperature of the filament, which increases the resistance. (Recall Resistance is related to Temperature). An increase in resistance lowers the current, which means less X-rays are produced.

None of these things have to do with the speed of the X-rays produced. Sure, it changes how many are produced, but they're still going at the same speed no matter how much you change it.

The HV portion increases the voltage across tube. Recall that Energy = Change in Voltage * Charge. Since the electrons are accelerating solely because of the voltage set up, this energy is also equal to the final kinetic energy of the electron. Thus, you see that the voltage of HV is related to the final kinetic energy.

On a side note, increasing the LV voltage actually decreases the current.