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OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012

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Original post by sweetascandy
What is a potentiometer ?! :O


essentially a variable resistor :cool:

Original post by sweetascandy
Also, what does a Zigzaggy line represent (if anything) in a circuit?!


pretty sure its also a potentiometer (us version i think) but a picture to clarify would be helpful
(edited 12 years ago)
Reply 341
Avatar for JazzyM94
JazzyM94
when working out the workfunction of a KE - Frequency graph, i got 1.657,
but according to the graph, work function should be negative
can i just make it -1.657, or an i doing something wrong?
Original post by JazzyM94
when working out the workfunction of a KE - Frequency graph, i got 1.657,
but according to the graph, work function should be negative
can i just make it -1.657, or an i doing something wrong?


Yeah you a right. How can a work function be negative? Just on that particular graph I think you have to change you answer to a negative.
This is the 3rd time im doing this unit.
First time got an 'e' and i knew i didnt do good after the exam :frown:
Second time got a 'd' - but this time i thought i did really well :confused::mad:

Hopefully 3rd time lucky? :colone:

Goodluck everyone!!

P.s We've never been taught Stopping potential and i dont seem to find it in the OCR Physics 1 textbook :confused:
Off to bed. Enjoy the exam guys.
Original post by JazzyM94
when working out the workfunction of a KE - Frequency graph, i got 1.657,
but according to the graph, work function should be negative
can i just make it -1.657, or an i doing something wrong?


No, because work function energy = the minimum energy required to release an electron, so it's always positive. I just did a question on this, where they told me to work out the work function from a graph and I didn't put the negative sign in but I was correct.

TL; DR: So don't put the negative sign infront of the work function energy. :smile:
Original post by The medjai
Off to bed. Enjoy the exam guys.


WAAIITTT...answer my question please on the page before please :biggrin:
Original post by JazzyM94
when working out the workfunction of a KE - Frequency graph, i got 1.657,
but according to the graph, work function should be negative
can i just make it -1.657, or an i doing something wrong?


the graph is rearranged to ke= hf - work function to fit in with y=mx+c so the y intercept of the graph is -ve work function. hope this helped
Original post by FristyKino
This is the 3rd time im doing this unit.
First time got an 'e' and i knew i didnt do good after the exam :frown:
Second time got a 'd' - but this time i thought i did really well :confused::mad:

Hopefully 3rd time lucky? :colone:

Goodluck everyone!!

P.s We've never been taught Stopping potential and i dont seem to find it in the OCR Physics 1 textbook :confused:


Got this off Yahoo answers but explains perfectly:

Stopping potential is the voltage required to stop an electron ejected by the metal. First find the energy of the incident light:
E=hf = hc/lamda
Then find the max kinetic energy of the ejected electrons:
K = hf - phi
where phi is the work function of the metal, and hf the energy of incident light. The max kinetic energy is also equal to:
K=eV
where e is the elementary charge and V is the stopping potential. So solve for V
Original post by sweetascandy
Also, what does a Zigzaggy line represent (if anything) in a circuit?!


Does it say in the question what it represents?
Reply 350
Avatar for SamXi
SamXi
7
Original post by The medjai
Got this off Yahoo answers but explains perfectly:

Stopping potential is the voltage required to stop an electron ejected by the metal. First find the energy of the incident light:
E=hf = hc/lamda
Then find the max kinetic energy of the ejected electrons:
K = hf - phi
where phi is the work function of the metal, and hf the energy of incident light. The max kinetic energy is also equal to:
K=eV
where e is the elementary charge and V is the stopping potential. So solve for V


You just find the voltage needed to make the current 0 then multiply the voltage by electron charge (e)

eV=E
Original post by The medjai
Got this off Yahoo answers but explains perfectly:

Stopping potential is the voltage required to stop an electron ejected by the metal. First find the energy of the incident light:
E=hf = hc/lamda
Then find the max kinetic energy of the ejected electrons:
K = hf - phi
where phi is the work function of the metal, and hf the energy of incident light. The max kinetic energy is also equal to:
K=eV
where e is the elementary charge and V is the stopping potential. So solve for V


Thanks alot!!
Original post by SamXi
You just find the voltage needed to make the current 0 then multiply the voltage by electron charge (e)

eV=E


That's a pretty cool way of working it out. The one I posted you can still use to check if answer is right.
Original post by The medjai
Does it say in the question what it represents?


Noooo, not that question. The question which I quoted you in. check post #355 PLEASE. I will love you forever. <3
Reply 354
Avatar for SamXi
SamXi
7
Goodluck fella's im off to bed !
Original post by SamXi
Goodluck fella's im off to bed !


Waaaiittt... ca you answer my question please?
The question which I quoted you in. check post #355 PLEASE. I will love you forever. <3
Original post by sweetascandy
Waaaiittt... ca you answer my question please?
The question which I quoted you in. check post #355 PLEASE. I will love you forever. <3


Reflected from the pulley. (and a node is produced at this point)
Original post by goerigi
i dont think we need to know why, just it does, and how they affect each other


We are meant to know; it's in the spec.

It's because:
the maximum kinectic energy of the electrons is independent of intensity because only 1 photon will interact with 1 electron... however, increasing intensity increases the number of photons that hit the photocell per second... if more photons are interacting with more electrons per second, then more electrons will be released per second and allowed to flow around the circuit... more charge carriers per second results in an increase in current (Dale's words)
Original post by sweetascandy
We are meant to know; it's in the spec.

It's because:
the maximum kinectic energy of the electrons is independent of intensity because only 1 photon will interact with 1 electron... however, increasing intensity increases the number of photons that hit the photocell per second... if more photons are interacting with more electrons per second, then more electrons will be released per second and allowed to flow around the circuit... more charge carriers per second results in an increase in current (Dale's words)



And the photons must have energy = work function
so that the electrons can be released. (yes?)
Original post by FristyKino
And the photons must have energy = work function
so that the electrons can be released. (yes?)


Yes. But not only must the electrons have the work function energy, the frequency of their radiation must also be equal to or greater than the threshold frequency.

Do you know this or would you like me to go into detail about what the threshold frequency is, etc.?

Hope it helps btw! :smile:

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