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# AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] watch

1. (Original post by Sbarron)
I'm really struggling to get to grips with these two pages 😔Attachment 426843
I'll try to explain it in a nutshell:1. binding energy is the energy you need to seperate a nucleus into its constituent nucleons. 2. the most stable isotopes have around 8.8MeV binding energy per nucleon e.g iron. 3. When you split a nucleus (fission) you get energy out because the products have a higher binding energy per nucleon than what you started with. The reason you actually get out this energy is because the mass of the product is less than the mass of its seperted nucleons therefore you have extra energy.4. You will get more energy from fusion because there is a larger difference in the binding energy per nucleon between the reactants and products (it helps if you look at the graph). hopefully this helps a little bit? this might help too:

2. I wish they'd release that stolen paper so we could use it as revision 😂
3. (Original post by CD223)
It's to do with binding energy. Individual neutrons don't have binding energy as they're bound to nothing else

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Ahh I see, but you would have to calculate neutrons if they were part of a nucleus. For binding energy what is the formula is it like the final mass - initial mass?

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4. (Original post by Jimmy20002012)
Ahh I see, but you would have to calculate neutrons if they were part of a nucleus. For binding energy what is the formula is it like the final mass - initial mass?

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Binding energy is the mass defect in atomic mass units times 931.3 MeV. Or you could use the actual mass defect in kg times c squared.

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5. (Original post by EHR223)
There is a fair chance of it coming up my friend, but don't put all your eggs in one basket!
From first principles? It's not examined, as it involved vectors in 3 dimensions, which cannot be examined. They may ask to substitute one ofthe gas law equations into the others, but they cannot ask for it from very first principles. If you're curious, it's done here.

http://www.sciwebhop.net/sci_web/phy...tion_of_pv.htm

6. I can do part 1. p = M/V which is fairly simple

part2?
7. (Original post by Amanzz)
From first principles? It's not examined, as it involved vectors in 3 dimensions, which cannot be examined. They may ask to substitute one ofthe gas law equations into the others, but they cannot ask for it from very first principles. If you're curious, it's done here.

http://www.sciwebhop.net/sci_web/phy...tion_of_pv.htm
The book uses the same strategy but i think a small error is assumed when finding the time during the collision of the atom/molecule of the gas and the wall of the container. It says t=2L/u but this t is the time for the whole journey of the atom/molecule and therefore cannot be substituted into the momentum formula to find the force during the collision. The time during the collision is so small to be measured as stated. If anyone thinks I'm mistaken i would be very happy if you could dispel my confusion.

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8. Does anyone know if we have to know the paradoxes for special relativity?

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9. (Original post by gcsestuff)
Does anyone know if we have to know the paradoxes for special relativity?

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I think they are only to arouse curiosity, I'm not sure lol

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10. I've found this derivation of the impulse of the molecule in a container more reasonable.

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11. (Original post by AR_95)

I can do part 1. p = M/V which is fairly simple

part2?
R=r0 * A^1/3
solve for r0 using radius of a gold nucleus.
Use this value of r0 for calculating radius of aluminium nucleus, using new value of nucleon number.
12. (Original post by Mehrdad jafari)
The book uses the same strategy but i think a small error is assumed when finding the time during the collision of the atom/molecule of the gas and the wall of the container. It says t=2L/u but this t is the time for the whole journey of the atom/molecule and therefore cannot be substituted into the momentum formula to find the force during the collision. The time during the collision is so small to be measured as stated. If anyone thinks I'm mistaken i would be very happy if you could dispel my confusion.

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It's due to the fact we want the average force exerted by the atom, so we say that it is the time taken to be 2//v. I too had to research this as I couldn't see the skipped step when I came across it, however, it makes sense. It is how often the collision arises rather than the time of the collision (which is negligible), and so it happens every time the molecule moves back and forth from it's original position. The original position is 2l, and it's travelling at a constant velocity of magnitude v, hence 2l/v. As I said, the derivation is 99.9% unlikely to come up.
13. For muon decay considering time dilation, what do we need to know?

Is it just that as the half life of a muon is only 1.46*10^-6 if we considered them about 10 km above the earth none should reach the surface according to the earths frame of reference but due to time dilation and as they are traveling near the speed of light time goes slower for them so we can actually detect muons at the surface.

I'm so confused, is that all we need to know?

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14. (Original post by Amanzz)
It's due to the fact we want the average force exerted by the atom, so we say that it is the time taken to be 2//v. I too had to research this as I couldn't see the skipped step when I came across it, however, it makes sense. It is how often the collision arises rather than the time of the collision (which is negligible), and so it happens every time the molecule moves back and forth from it's original position. The original position is 2l, and it's travelling at a constant velocity of magnitude v, hence 2l/v. As I said, the derivation is 99.9% unlikely to come up.
Well, when considering the force exerted by the molecule in a time t(very small t), we are already assuming the average force since the magnitude of the forced exerted by the molecule each time (t) is not necessarily constant. Also assuming t=2L/u doesn't give us the average time of the impulse because if it was giving us the average time of the impulse then it would basically mean that the molecule is constantly colliding with other molecules of gas in the container on its way 2L. This is simply a wrong assumption as the the collisions between the molecules are within the system and cannot affect the pressure of the container(inertial frame of reference).

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15. It's really unnerving when I see a question on part of the spec I had no idea existed, then I realise it applied to an optional topic 😂

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17. (Original post by xela238)

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Yep.

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18. So basically you know that density of a nucleus is pretty much the same for all nuclei.

Use that fact
19. (Original post by AR_95)

I can do part 1. p = M/V which is fairly simple

part2?
Density of a nucleus is the same for all nuclei (pretty much).

so you know the mass of the new nucleus and using density you can work out radius of the nucleus

density = mass/(4/3 * pi * R^3)

where mass is the mass of 27 nucleons

I think the final answer is like 3.54 x 10^-15

One thing I don't know is if you use nucleon mass ( U or AMU 1.661) or mass of each proton (1.673) and each neutron (1.675) to calculate the mass of a nucleus since in reality it is more closer to nucleon in my opinion so I use that.
20. (Original post by betbi3etwerrd)
Density of a nucleus is the same for all nuclei (pretty much).

so you know the mass of the new nucleus and using density you can work out radius of the nucleus

density = mass/(4/3 * pi * R^3)

where mass is the mass of 27 nucleons

I think the final answer is like 3.54 x 10^-15

One thing I don't know is if you use nucleon mass ( U or AMU 1.661) or mass of each proton (1.673) and each neutron (1.675) to calculate the mass of a nucleus since in reality it is more closer to nucleon in my opinion so I use that.
how comes density here is 2.4 where as in my nelson thornes book it says 3.4

Clearly a massive typo on one or the other?

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