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PHYA5 ~ 20th June 2013 ~ A2 Physics

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Maybe you could look at it as each particle exerting an AVERAGE impulse of 2mv where t is the time between collisions

obviously in reality, most of that impulse is concentrated during the time when it's colliding, but for our purposes we can spread it across the whole time between collisions and including collisions (because they happen 'instantaneously'). So you average that out with all of the time between collisions, when the impulse the molecule exerts is zero, to get a value which is kind of the force the molecule contributes all the time

Reply 641

Pinkhead

Original post by Jack93o

still don't get it

for one collision, how can force simply be equal to change in momentum?

change in momentum is impulse, whereas force is change in momentum divided by time

and just to be clear, when you say, 'average force', this is for a collision between a molecule and a wall, right?

I think I was supposed to say ''The average force = change of momentum per collision × number of collisions per second.''
I probably overcomplicated it.

We're thinking about the force transferred to the wall by the molecule. So gain in momentum of the wall in one second because of the collision = momentum transferred in one collision x number of collisions per second.

Reply 642

Minumi

Original post by saba146

the CGP textbook is quite good :smile:

I have that, it's not bad, but I find CGP better for cramming/reviewing just before the exam, as it's so concise!

Original post by Pinkhead

I used this alongside the AQA textbook:
http://www.antonine-education.co.uk/Pages/Physics_5/Overview.htm#Thermal_Physics

And I have the same problem. I keep using methods I use in Chemistry :frown:

. That's probably the only annoying part of the core unit.

Ah that's brilliant, thanks! I've been looking for something like this.
Yeah I keep on trying to be clever and linking them, but it always ends in disaster :colonhash:

Reply 643

cooldudeman

In thermal reactor, the neutron is fired at the U238 so it becomes 239 but where does U235 come from??

Posted from TSR Mobile

Reply 644

crc290

Original post by cooldudeman

In thermal reactor, the neutron is fired at the U238 so it becomes 239 but where does U235 come from??

Posted from TSR Mobile

The fuel rods consist of U238 which is enriched with U235. The U235 undergoes fission when it is bombarded with neutrons. U238 is not fissile. Is that what you meant?

Reply 645

cooldudeman

Original post by crc290

The fuel rods consist of U238 which is enriched with U235. The U235 undergoes fission when it is bombarded with neutrons. U238 is not fissile. Is that what you meant?

So do the fired neutrons colide with the fuel ro?
Posted from TSR Mobile

Reply 646

crc290

Original post by cooldudeman

So do the fired neutrons colide with the fuel ro?
Posted from TSR Mobile

Well yeah, they collide with the U235 nuclei in the fuel rods

Reply 647

Sabr

guys can someone explain the balmer series and why it starts at n = 2?
(thats if ur doing the astrophysics)

Reply 648

JRP95

If u235 is the one that undergoes fission then why are the fuel rods mainly u238 (which is non fissionable and so is pointless?) and only 2-3% u235 when that is fissionable (surely more fission would occur or you'd get a longer chain reaction with a greater mass of fissile material right?)

Reply 649

bugsuper

Original post by JRP95

A number of reasons:

- Uranium is naturally 99% U-238, the non-fissionable isotope. You have to get three times as much Uranium and then go through the complicated process of enriching it to get fuel rod material that only has 3% of the fissionable isotope, so getting a "pure" fuel rod would be prohibitively expensive. (They do something similar to this for nuclear weapons, but I imagine that they're much more expensive to build then maintaining a nuclear reactor is.) You can't just selectively mine the Uranium ore.

- The second, far more important reason, is that if you had a rod that was made of purely U-235, it would almost certainly blow you to hell. The point is not a "longer chain reaction", but instead a CONTROLLED chain reaction. The non-fissionable isotope actually helps, because it absorbs some of the neutrons that are produced by fission - as well as the control rods. If it didn't do this, then the reaction would quickly spiral out of control and you have the workings of a nuclear bomb.

To illustrate the point, U-235 does naturally decay by fission - although with a long half-life - without needing to be initiated or bombarded by neutrons. So if you had a pure fuel rod of the stuff, one of the nuclei would probably decay on its own, almost certainly spark a chain reaction, and kill you before you were able to put it in the semi-safety of a reactor.

(edited 10 years ago)

Reply 650

Pinkhead

Original post by bugsuper

To illustrate the point, U-235 does naturally decay by fission - although with a long half-life - without needing to be initiated or bombarded by neutrons. So if you had a pure fuel rod of the stuff, one of the nuclei would probably decay on its own, almost certainly spark a chain reaction, and kill you before you were able to put it in the semi-safety of a reactor.

To be honest, spontaneous fission occurs so rarely that it's negligible. You can get a pure rod of U-235 and it probably won't blow up any time soon.

Reply 651

posthumus

Original post by Pinkhead

To be honest, spontaneous fission occurs so rarely that it's negligible. You can get a pure rod of U-235 and it probably won't blow up any time soon.

Would it blow up as soon as you make it?
I mean if you made the fuel rod with U-235 being above it's critical mass

Reply 652

Allahu Akbar

Does anyone know where you can get mark schemes for the nuclear and thermal physics? I've looked on the aqa website and I can't find any, just for the extra topics.

Reply 653

bugsuper

Original post by Pinkhead

To be honest, spontaneous fission occurs so rarely that it's negligible. You can get a pure rod of U-235 and it probably won't blow up any time soon.

Fair enough! I still wouldn't want to be in a room with a pure rod of U-235

Reply 654

Pinkhead

Original post by posthumus

Would it blow up as soon as you make it?
I mean if you made the fuel rod with U-235 being above it's critical mass

Well you would need a neutron with the right speed being released from the nuclide. Critical mass only matters after a neutron starts the reaction right?

Reply 655

Allahu Akbar

Original post by Pinkhead

To be honest, spontaneous fission occurs so rarely that it's negligible. You can get a pure rod of U-235 and it probably won't blow up any time soon.

All you need though is one stray neutron and 'Allahu Akbar' ... you're history. Wouldn't catch me hanging around billions upon billions of angry U-235 nuclei

Reply 656

ThatRandomGuy

The more negative the absolute magnitude, the brighter it is right? And likewise the more negative the apparent magnitude the brighter it appears?

Reply 657

bugsuper

Original post by ThatRandomGuy

The more negative the absolute magnitude, the brighter it is right? And likewise the more negative the apparent magnitude the brighter it appears?

Yes.

Re: the nuclear fission thing - even if it wouldn't spontaneously blow up, it's still going to be horribly dangerous to actually put 100% U-235 in a reactor and bombard it with thermal neutrons

Reply 658

ThatRandomGuy

This AQA PDF thing is crap for self teaching.

Why is a star a black body? As in how does a star absorb and then emit 100% of all radiation that lands on it?

Reply 659

bugsuper

Original post by ThatRandomGuy

This AQA PDF thing is crap for self teaching.

Why is a star a black body? As in how does a star absorb and then emit 100% of all radiation that lands on it?

Stars aren't perfect blackbodies, I don't think. They're just such good emitters of radiation that they can be modelled as blackbodies, and the theories that apply to these objects can be used to model stars to a high degree of accuracy.

https://en.wikipedia.org/wiki/Black_body#Stars_and_planets

That might help? I don't think we necessarily need to be able to explain why stars can be modelled as blackbodies beyond "they're good emitters and absorbers of radiation"