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Particle Physics...

Hello...

First:

don't you think we should have a "Post your physics questions here" thread?

Ok...I'm doing some questions on particle physics and accelerators and I just want to clear a few things up:

1)One questions reads:
A proton crossing the "D's" of a cyclotron faces a negative potential of 30kV at the D opposite. How many revolutions must the proton make to reach a kinetic energy of 25 Mev?

What is the best way to go about this quesiton? I've worked out the energy, and velocity at a KE of 25 MeV...but I'm not sure how to translate these figures into times and revolutions...

2) In another question I have to work out how many revolutions a proton makes before leaving the cyclotron. I have figures for the radius, magnetic field strength, acceleration voltage, frequency and KE.

How do I use the accelerating voltage figure, 35 kV, to help me with this question and what is the best way to go about it. At the moment, I've simply used the v^2 = u^2 +2as equation but I don't think this is right...

Thank you very much for your time!
I do think i Physics question thread would be great... Unfortunately the physics forum is the most unhelpful sub forum, so i doubt it'd have much success.
Reply 2
LearningMath
I do think i Physics question thread would be great... Unfortunately the physics forum is the most unhelpful sub forum, so i doubt it'd have much success.



:frown:
Reply 3
I haven't done any of this stuff yet (so have little-to-no idea how a cyclotron works), so take my advice with at least one pinch of salt.

1) I would find out how much kinetic energy is transferred to the particle with each crossing of the potential gap (in eV), and find how many times it needs to do that to get 25 MeV. I don't think velocity will come into it, since it's going to gain more velocity with earlier crossings than it does with later crossings (since the longer it's been crossing the faster it'll be going).

2) I'll think about this and get back to you.
Reply 4
Adje
I haven't done any of this stuff yet, so take my advice with at least one pinch of salt.

1) I would find out how much kinetic energy is transferred to the particle with each crossing of the potential gap (in eV), and find how many times it needs to do that to get 25 MeV. I don't think velocity will come into it, since it's going to gain more velocity with earlier crossings than it does with later crossings (since the longer it's been crossing the faster it'll be going).

2) I'll think about this and get back to you.

Thanks Adje for this.

I'll go try that now...
Reply 5
2) Hmm. How about finding the instantaneous linear velocity required for the particle to be travelling in a circle with whatever radius it is you require when it is subject to a particular centripetal force (does this have something to do with the magnetic field strength?), and then calculating the associated kinetic energy for that velocity, which leaves you with the same stuff to work out as in the other question?
Reply 6
Adje
2) Hmm. How about finding the instantaneous linear velocity required for the particle to be travelling in a circle with whatever radius it is you require when it is subject to a particular centripetal force (does this have something to do with the magnetic field strength?), and then calculating the associated kinetic energy for that velocity, which leaves you with the same stuff to work out as in the other question?

I'll try this too...

...using your advice for the other question I get 5.21 x10^21 revolutions. Does that seem feasible?
Reply 7
Graphix
Hello...

First:

don't you think we should have a "Post your physics questions here" thread?

Ok...I'm doing some questions on particle physics and accelerators and I just want to clear a few things up:

1)One questions reads:
A proton crossing the "D's" of a cyclotron faces a negative potential of 30kV at the D opposite. How many revolutions must the proton make to reach a kinetic energy of 25 Mev?

What is the best way to go about this quesiton? I've worked out the energy, and velocity at a KE of 25 MeV...but I'm not sure how to translate these figures into times and revolutions...

2) In another question I have to work out how many revolutions a proton makes before leaving the cyclotron. I have figures for the radius, magnetic field strength, acceleration voltage, frequency and KE.

How do I use the accelerating voltage figure, 35 kV, to help me with this question and what is the best way to go about it. At the moment, I've simply used the v^2 = u^2 +2as equation but I don't think this is right...

Thank you very much for your time!


The whole forum is about answering questions!


The way to appraoch these questionsis to think about how much energy the prticle gains in each lap.

In a cyclotron, we flip the PD so that every timeit crosses a gap it gains energy. If it moves through a PD of 1V, it gains an evergy 1eV.

It crosses the gap twice per cycle so gains 2x30keV each time it goes round.

That should help.
Reply 8
teachercol
The whole forum is about answering questions!


The way to appraoch these questionsis to think about how much energy the prticle gains in each lap.

In a cyclotron, we flip the PD so that every timeit crosses a gap it gains energy. If it moves through a PD of 1V, it gains an evergy 1eV.

It crosses the gap twice per cycle so gains 2x30keV each time it goes round.

That should help.

I know...but in case people only have one question (which I don't at this time) it's easier than starting a thread...

...anyway...

so the electron only gains 1eV per volt? Isn't then energy = Q x V? Or am I missing something?

Thanks for your help!
Reply 9
No - its that easy.
Graphix
I know...but in case people only have one question (which I don't at this time) it's easier than starting a thread...

...anyway...

so the electron only gains 1eV per volt? Isn't then energy = Q x V? Or am I missing something?

Thanks for your help!

Yes, that's how you define a voltage. The energy per unit charge.
Reply 11
teachercol
No - its that easy.

Okay...so if an electron passes through say 35kV, then it will gain 35keV of energy???

Thanks very much...and yes, you've just made the whole thing very simple!!
Graphix
Okay...so if an electron passes through say 35kV, then it will gain 35keV of energy???

Thanks very much...and yes, you've just made the whole thing very simple!!

yes, but it gains that each time it passes through, because of the fact that the voltage reverses.
Reply 13
F1 fanatic
yes, but it gains that each time it passes through, because of the fact that the voltage reverses.

Oui...je sais...

I was just checking.

I've worked them out now and it's quite annoying that I was making them way harder than they needed to be...

...well at least I know how to handle them next time :wink:
Graphix
Oui...je sais...

I was just checking.

I've worked them out now and it's quite annoying that I was making them way harder than they needed to be...

...well at least I know how to handle them next time :wink:

you'll often find that dealing with things from the perspective of energy is a lot simpler than dealing with them through equations of motion. It is partly why energy is such a useful concept.
Reply 15
F1 fanatic
you'll often find that dealing with things from the perspective of energy is a lot simpler than dealing with them through equations of motion. It is partly why energy is such a useful concept.

Yes, I'll keep this in mind...
Reply 16
F1 fanatic
you'll often find that dealing with things from the perspective of energy is a lot simpler than dealing with them through equations of motion. It is partly why energy is such a useful concept.

And thank you for your help F1 fanatic!

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