# PHYA5 ~ 20th June 2013 ~ A2 Physics Watch

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#101

(Original post by

Can't believe there's a thread on this already.. to be fair i'm busy concentrating on re-sitting Unit 4, the joys.

Anyone else doing the 'turning points' optional module?

**Bobilina**)Can't believe there's a thread on this already.. to be fair i'm busy concentrating on re-sitting Unit 4, the joys.

Anyone else doing the 'turning points' optional module?

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#102

**Bobilina**)

Can't believe there's a thread on this already.. to be fair i'm busy concentrating on re-sitting Unit 4, the joys.

Anyone else doing the 'turning points' optional module?

I'm doing Turning Points Just about to finish it, have you started?

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#103

(Original post by

yeah I'm doing it school hasn't started on it yet though, electron stuff is pretty easy, just memorising cathode ray stuff erm then there's photo-electric effect and wave particle duality and then special relativity. So it's pretty much special relativity and the odd thing here and there to look up on how's the rest of the course going ?

**PhilosopherKing**)yeah I'm doing it school hasn't started on it yet though, electron stuff is pretty easy, just memorising cathode ray stuff erm then there's photo-electric effect and wave particle duality and then special relativity. So it's pretty much special relativity and the odd thing here and there to look up on how's the rest of the course going ?

The rest of the course is going quite well, I actually enjoy it which motivates me to revise and work for it. Whereas with Unit 4 I had no motivation or interest in the topics and is therefore a re-sit

Sorry for the super late reply i've been in Switzerland for 4 days... #cern

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#104

(Original post by

I'm doing Turning Points Just about to finish it, have you started?

**Technetium**)I'm doing Turning Points Just about to finish it, have you started?

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#105

Lucky you! At least you are being taught the topic. In my school it is an 'independent study' option It is not that hard

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#106

(Original post by

Lucky you! At least you are being taught the topic. In my school it is an 'independent study' option It is not that hard

**Technetium**)Lucky you! At least you are being taught the topic. In my school it is an 'independent study' option It is not that hard

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#107

(Original post by

Same here ! It doesn't seem too long though... Special Relativity looks disappointingly short, nothing on Space-time and stuff from what I've seen

**posthumus**)Same here ! It doesn't seem too long though... Special Relativity looks disappointingly short, nothing on Space-time and stuff from what I've seen

There is a little bit on the Twin Paradox, which I am finding quite confusing. I love the chapter on Newton and Huygens

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#108

(Original post by

There is a little bit on the Twin Paradox, which I am finding quite confusing. I love the chapter on Newton and Huygens

**Technetium**)There is a little bit on the Twin Paradox, which I am finding quite confusing. I love the chapter on Newton and Huygens

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#110

(Original post by

Is this in the Nelson Thornes book?? I only use the Collins Revision Guide

**posthumus**)Is this in the Nelson Thornes book?? I only use the Collins Revision Guide

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#111

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#112

Fairly silly question, but yeah...

I don't seem to understand the inverse square law with respect to gamma radiation very well? How exactly do I use I=k/r^2, as they don't give you the value of k on the data sheet?

I was doing the Summary Questions at the end of section 9.5 in the Nelson Thornes book, and got stuck on question 4 (b);

"A point source of gamma radiation is placed 200mm from the end of a Geiger tube. The corrected count rate was measured at 12.7 counts per second. Calculate the distance between the source and the tube for a corrected count rate of 20 counts per second."

The correct answer is 160mm.

I don't seem to understand the inverse square law with respect to gamma radiation very well? How exactly do I use I=k/r^2, as they don't give you the value of k on the data sheet?

I was doing the Summary Questions at the end of section 9.5 in the Nelson Thornes book, and got stuck on question 4 (b);

"A point source of gamma radiation is placed 200mm from the end of a Geiger tube. The corrected count rate was measured at 12.7 counts per second. Calculate the distance between the source and the tube for a corrected count rate of 20 counts per second."

The correct answer is 160mm.

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#113

(Original post by

Fairly silly question, but yeah...

I don't seem to understand the inverse square law with respect to gamma radiation very well? How exactly do I use I=k/r^2, as they don't give you the value of k on the data sheet?

I was doing the Summary Questions at the end of section 9.5 in the Nelson Thornes book, and got stuck on question 4 (b);

"A point source of gamma radiation is placed 200mm from the end of a Geiger tube. The corrected count rate was measured at 12.7 counts per second. Calculate the distance between the source and the tube for a corrected count rate of 20 counts per second."

The correct answer is 160mm.

**Xiomara**)Fairly silly question, but yeah...

I don't seem to understand the inverse square law with respect to gamma radiation very well? How exactly do I use I=k/r^2, as they don't give you the value of k on the data sheet?

I was doing the Summary Questions at the end of section 9.5 in the Nelson Thornes book, and got stuck on question 4 (b);

"A point source of gamma radiation is placed 200mm from the end of a Geiger tube. The corrected count rate was measured at 12.7 counts per second. Calculate the distance between the source and the tube for a corrected count rate of 20 counts per second."

The correct answer is 160mm.

So firstly, you'd do k=Ir^2 where I=12.7 and r=0.2 and you will get the constant.

The use r^2= k/I with the value of k you just calculated and the new I, 20.

Finally route that answer to get your final answer.

Hope that helps

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#114

(Original post by

Right, so what you first have to work out is the constant for that particular source of radiation.

So firstly, you'd do k=Ir^2 where I=12.7 and r=0.2 and you will get the constant.

The use r^2= k/I with the value of k you just calculated and the new I, 20.

Finally route that answer to get your final answer.

Hope that helps

**Seb Hudson**)Right, so what you first have to work out is the constant for that particular source of radiation.

So firstly, you'd do k=Ir^2 where I=12.7 and r=0.2 and you will get the constant.

The use r^2= k/I with the value of k you just calculated and the new I, 20.

Finally route that answer to get your final answer.

Hope that helps

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#115

hey guys i was just wondering how come i cant find the markschemes for section a, which is nuclear and thermal physics, if anyone can send them to me i would be really helpful.

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#116

(Original post by

hey guys i was just wondering how come i cant find the markschemes for section a, which is nuclear and thermal physics, if anyone can send them to me i would be really helpful.

**ehtisham_1**)hey guys i was just wondering how come i cant find the markschemes for section a, which is nuclear and thermal physics, if anyone can send them to me i would be really helpful.

The reason they don't publish Section A (nuclear and thermal physics) on its own is because it isn't standalone.

Hope that helps!

Edit: Obviously didn't help.

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#117

Can anyone please explain how to derive the pressure of an ideal gas in stupid people terms? I missed the lesson on it and the book is quite confusing :/ And does anyone know how they might ask us about it in the exam?

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#118

(Original post by

Can anyone please explain how to derive the pressure of an ideal gas in stupid people terms? I missed the lesson on it and the book is quite confusing :/ And does anyone know how they might ask us about it in the exam?

**Xiomara**)Can anyone please explain how to derive the pressure of an ideal gas in stupid people terms? I missed the lesson on it and the book is quite confusing :/ And does anyone know how they might ask us about it in the exam?

consider a particle with 1 unit mass (m) travelling from one side of a cube (l) container to the other (therefore total distance travelled is 2l)

We know the change in momentum will be -2mv

F=-2mv/t

we know t is basically v=

**2**l/t rearranged... t=v/

**2**l so sub that in

F=-2mv/2l/v = -mv

^{2}/l

Pressure is force/area therefore... since you know that area is l

^{2}:

p=mv

^{2}/l

^{3}

l^3 is the same as the volume so you can rewrite it...

p=mv

^{2}/V

The total pressure will be all the pressures of each atom added up, because each will not be travelling at the same velocity...

P= mv

^{21}/V + mv

^{22}/V + mv

^{23}/V + mv

^{24}/V......

= m(v

^{21}+ v

^{22 }+ v

^{23}+ v

^{24})/V

Okay now instead of writing velocities over and over again you can put a line on top of v to indicate 'the mean squared speed'...which is basically like an average. I don't know how to do that on here so I'm just going to strike it out like so

^{2}Now you also still an N infront to indicate how many molecules you are dealing with. You should end up with this:

P=Nm

^{2}/V

Now lets convert this from velocity to speed... speed has no direction. Since in the box the particles can go in x y or z (because its 3D ).

We will indicate speed with 'c'.

c

^{2}=v

^{2x }+v

^{2}

_{y }+ v

^{2z}

Particles are like to travel in all 3 directions, therefore we can just simplify this by using the mean velocity of all three and multiplying it by 3... we will also indicate mean speed with

^{2}= 3

^{2}

Now finally just sub this into your previous equation

P=Nm

^{2}/V

you should get P=Nm

^{2}/3V

Rearrange to make PV the subject on one side and you will get the formula you get in the exam

Times like this I really wish I knew how to use 'latex'

Hope that helps

EDIT: Sorry I saw a slight mistake when explaining the second step, I've put the change in bold

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#119

(Original post by

I really don't think they will ask us to derive it but I guess its good to know ...

consider a particle with 1 unit mass (m) travelling from one side of a cube (l) container to the other (therefore total distance travelled is 2l)

We know the change in momentum will be -2mv

F=-2mv/t

we know t is basically v=

F=-2mv/2l/v = -mv

Pressure is force/area therefore... since you know that area is l

p=mv

l^3 is the same as the volume so you can rewrite it...

p=mv

The total pressure will be all the pressures of each atom added up, because each will not be travelling at the same velocity...

P= mv

= m(v

Okay now instead of writing velocities over and over again you can put a line on top of v to indicate 'the mean squared speed'...which is basically like an average. I don't know how to do that on here so I'm just going to strike it out like so~~v~~

P=Nm~~v~~

Now lets convert this from velocity to speed... speed has no direction. Since in the box the particles can go in x y or z (because its 3D ).

We will indicate speed with 'c'.

c

Particles are like to travel in all 3 directions, therefore we can just simplify this by using the mean velocity of all three and multiplying it by 3... we will also indicate mean speed with~~c~~

~~c~~~~v~~

Now finally just sub this into your previous equation

P=Nm~~v~~

you should get P=Nm~~c~~

Rearrange to make PV the subject on one side and you will get the formula you get in the exam

Times like this I really wish I knew how to use 'latex'

Hope that helps

EDIT: Sorry I saw a slight mistake when explaining the second step, I've put the change in bold

**posthumus**)I really don't think they will ask us to derive it but I guess its good to know ...

consider a particle with 1 unit mass (m) travelling from one side of a cube (l) container to the other (therefore total distance travelled is 2l)

We know the change in momentum will be -2mv

F=-2mv/t

we know t is basically v=

**2**l/t rearranged... t=v/**2**l so sub that inF=-2mv/2l/v = -mv

^{2}/lPressure is force/area therefore... since you know that area is l

^{2}:p=mv

^{2}/l^{3}l^3 is the same as the volume so you can rewrite it...

p=mv

^{2}/VThe total pressure will be all the pressures of each atom added up, because each will not be travelling at the same velocity...

P= mv

^{21}/V + mv^{22}/V + mv^{23}/V + mv^{24}/V......= m(v

^{21}+ v^{22 }+ v^{23}+ v^{24})/VOkay now instead of writing velocities over and over again you can put a line on top of v to indicate 'the mean squared speed'...which is basically like an average. I don't know how to do that on here so I'm just going to strike it out like so

^{2}Now you also still an N infront to indicate how many molecules you are dealing with. You should end up with this:P=Nm

^{2}/VNow lets convert this from velocity to speed... speed has no direction. Since in the box the particles can go in x y or z (because its 3D ).

We will indicate speed with 'c'.

c

^{2}=v^{2x }+v^{2}_{y }+ v^{2z}Particles are like to travel in all 3 directions, therefore we can just simplify this by using the mean velocity of all three and multiplying it by 3... we will also indicate mean speed with

^{2}= 3^{2}Now finally just sub this into your previous equation

P=Nm

^{2}/Vyou should get P=Nm

^{2}/3VRearrange to make PV the subject on one side and you will get the formula you get in the exam

Times like this I really wish I knew how to use 'latex'

Hope that helps

EDIT: Sorry I saw a slight mistake when explaining the second step, I've put the change in bold

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