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# G495 - Field and Particle Pictures: June 11th Tweet

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1. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Oromis263)
Haha, I feel the same! :P Have you got the ones off of the Matthew Moodle site? If it asks you to login, just hit login as guest.
Thankyou very much for this, should have noticed these before hand, it's one of my local schools!

(Original post by Oromis263)
Ok, well I reckon we should start testing each other on our written answers, as usually they're the easiest to lose marks on! Have these few questions to get us started:

1. Explain why a hydrogen atom cannot take all of the energy from an electron that has 12.0eV of energy (given that the hydrogen is in its ground state when the electron collides with it).
Presumably something to do with an atom only being allowed to sit at a certain energy level?
2. Explain how a rotating magnet causes an emf in a coil of a simple AC generator.
An emf is induced whenever relative motion occurs between a conductor and a magnet; any region of which flux 'lines' are getting cut.
3. Given
i) Explain why one of the particles X or Y could be an electron.
Conservation of nucleon and proton numbers needs to occur; must have nucleon number zero and a proton number of -1. Since proton number is also the relative charge, X or Y must be an electron.
ii) If X is an electron, suggest why Y could be an antineutrino
Must be a lepton to balance lepton number, must be chargeless and massless, hence nucleon number and proton number is zero. Hence an antineutrino.

Just some to get us started.
(Original post by Oromis263)
I have a few more, nobody answered my first set. These aren't challenging, just quite nice ones for putting together good word structure and practise being coherent. :P

1. Explain why the intensity of gamma photons decreases with increasing distance from the source (the source being a radioactive sample).
Photons get emitted in all directions so as you get further away from the source the amount of photons per unit area decreases. (ie they get spread out) Since there are less per unit area, the intensity is clearly less.
2. In an experiment, protons and antiprotons travel in opposite directions through an evacuated tube which has a uniform magnetic flux density.

i) Suggest why protons and antiprotons travel in opposite directions in the evacuated tube.
Opposite charges
Sometimes when a proton and antiproton collide, a particle called the Z is created. The Z particle is unstable and decays quickly into a positron and an electron.

ii) Suggest the equation for the decay of the Z particle, showing nucleon and charge numbers.
(0,0)Z --> (0,-1)β+ + (0,1)β- (Can't be bothered to latex; sorry)

iii) The experiment shows that the Z particle has a rest energy of 93 GeV. Suggest how this is determined by experiment.
Find the energy of the annihilation between the particles and antiparticles? This should be equal to the rest energy? (Please help on this; no idea!)
3. Explain why the core of a pulse counter is made from thin iron sheets glued together rather than being solid iron.
Lamination is used to help reduce eddy currents by creating a slight resistant air gap between the sheets. These eddy currents release energy by heat and are wasteful, hence why in this case they are not wanted.
4. The nucleons in a nucleus are bound to each other by the strong force. Explain why this requires the mass of the nucleus to be less than the mass of its separate nucleons.
In order to seperate a nucleus (ie split the nucleons up) a certain amount of extra energy is needed to help overcome the nuclear strong force. Since mass is closely related to the energy of a particle (ie E=mc^2), less mass of the nucleus is required to help hold the nucleus together. (Not sure about wording? Any help?)
Here's a fairly mathsy one:
An oil drop of mass 1.63x10^-14kg is held stationary in the space between two charged plates 3cm apart. The potential difference between the plates is 5000V. The density of the oil used is 880kgm^-3.
(a)Describe the relative magnitude and direction of the forces acting on the oil drop. (2)
(b)Calculate the charge on the oil drop using g=9.81 (3)

The electric field is switched off and the oil drop falls towards the bottom plate.
(c)Explain why the oil drop reaches terminal velocity as it falls. (3)

Think this may be an old specification question, certainly don't remember learning about it, but interesting nevertheless, and it comes from the revision guide, so who knows?
2. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Oromis263)
Have a go first, and quote this one back.
4. The nucleons in a nucleus are bound to each other by the strong force. Explain why this requires the mass of the nucleus to be less than the mass of its separate nucleons.

Because mass gets converted into energy to keep the nucleus bound so the mass of it is less than the particles that make it up.
3. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Revent)
4. The nucleons in a nucleus are bound to each other by the strong force. Explain why this requires the mass of the nucleus to be less than the mass of its separate nucleons.

Because mass gets converted into energy to keep the nucleus bound so the mass of it is less than the particles that make it up.
Yeah pretty much, my answer was:

Less mass equates to less energy in the nucleons when they are a part of the nucleus (by the relationship E = mc^2). This occurs because some energy is needed to provide the binding energy to keep the nucleons together.

I also talked about why the binding energy was required (ie, the strong force must overcome the electromagnetic repulsion between protons), but the critical marking points are in bold.
4. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Oromis263)
Yeah pretty much, my answer was:

Less mass equates to less energy in the nucleons when they are a part of the nucleus (by the relationship E = mc^2). This occurs because some energy is needed to provide the binding energy to keep the nucleons together.

I also talked about why the binding energy was required (ie, the strong force must overcome the electromagnetic repulsion between protons), but the critical marking points are in bold.
Ah right. I knew what the answer was, just had some trouble of wording it

You got any questions to do with chap 15 & 16? This is quite nice.
5. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Revent)
Ah right. I knew what the answer was, just had some trouble of wording it

You got any questions to do with chap 15 & 16? This is quite nice.
The amplitude of the EMF for a given rotational speed can be increased by using a stronger magnet which increases the flux density in the coil.

Describe and explain two other modifications to the pulse counter to increase the amplitude of the EMF for a given rotational speed.

6. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Oromis263)
The amplitude of the EMF for a given rotational speed can be increased by using a stronger magnet which increases the flux density in the coil.

Describe and explain two other modifications to the pulse counter to increase the amplitude of the EMF for a given rotational speed.

More coils! emf= N d(phi)/dt. so increasing N helps.

I was gonna say stronger magnet...but. lol No idea what the second would be. Thicker magnet maybe? This gives more flux lines/higher flux density?

Edit: Also, which formulae do we NEED to know for this exam that aren't synoptic? ie. F=BIL, B=Phi/A, etc...
Last edited by Revent; 05-06-2012 at 21:23.
7. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Revent)
More coils! emf= N d(phi)/dt. so increasing N helps.

I was gonna say stronger magnet...but. lol No idea what the second would be. Thicker magnet maybe? This gives more flux lines/higher flux density?
You're correct, I would just work on structuring it a bit better. Obviously you would probably write in proper sentences in the exam, but it doesn't hurt to get the practise beforehand.

By increasing the number of coils, the flux linkage is increased, thus leading to a larger EMF (by the relationship EMF = d(Nphi)/dt)

By decreasing the air gap between the stator and the rotor (magnet and core), the magnetic circuit is improved, thus increasing the flux, thus leading to a larger EMF (by same relationship)

Increasing the dimensions of the apparatus will increase the flux linkage, thus leading to a larger EMF (again, same relationship)

Increasing the permeability of the (iron) core will improve the magnetic circuit, thus increasing flux linkage, etc etc

By laminating the core, eddy currents (that oppose the flux) are reduced, leading to a larger flux linkage, etc etc again.

Basically, on questions like that, just take your pick of anything that will lead to affecting the relationship EMF = d(Nphi)/dt, and your sure to get the marks.
8. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Oromis263)
You're correct, I would just work on structuring it a bit better. Obviously you would probably write in proper sentences in the exam, but it doesn't hurt to get the practise beforehand.

By increasing the number of coils, the flux linkage is increased, thus leading to a larger EMF (by the relationship EMF = d(Nphi)/dt)

By decreasing the air gap between the stator and the rotor (magnet and core), the magnetic circuit is improved, thus increasing the flux, thus leading to a larger EMF (by same relationship)

Increasing the dimensions of the apparatus will increase the flux linkage, thus leading to a larger EMF (again, same relationship)

Increasing the permeability of the (iron) core will improve the magnetic circuit, thus increasing flux linkage, etc etc

By laminating the core, eddy currents (that oppose the flux) are reduced, leading to a larger flux linkage, etc etc again.

Basically, on questions like that, just take your pick of anything that will lead to affecting the relationship EMF = d(Nphi)/dt, and your sure to get the marks.
Ooh right. Thanks!
So what formulae do we need to know that aren't synoptic? Like F=BIL, B=Phi/A, etc...
9. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Oromis263)
Any chance of you being able to check my answers from earlier; hope it's not too much to ask I think they were in the first post on page 4. Cheers
10. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Miller693)
Thankyou very much for this, should have noticed these before hand, it's one of my local schools!

Here's a fairly mathsy one:
An oil drop of mass 1.63x10^-14kg is held stationary in the space between two charged plates 3cm apart. The potential difference between the plates is 5000V. The density of the oil used is 880kgm^-3.
(a)Describe the relative magnitude and direction of the forces acting on the oil drop. (2)
(b)Calculate the charge on the oil drop using g=9.81 (3)

The electric field is switched off and the oil drop falls towards the bottom plate.
(c)Explain why the oil drop reaches terminal velocity as it falls. (3)

Think this may be an old specification question, certainly don't remember learning about it, but interesting nevertheless, and it comes from the revision guide, so who knows?
Regarding your questions, an antineutrino is an anti lepton I believe. Lepton numbers have to balance. Electron is +1 and antineutrino is -1

For 2ii) It shouldn't be beta, it should be e because an electron and positron are formed. This is not beta decay.
For 2iii) You can do it by measuring the energy levels of the positron and electron (because that is all that is formed) and that will give you the energy of the Z. (conservation of energy)

but the rest seem fine Although I'm not really the physics pro here I only started revising for this exam two days ago
11. Re: G495 - Field and Particle Pictures: June 11th
How is everyone doing this exam preparing for section C? :/
12. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Revent)
Ooh right. Thanks!
So what formulae do we need to know that aren't synoptic? Like F=BIL, B=Phi/A, etc...
I'm not sure, but I think we're expected to know the two formulas for permeance;
permeance= flux/ current turns
and permeance = permeability * area / length (this one's fairly easy to remember by looking at the formula for conductivity, which is in the formula book)
13. Re: G495 - Field and Particle Pictures: June 11th
(Original post by PhysicsGirl)
I'm not sure, but I think we're expected to know the two formulas for permeance;
permeance= flux/ current turns
and permeance = permeability * area / length (this one's fairly easy to remember by looking at the formula for conductivity, which is in the formula book)
You probably remember it, there's the one that looks something like:

Involved when looking at the diffraction minima of electron scattering, or something like that...
14. Re: G495 - Field and Particle Pictures: June 11th
also the formula used for calculating the electron charge in the millikan oil drop experiment might come up:

(q*V)/d = mg
where q=charge, V=potential difference, d=distance of the plates, m=mass, g=9.8

the might even ask you to work out the radius of the oil drop, where:
m=pV
-> m=4/3*pi*p*r^3
15. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Revent)
Ooh right. Thanks!
So what formulae do we need to know that aren't synoptic? Like F=BIL, B=Phi/A, etc...
And I'm not entirely sure if we are required to know this or not, and it's quite easily remembered by comparing it to the number decayed formula, but the formula for intensity of radiation passing through a material is;
I=I0e-mu(x), where half thickness, x1/2 is ln2/ mu

(Original post by Oromis263)
You probably remember it, there's the one that looks something like:

Involved when looking at the diffraction minima of electron scattering, or something like that...
Indeed, I think we're supposed to remember that one too Thanks!
16. Re: G495 - Field and Particle Pictures: June 11th
This may be a really silly question so I apologise...

In nuclear equations, the bottom number is proton number right? So when you get an electron being -1 on the bottom, its not -1 proton obviously, just -1 charge, so is it better to think of the number as just the charge or what?

I'm really confusing myself over this.
17. Re: G495 - Field and Particle Pictures: June 11th
(Original post by PhysicsGirl)
And I'm not entirely sure if we are required to know this or not, and it's quite easily remembered by comparing it to the number decayed formula, but the formula for intensity of radiation passing through a material is;
I=I0e-mu(x), where half thickness, x1/2 is ln2/ mu

Indeed, I think we're supposed to remember that one too Thanks!
It is usually a hell of a lot easier in most of the half thickness questions I've encountered just to do (0.5)^x where x is the number of half thicknesses passed through, then multiply the activity by that, and most of the reports I've read say similar.

Ah, I got it right, I wasn't 100% sure, d is the diameter, no?
18. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Amy-Rose)
This may be a really silly question so I apologise...

In nuclear equations, the bottom number is proton number right? So when you get an electron being -1 on the bottom, its not -1 proton obviously, just -1 charge, so is it better to think of the number as just the charge or what?

I'm really confusing myself over this.
When describing a nucleus, eg U-235, the 92 represents the number of protons in the nucleus. However, because a proton has a charge of +1, it also represents the charge of the nucleus, so you can think of it either way, as proton number or charge number. If it makes it easier, you can just think of it as the charge of the particle (remembering that you're only looking at the nucleus of an atom when doing so. Atoms have an overall charge of 0)
19. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Oromis263)
When describing a nucleus, eg U-235, the 92 represents the number of protons in the nucleus. However, because a proton has a charge of +1, it also represents the charge of the nucleus, so you can think of it either way, as proton number or charge number. If it makes it easier, you can just think of it as the charge of the particle (remembering that you're only looking at the nucleus of an atom when doing so. Atoms have an overall charge of 0)
Hmm, that does make a bit more sense yes. It's just thinking of proton number I know they don't have protons... I'll probably just think of it as charge, like you said. Thank you.

Also, does anyone have any questions on absorbed dose/dose equivalent? We didn't do any in class and with few past papers I'm struggling for material.
20. Re: G495 - Field and Particle Pictures: June 11th
(Original post by Amy-Rose)
Hmm, that does make a bit more sense yes. It's just thinking of proton number I know they don't have protons... I'll probably just think of it as charge, like you said. Thank you.

Also, does anyone have any questions on absorbed dose/dose equivalent? We didn't do any in class and with few past papers I'm struggling for material.
Follow this link here and just click login as a guest if it asks you. There you will find a plethora of legacy papers, which will easily keep you busy up until the exam. I can't remember any specific years with questions on absorbed dose/dose equivalent, but there will be some in there.
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