AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread]

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Has the gas equation proof ever come up? Ive just learned it and im wondering if ive wasted my time....

Apparently not. Our teachers just claim AQA want us to be aware of it? But that's so dodgy. Why make us learn that bs just to not test us on it.

I'm genuinely considering never learning it but I can tell if I don't it'll come up as a 6 mark question or something. But then if I do learn it then watch how it doesn't come up lol

Reply 301

rory58824

I'm expecting a derivation on the kinetic energy equation for the 6 marker.

Reply 302

philo-jitsu

Okay this is driving me insane, how would people answer this question, I swaer the mark scheme is wrong

u-238 undergoes alpha emmision, work out how much energy is released in MeV?

how do you work out the mass defect?

I though you do neutron number*1.00867 + proton number *1.00728. for every side of equation then subtract? is this wrong?

Reply 303

Yo12345

Original post by rory58824

I'm expecting a derivation on the kinetic energy equation for the 6 marker.

How is it done?

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Reply 304

CourtlyCanter

Anyone else doing Medical Physics?

Reply 305

Yo12345

Original post by rory58824

I'm expecting a derivation on the kinetic energy equation for the 6 marker.

Is that the PV = Nm(crms)^2/3 deravition and then re-arrange?

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Reply 306

rory58824

Original post by Yo12345

Is that the PV = Nm(crms)^2/3 deravition and then re-arrange?

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Yeah that one. I should have said the GAS kinetic energy equation derivation lol.

Reply 307

Franckenstar

Original post by philo-jitsu

1u = 1.661 x 10^-27. You have to convert to of before using E=mc^2. mass defect = left hand side - right hand side.

Reply 308

treeporn

Original post by lucabrasi98

Do we ACTUALLY need to know that disgusting gas equation proof? I've never even bothered with it. Apparently we just "need to be aware of it". But then what's the point of it being in the book?
For some reason,

t= rest frame.

T0= proper time

However,

L0= rest frame.

And

L0= proper length

It makes no sense but I've learnt to accept it. It gets me the right answer afterall

Yeah, that's what I thought it was! It's been getting me the correct answer for them now except in the June 2010 paper it said that L0 was in the moving frame which is obviously incorect lmao. But thank you for your confirmation!

*edit: It wasn't in the June 2010 paper, dw it was in a collective set of past questions on special relativity, don't matter!

(edited 7 years ago)

Reply 309

philo-jitsu

Original post by Franckenstar

1u = 1.661 x 10^-27. You have to convert to of before using E=mc^2. mass defect = left hand side - right hand side.

I have been doing that, well I just use 931.5 to find MeV but I am still getting the mass defect wrong...

Its number 14 on this

https://e4cf8bb391554b7c9d8e0fc4226948a24b1d6886.googledrive.com/host/0B1ZiqBksUHNYZnZFc3hVb3c4WXM/From-AQA-Set-1/5.2%20Nuclear%20Energy.pdf

Reply 310

Yo12345

Original post by rory58824

Yeah that one. I should have said the GAS kinetic energy equation derivation lol.

That would be a nice 6 marker

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Reply 311

cowie

If we work out the amount of marks which would be given for derivation we may be able to work out if a question regarding it is plausible. I reckon the whole derivation would be far more than 6 marks so hence shouldn't be asked as 6 markers usually involve 4 marks worth of information with 2 marks for SPG..
Also I wouldn't imagine they would ask consecutive questions regarding it considering it would take up probably about 10 marks meaning a quarter of part A which is not usually done regarding one small aspect of the spec?
If you see any flaws in this logic then please yell as loud as you can and I can get to memorising it ahah!

Reply 312

lucabrasi98

Original post by cowie

Also I wouldn't imagine they would ask consecutive questions regarding it considering it would take up probably about 10 marks meaning a quarter of part A which is not usually done regarding one small aspect of the spec?

I'd break down and cry if that happened. Jesus christ.

I don't see how it could come up because there's no possible way that they could logically ask it. The only calculation based topic with 10+ single questions is M1 lol.

I hope my risk to not learn it pays off. I might get up at 4AM the day before the exam and learn it then, but probably not

Reply 313

Yo12345

Original post by Yo12345

For Intensity, is the corrected count rate the intensity? If not, what is the corrected count rate in calculation terms. How did you find activity from corrected count rate?

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Bump

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Reply 314

cowie

Original post by lucabrasi98

I've half committed it to memory already nonetheless.. Nothing will go wrong! (hopeful statement which I use too much and too optimistically..)
https://www.youtube.com/watch?v=Qsa4aAdpHfY - VERY good youtube vid explaining the derivation!

Reply 315

particlestudent

Do we need to know the density of an ideal gas?

Reply 316

ChillGod

Original post by philo-jitsu

Turning points questions

In the cathode ray experiment, cathodes rays were produced in an evacuated glass tube, what produced the 'rays' if the electrons couldn't collide with anything as it was evacuated?

Also in the jj Thompson experiment where there was a small amount of hydrogen in the tube, what causes the production of hydrogen ions? is it the electron ray produced by the cathode/anode knocking them off? Are the electrons colliding with the proton of the hydrogen nucleus and creating a neutron? I'm confused as to what causes the production of hydrogen ions? Is it multiple factors

Also it is said that Thompson concluded that the cathode ray consisted of particles that made up an atom, how did he come to this conclusion based on this experiment?

Thanks in advance for any help

You've got a few misconceptions here about the experiment.

First lets talk about how the electron gun works.

Electrons are made by heating up a metal coil on the negative electrode (cathode). Because you are giving the thermal energy, they increase in KE until they have enough energy to leave - this is called thermionic emission. Once they leave the surface of the coiled metal, they are accelerated to the positive electrode (anode). Why? There's a electric field set up by the high potential difference. In a electric field -ve charges accelerate to the +ive plate.

The anode has a hole in it, so some electrons collide, but the ones that go through make a narrow beam. Think of putting a hole through a bottle cap and putting water through, the cap will fill up with water but water will still come out from the hole in a nice straight beam.

Anyway. Once they've made it past the anode they are no longer in an electric field so have constant velocity. They then leave the evacuated tube at that same constant velocity.

What causes hydrogen ions? I don't think hydrogen ions are produced. The electron gun I was talking about is inside a glass bulb filled with hydrogen gas. So once the electrons left the electron gun they would find themselves in the glass bulb. Inside the electron gun is evacuated but outside it is the glass bulb, you with me?

Anyway on either side of the bulb there are two coils producing a magnetic field. If the gun is arranged at right angles to the magnetic field, when the beam of electrons leave the gun they will also be at right angles to the field. From Unit 4 physics you should know that the beam will curve in a circle.

Cool so now we have the beam of electron travelling in a circle but we can't see them. That's where the hydrogen atoms come in. The electron collide with them and excite the electrons in the hydrogen atoms. From Unit 1 you should remember if an electron is excited, it moves up an energy level but doesn't leave the atom, when it dexcites it releases a photon seen as light.

that is the only reason the hydrogen atoms are needed, the help illuminate the path of electrons when they're excited. Once he could see the circle, thompson could measure the radius and equal centripetal force to BeV. Did some rearranging to get e/m.

To your last question, electrons make up some of the atom but not the nucleus. The cathode ray is made from electrons not particles. How did he come to that conclusion? He found out the properties of the electron beam was the same, no matter the material the coil of wire in the electron gun was made of. It would always have the same mass charge ratio. Since the ratio was 1800 times bigger than a protons (they knew the e/m of protons before hand) it's mass would have to be very small.

Anything I haven't got through properly? Nice source of impromptu revision