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

Reply 26

handsome7654

any1 have good notes for unit 5? :biggrin:

Reply 27

number23

15

can someone explain stopping potential to me? i really dont get it :frown:

Reply 28

Onee-chan

1

Original post by number23

can someone explain stopping potential to me? i really dont get it :frown:

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)" :rolleyes:

. 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 :biggrin:

Reply 29

number23

15

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)" :rolleyes:

. 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 :biggrin:

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:

KE=hf-work function-eV_s=0[br]V_s=(hf)/e-(work function)/e

And a graph of

V_s

against f would give gradient

h/e

and y-intercept

-(work function)/e

okay think i get it now :smile:

(edited 11 years ago)

Reply 30

kimmey

7

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 :frown:

Reply 31

don'tTRIP.

2

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 :frown:

I'm doing Turning Points, but I came across this video earlier...

http://www.brightstorm.com/science/physics/light/lens-equation/

Hope this is what you're looking for :smile:

Reply 32

number23

15

For people doing turning points, is this correct for relativity?

is

1/(1-v^2/c^2)

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

t_0

the proper time? ie the time calculated without considering special relativity?

Thanks, if anyone could add to this it would be great

Reply 33

number23

15

also how do muons help prove time dilation? :confused:

Reply 34

Onee-chan

1

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:

KE=hf-work function-eV_s=0[br]V_s=(hf)/e-(work function)/e

And a graph of

V_s

against f would give gradient

h/e

and y-intercept

-(work function)/e

okay think i get it now :smile:

No problem! Glad it helped. :biggrin:

Original post by number23

For people doing turning points, is this correct for relativity?

is

1/(1-v^2/c^2)

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

t_0

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? :confused:

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?

Reply 35

number23

15

Original post by Onee-chan

No problem! Glad it helped. :biggrin:

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 :smile:

i think we might need to know the definitions for proper time, work function and inertial reference frame

Reply 36

Onee-chan

1

Original post by number23

so there is time dilation of the muons travel time between the muon and the observer
thanks for that :smile:

i think we might need to know the definitions for proper time, work function and inertial reference frame

You're welcome! :biggrin:

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! :frown:

Reply 37

kimmey

7

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 :smile:

Reply 38

kimmey

7

Original post by don'tTRIP.

I'm doing Turning Points, but I came across this video earlier...

http://www.brightstorm.com/science/physics/light/lens-equation/

Hope this is what you're looking for :smile:

thanks, this was very helpful :smile:

Reply 39

callmenighthawk

2

can someone PLEASE help me with the

derivation.

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.

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

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