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AQA Physics A - PHYA5 (18/06/12) - Exam thread

Physics exam discussion - share revision tips in preparation for GCSE, A Level and other physics exams and discuss how they went afterwards.

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1. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by number23)
thanks, how are you finding it?
No worries. Turning Points is fairly straightforward IMO, just a few topics I need to brush up on, should be okay. You?
2. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by don'tTRIP.)
No worries. Turning Points is fairly straightforward IMO, just a few topics I need to brush up on, should be okay. You?
Well atm im just finishing Unit 5 section a notes, then ill do notes for turning points.. hopefully should be okay after that
3. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
You know the molecular theory model in unit 5, how much do we need to know about the equation: ? do we need to derive it? thanks
4. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
for people doing turning points:

which method for finding the e/m ratio do we need to know?
5. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by number23)
You know the molecular theory model in unit 5, how much do we need to know about the equation: ? do we need to derive it? thanks
Yes, unfortunately you do need to be able to derive it.
6. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by leeandrewarmstrong)
Yes, unfortunately you do need to be able to derive it.
thanks. also, in my textbook there's abit about measuring the wavelength of radio waves, but its not in the specification. do we need to know it ? thanks
7. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
any1 have good notes for unit 5?
8. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
can someone explain stopping potential to me? i really dont get it
9. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by number23)
can someone explain stopping potential to me? i really dont get it
The stopping potential is basically the p.d. you need to apply in order to do sufficient work to make the electrons lose all its kinetic energy.

Detailed explanation (really long winded sorry!):

Okay, the circuit is basically a photocell (consisting of an anode and the metal plate acting as a cathode) and a microammeter all connected to a potential divider with a voltmeter measuring p.d. across potential divider.

At first, the potential divider is adjusted so there is no current through the microammeter.

So say light is shone and we have electrons with KE = hf - work function. As this is happening in a photocell, as soon the electron is emitted from the metal surface, it moves towards the anode (in the photocell), because it's acted on by an electric force. At this point there is a current through the microammeter.

But, if we turn the potential divider up so it inputs a p.d. into the circuit, which acts in the opposite direction to the p.d./ current generated by the photocell. I.e.: it makes the metal cathode increasingly positive - the electrons need increasing energy to escape the metal surface.

In equation form:

KE of electrons = hf - work function (as before) - additional energy due to work done by the p.d. (= eV since W=qV)

The potential difference at which the kinetic energy of the electrons is zero is, well, imaginatively called the "stopping potential (Vs)". At this point the microammeter should again have zero reading.

Therefore in equation form:

KE = hf - work function - eVs = 0

Rearrange gives:

eVs = hf - work function.

You can then divide both sides by e to give you Vs = hf/e - work function/e and you can plot a graph of Vs against f to get h and work function.

That's it I think? If there are any further bits you don't understand feel free to ask
10. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by Onee-chan)
The stopping potential is basically the p.d. you need to apply in order to do sufficient work to make the electrons lose all its kinetic energy.

Detailed explanation (really long winded sorry!):

Okay, the circuit is basically a photocell (consisting of an anode and the metal plate acting as a cathode) and a microammeter all connected to a potential divider with a voltmeter measuring p.d. across potential divider.

At first, the potential divider is adjusted so there is no current through the microammeter.

So say light is shone and we have electrons with KE = hf - work function. As this is happening in a photocell, as soon the electron is emitted from the metal surface, it moves towards the anode (in the photocell), because it's acted on by an electric force. At this point there is a current through the microammeter.

But, if we turn the potential divider up so it inputs a p.d. into the circuit, which acts in the opposite direction to the p.d./ current generated by the photocell. I.e.: it makes the metal cathode increasingly positive - the electrons need increasing energy to escape the metal surface.

In equation form:

KE of electrons = hf - work function (as before) - additional energy due to work done by the p.d. (= eV since W=qV)

The potential difference at which the kinetic energy of the electrons is zero is, well, imaginatively called the "stopping potential (Vs)". At this point the microammeter should again have zero reading.

Therefore in equation form:

KE = hf - work function - eVs = 0

Rearrange gives:

eVs = hf - work function.

You can then divide both sides by e to give you Vs = hf/e - work function/e and you can plot a graph of Vs against f to get h and work function.

That's it I think? If there are any further bits you don't understand feel free to ask
thanks, this is the main thing im struggling with in turning points >.<

so the kinetic energy of the photons is reduced by the work function.. and when you apply a potential difference it is reduced even more. When the kinetic energy=0, the corresponding potential=stopping potential?

so:

And a graph of against f would give gradient and y-intercept

okay think i get it now
Last edited by number23; 29-05-2012 at 15:42.
11. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
im doing medical option, i havent been taught it at all and i have to self learn it, its going okay, i use the purple revision guide. does anyone doing this know how to do lens equation questions, i jus cant do it
12. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by kimmey)
im doing medical option, i havent been taught it at all and i have to self learn it, its going okay, i use the purple revision guide. does anyone doing this know how to do lens equation questions, i jus cant do it
I'm doing Turning Points, but I came across this video earlier...

http://www.brightstorm.com/science/p...lens-equation/

Hope this is what you're looking for
13. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
For people doing turning points, is this correct for relativity?

is the factor you multiply your 'stationary' lengths/times by to get the new lengths/times as a result of the effects of special relativty?

Also, is the proper time? ie the time calculated without considering special relativity?

Thanks, if anyone could add to this it would be great
14. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
also how do muons help prove time dilation?
15. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by number23)
thanks, this is the main thing im struggling with in turning points >.<

so the kinetic energy of the photons is reduced by the work function.. and when you apply a potential difference it is reduced even more. When the kinetic energy=0, the corresponding potential=stopping potential?

so:

And a graph of against f would give gradient and y-intercept

okay think i get it now

(Original post by number23)
For people doing turning points, is this correct for relativity?

is the factor you multiply your 'stationary' lengths/times by to get the new lengths/times as a result of the effects of special relativty?

Also, is the proper time? ie the time calculated without considering special relativity?

Thanks, if anyone could add to this it would be great
This factor is what you use to multiply the proper time and mass to get the relativistic time/ mass. But for length, it's the other way around. I.e.: you times the relativistic length by this to get the proper mass.

T0 is the proper time.

(Original post by number23)
also how do muons help prove time dilation?
I think it's because experiments have shown them to have an extremely short half-life, so if relativity is not taken into account, they would only travel about 500m or some other figure from cosmic rays. Yet scientists have measured them much further than that.

Relativity explains it because the time measured is the relativistic time so it's much longer than the proper time/ half life of the muons. Speed is unchanged so the distance travelled when measured by scientists is much longer than by an observer travelling at the same speed as the muons.

Therefore this supports relativity.

Do you guys know if we need to learn any definitions for Turning Points?
16. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by Onee-chan)

This factor is what you use to multiply the proper time and mass to get the relativistic time/ mass. But for length, it's the other way around. I.e.: you times the relativistic length by this to get the proper mass.

T0 is the proper time.

I think it's because experiments have shown them to have an extremely short half-life, so if relativity is not taken into account, they would only travel about 500m or some other figure from cosmic rays. Yet scientists have measured them much further than that.

Relativity explains it because the time measured is the relativistic time so it's much longer than the proper time/ half life of the muons. Speed is unchanged so the distance travelled when measured by scientists is much longer than by an observer travelling at the same speed as the muons.

Therefore this supports relativity.

Do you guys know if we need to learn any definitions for Turning Points?
so there is time dilation of the muons travel time between the muon and the observer
thanks for that

i think we might need to know the definitions for proper time, work function and inertial reference frame
17. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by number23)
so there is time dilation of the muons travel time between the muon and the observer
thanks for that

i think we might need to know the definitions for proper time, work function and inertial reference frame
You're welcome!

Is there a definition page like the other topics? Thanks.

How are the other topics going? I realised that I've basically forgotten the radioactivity topic and the grade boundaires are ridiculously high!
18. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
hi does anyone have the mark scheme to the exam style questions right at the end of the book. i have chapters 9-12 but i dont have the ones right at the end. if anyone could post a link to them i would be eternally grateful
19. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread
(Original post by don'tTRIP.)
I'm doing Turning Points, but I came across this video earlier...

http://www.brightstorm.com/science/p...lens-equation/

Hope this is what you're looking for
20. Re: AQA Physics A - PHYA5 (18/06/12) - Exam thread

my teacher said we won't need to know how to derive it fully, just understand why it is derived like that (what laws and principles are happening) as it's never come up as fully deriving it.

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