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Proton Energy Question

heyy can someone please answer me..pls plsss

protones are accelerated to an energy of 10Gev and the beam is made to collide head on with a beam of anti protons, aslo with enery of 10Gev. how much energy is avaiabe from each collision to produce new particles
take the photon mass as 940Mev/c2


someone please explain me how to do this

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Original post by iwantopas19
heyy can someone please answer me..pls plsss

protones are accelerated to an energy of 10Gev and the beam is made to collide head on with a beam of anti protons, aslo with enery of 10Gev. how much energy is avaiabe from each collision to produce new particles
take the photon mass as 940Mev/c2


someone please explain me how to do this


You're saying that each single proton is accelerated to a K.E. of 10Gev, right? And also that the energy equivalence of the mass is 940Mev (because mass can't be expressed in energy units!)

If so... the anti-matter and matter will always annihilate to produce 'pure' energy/photons. Therefore, it's the total energy of the two protons that will be the total energy of the new photons/'particles' (which is the sum of their KE and mass energy) = 10Gev + 10Gev +940 Mev +940 Mev...
Reply 2
the correct answer is 6.5Mev..
I have no idea how..
I just dont understand particle physics...not at all :/
Reply 3
Original post by iwantopas19
the correct answer is 6.5Mev..
I have no idea how..
I just dont understand particle physics...not at all :/


6.5MeV is definitely not correct... (esoecially with particles accelerated in the GeV range...)

Is 10GeV the kinetic or total energy of each beam?

If it is the total energy then we have 20GeV with which to make new particles as we are sitting in the centre of mass frame so we can make particles with no kinetic energy.

If it is the kinetic energy then we have 20GeV+2(940)MeV~22GeV with which to make new particles....
Reply 4
Original post by natninja
6.5MeV is definitely not correct... (esoecially with particles accelerated in the GeV range...)

Is 10GeV the kinetic or total energy of each beam?

If it is the total energy then we have 20GeV with which to make new particles as we are sitting in the centre of mass frame so we can make particles with no kinetic energy.

If it is the kinetic energy then we have 20GeV+2(940)MeV~22GeV with which to make new particles....



Its the kinetic energy of the beam.
but look at the question, it says ''head on collision'' which means all the energy should be needed to produce new particles right...
I have no idea. this question confused everything i know. its a question from revision guide. :/ and thats the answer they gave. 6.5Mev


I
Reply 5
Original post by iwantopas19
Its the kinetic energy of the beam.
but look at the question, it says ''head on collision'' which means all the energy should be needed to produce new particles right...
I have no idea. this question confused everything i know. its a question from revision guide. :/ and thats the answer they gave. 6.5Mev


I


Head on collision means that the centre of mass will be stationary and so all the energy that the particles had can be used to create new particles. i.e. the new particles are allowed a kinetic energy of zero and only have rest energy.

Ok I'll work through the question. (probably at a higher level than you require...)

So the total energy of each particle is Ke plus rest energy ~11GeV

so we get our momentum 4-vector for each particle (E,p) and (E,-p)

adding and squaring our 4-momenta obtains

(p1+p2)2=p1.p1+p2.p2+2p1.p2 and the square of a 4 vector is simply it's mass...

so Ef2=2m2+2(Ei2+p2) but m2+p2=E2

so Ef=2Ei=22GeV to make new particles

The answer in the revision guide is wrong - have confidence in what you know :smile:
(edited 10 years ago)
Reply 6
Original post by natninja
Head on collision means that the centre of mass will be stationary and so all the energy that the particles had can be used to create new particles. i.e. the new particles are allowed a kinetic energy of zero and only have rest energy.

Ok I'll work through the question. (probably at a higher level than you require...)

So the total energy of each particle is Ke plus rest energy ~11GeV

so we get our momentum 4-vector for each particle (E,p) and (E,-p)

adding and squaring our 4-momenta obtains

(p1+p2)2=p1.p1+p2.p2+2p1.p2 and the square of a 4 vector is simply it's mass...

so Ef2=2m2+2(Ei2+p2) but m2+p2=E2

so Ef=2Ei=22GeV to make new particles

The answer in the revision guide is wrong - have confidence in what you know :smile:




wow ::smile::smile::smile::smile::smile::smile::smile::smile::smile::smile::smile::smile: u seem really smart.

''So the total energy of each particle is Ke plus rest energy ~11GeV''

but can u please tell me how u found out the rest energy as 1? are u just assuming here?
Reply 7
Original post by iwantopas19
wow ::smile::smile::smile::smile::smile::smile::smile::smile::smile::smile::smile::smile: u seem really smart.

''So the total energy of each particle is Ke plus rest energy ~11GeV''

but can u please tell me how u found out the rest energy as 1? are u just assuming here?


rest energy for a proton is round about 940MeV (which was given in your question) which I rounded to a GeV (it's close enough - why I had wiggles instead of equals :tongue: ) you can work this out from converting it's mass to energy via E=mc2 for a particle at rest n SI and then converting to natural units by dividing by the charge on an electron.

I'll spoiler some nice factoids for you :tongue:

Spoiler

Reply 8
Original post by natninja
rest energy for a proton is round about 940MeV (which was given in your question) which I rounded to a GeV (it's close enough - why I had wiggles instead of equals :tongue: ) you can work this out from converting it's mass to energy via E=mc2 for a particle at rest n SI and then converting to natural units by dividing by the charge on an electron.

I'll spoiler some nice factoids for you :tongue:

Spoiler





:eek: thank you sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo muchhhhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

how do u know so well :redface: god bless u! :biggrin: :biggrin: :biggrin:
issit ok if i ask some other doubts tooo? if u dont mind?
Reply 9
Original post by iwantopas19
:eek: thank you sooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo muchhhhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

how do u know so well :redface: god bless u! :biggrin: :biggrin: :biggrin:
issit ok if i ask some other doubts tooo? if u dont mind?


Well I'm doing Physics at uni so I kinda have to know this stuff :tongue:

sure, ask away :smile:
Reply 10
hehe thats nice
Reply 11
Original post by natninja
Well I'm doing Physics at uni so I kinda have to know this stuff :tongue:

sure, ask away :smile:




can u please explain this to me?
Reply 12
Original post by iwantopas19
can u please explain this to me?


Ok so we can immediately discount A as the de Broglie wavelength is far less than the radius of even a nucleus. (about an order of 1018 less...) So we can safely say that the it does not display any wavelike properties.

We can probably be safe in saying it doesn't move at the speed of light but we can do a caluculation to be sure.

Say a tennis ball has a mass of about 0.5Kg so h/(10-33x2)=v so it goes at around about not very fast at all :tongue: (besides, it can't go at the speed of light as that would require infinite energy - things starting below the speed of light can never exceed it and things starting faster than the speed of light [called tachyons] can never get below it)

So that leaves that it behaves like a particle which we are not happy with. There is no reason why it shouldn't and as a macroscopic object we should observe it acting like a classical particle would anyway :smile:

So the answer is C: it does not display wavelike properties.

Also... just to add because it's interesting but nobody actually knows WHY quantum effects are not observed on a macroscopic scale or exactly what scale they stop being observable at...
(edited 10 years ago)
Reply 13
Original post by natninja
Ok so we can immediately discount A as the de Broglie wavelength is far less than the radius of even a nucleus. (about an order of 1018 less...) So we can safely say that the it does not display any wavelike properties.

We can probably be safe in saying it doesn't move at the speed of light but we can do a caluculation to be sure.

Say a tennis ball has a mass of about 0.5Kg so h/(10-33x2)=v so it goes at around about not very fast at all :tongue: (besides, it can't go at the speed of light as that would require infinite energy - things starting below the speed of light can never exceed it and things starting faster than the speed of light [called tachyons] can never get below it)

So that leaves that it behaves like a particle which we are not happy with. There is no reason why it shouldn't and as a macroscopic object we should observe it acting like a classical particle would anyway :smile:

So the answer is C: it does not display wavelike properties.

Also... just to add because it's interesting but nobody actually knows WHY quantum effects are not observed on a macroscopic scale or exactly what scale they stop being observable at...



:biggrin::biggrin::biggrin::biggrin::biggrin::biggrin::biggrin::biggrin: thank u thank u thankkk uuuuuuuuuuuuuuu sooooooooooooooooooooo muchhhhhhhhhhhhhhhhhhhhhhhhh.

u are smarter than my teachersssssssssssssss :o:o:smile:
Reply 14
http://www.thestudentroom.co.uk/showthread.php?t=1597711
1.Lawrence’s second cyclotron, built with the assistance of Stanley Livingston was 25#cm in diameter and accelerated protons to 1#MeV.
a Calculate the speed of these protons.
b Calculate the momentum of these protons.
c Calculate the magnetic field strength Lawrence used in this cyclotron, assuming the protons move in a circle around the very edge of the machine.
d Calculate the frequency of the voltage that Lawrence and Stanley had to apply to achieve this proton acceleration.

http://uk.answers.yahoo.com/question/index?qid=20111218044932AAAG3MM
I got the same ans. don't know what is wrong :frown:
have tried the relativistic KE as well
Reply 15
Original post by ccctmc
http://www.thestudentroom.co.uk/showthread.php?t=1597711
1.Lawrence’s second cyclotron, built with the assistance of Stanley Livingston was 25#cm in diameter and accelerated protons to 1#MeV.
a Calculate the speed of these protons.
b Calculate the momentum of these protons.
c Calculate the magnetic field strength Lawrence used in this cyclotron, assuming the protons move in a circle around the very edge of the machine.
d Calculate the frequency of the voltage that Lawrence and Stanley had to apply to achieve this proton acceleration.

http://uk.answers.yahoo.com/question/index?qid=20111218044932AAAG3MM
I got the same ans. don't know what is wrong :frown:
have tried the relativistic KE as well


Ok I'll have a go... but if you've got this wrong then I may well too... relativistic dynamics is not my strong point...

So the mass of a proton is approximately 1000MeV (940MeV) so I'll start by assuming that the total energy is now 941 MeV. so our lorentz factor is 941/940=1/sqrt(stuff) so from rearranging this we get that the answer is 1.38x107

Ok so lets assume the question is wrong and the proton is accelerated to a total energy of a GeV so now our lorentz factor is 1000/940 but then we get around 0.3c... and the internet says definitely 1MeV in fact it should be 1.2 MeV which gives... 1.51 x107...

from the actual answer we can see that classical physics should give a good approximation so we'll try that. So 1000000eV=1.6x10-13J so then the speed is again 1.38x107... so we'll conclude (for now that the provided answer is wrong).

so our momentum is gamma.m.v=941/940 x m x v =2.3x10-20 so if this is correct then the provided answer for the first part is not and this agrees with our classical momentum as out lorentz factor is essentially 1 with a discrepancy of about a part in a thousand...

The other fact of the matter is that if it is relativistic then there is no way you could do part (c) without undergraduate knowledge... so we'll assume the classical model and the answers should follow...

also - I did try googling to check... I found this thread on the second page...
Reply 16
thanks :smile:

Is the lorentz factor important in the edexcel A2?

the expression appears in the section of synchrotron :s-smilie:

Original post by natninja
Ok I'll have a go... but if you've got this wrong then I may well too... relativistic dynamics is not my strong point...

Reply 17
Original post by ccctmc
thanks :smile:

Is the lorentz factor important in the edexcel A2?

the expression appears in the section of synchrotron :s-smilie:




Sorry to pop in unannounced , but I dont think u have to know that so well..
check specification :smile:
Reply 18
Original post by ccctmc
thanks :smile:

Is the lorentz factor important in the edexcel A2?

the expression appears in the section of synchrotron :s-smilie:


It's important if you do any relativity at all... it's kinda the basis for time dilation, length contraction and relativistic energy

and the lorentz factor (gamma) may appear in your formulae as 1/sqrt(1-v^2/c^2) on another note I just had a relativity and mechanics exam that I'm fairly sure was written by beelzebub...
Reply 19
Original post by natninja
Well I'm doing Physics at uni so I kinda have to know this stuff :tongue:

sure, ask away :smile:





Can u please explain why the answer is C, why not B??

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