Physics: Post your doubts here!

[Shreeram]

Hi can anyone please help me with O/N 2010 P23 Q 4 c i, ii Thanks

 

[1357913579]

Forget the photoelectric effect for a moment. Think about what is meant by the intensity of the radiation of the target surface.

Intensity = power / [target area]
Rearranges to:
power = intensity * area

If you are keeping the intensity constant, and the area of the photo-target is constant, then the power of the incident radiation must also be constant.

Each photon brings with it a certain amount of energy. If you reduce the wavelength, you increase the amount of energy carried by each photon.

The energy delivered = (energy of a photon) * (number of photons)

power = energy / time

So:
power = (energy of a photon) * (number of photons) / time

If the power is constant, and the energy of a photon is increased, then (number of photons / time) must decrease.

Now we are back to the photoelectric effect.
Fewer photons/s = fewer electrons/s = smaller current.

EDIT:

Nope!

The charge on an electron is fixed.
The total charge delivered to the photo-anode = (charge on electron ) * (number of electrons)

Electric current = [charge] / [time]

Current = (charge on electron ) * (number of electrons)/(time)

If you have fewer electrons/second emitted by the photo-surface, you have a smaller current.

How quickly the electrons travel does not change that. (this is what my friend said i got no idea about this)

I have a doubt in this
isnt your explanation valid if electron behave like wave but here we are talking abut photo electric effect

 

[1357913579]

I have a doubt in this
isnt your explanation valid if electron behave like wave but here we are talking abut photo electric effect

More intense radiation simply means that more packets of energy (photons) are delivered each second. But the energy of each packet is unchanged. So if there wasn't enough energy to cause photoelectric emission, making it brighter won't change anything,

 

[daredevil]

my teacher taught me tht...
it isnt thr in d book??

nope it isn't in the book and its the first time i've heard of it as far as i know
snowbrood hav u heard of this before??

 

[daredevil]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_22.pdf

in question 5b
why do we take the voltage to be 12V (as per the ms) .... shouldn't we take it as (12-3) = 9V??
I was calculated correctly in the previous part to be 2.5A
and then
P=IV
=(2.5)(9)
=22.5W ??
the answer given in the ms is 30W
what am i not getting at here?

 

[snowbrood]

nope it isn't in the book and its the first time i've heard of it as far as i know
snowbrood hav u heard of this before??

well i just sort it out e is elementary charge we know V=w/q so W=QV=eV simple as that thats how i just find it out dont panic just think and u will go way too ahead

 

[snowbrood]

More intense radiation simply means that more packets of energy (photons) are delivered each second. But the energy of each packet is unchanged. So if there wasn't enough energy to cause photoelectric emission, making it brighter won't change anything,

well i am in AS level this is the answer that my friend gave me i dont know what it means lol

 

[daredevil]

well i just sort it out e is elementary charge we know V=w/q so W=QV=eV simple as that thats how i just find it out dont panic just think and u will go way too ahead

wooaah fancy working out to be done in a paper where u have got just one hour -_-
thanks btw,... i do get it now but how r v supposed to think it over in the paper is what bugs me :O

 

[Tabi Sheikh]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_22.pdf

in question 5b
why do we take the voltage to be 12V (as per the ms) .... shouldn't we take it as (12-3) = 9V??
I was calculated correctly in the previous part to be 2.5A
and then
P=IV
=(2.5)(9)
=22.5W ??
the answer given in the ms is 30W
what am i not getting at here?

The question specifically asked for the power transformed by battery B that is why we will use 12V in calculation.

 

[Soldier313]

Aoa wr wb
Can someone please explain
5 b ii - ?
10 b - please provide a detailed explanation for the diagram
11 - i don't understand the last phrase in the ms where it says the crystal is cut.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_42.pdf

Thank you

 

[daredevil]

The question specifically asked for the power transformed by battery B that is why we will use 12V in calculation.

yeah sorry... i just realized my mistake after i posted the question thanks anyways

 

[Tabi Sheikh]

Hi can anyone please help me with O/N 2010 P23 Q 4 c i, ii Thanks

since he measured the distance from the top f ball but when the ball hit the table only the bottom of the ball hits the table therefore the y intercept of graph is the diameter of the ball that is approximately 1.75cm according to my line.

 

[Tabi Sheikh]

yeah sorry... i just realized my mistake after i posted the question thanks anyways

ur welcome

 

[daredevil]

The question specifically asked for the power transformed by battery B that is why we will use 12V in calculation.

ok but wat about the last part of the question... how do we know that the circuit can be used to store energy in A?
i mean the marking scheme does say that because power in B is greater than energy dissipated per unit time but i don;t get it.... can u explain it to me please?

 

[Tabi Sheikh]

The key to the answer is in part b iii where we calculate the "power dissipated by the resistors" as it is less than the power supplied by the battery B therefore we can say that we can store rest of the energy in battery A.

 

[Tabi Sheikh]

Aoa wr wb
Can someone please explain
5 b ii - ?
10 b - please provide a detailed explanation for the diagram
11 - i don't understand the last phrase in the ms where it says the crystal is cut.

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_42.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_ms_42.pdf

Thank you

10 b)


since we only want the current to flow from positive to negative therefore we use the diode to forward bias he circuit

 

[Soldier313]

10 b)

View attachment 24228
since we only want the current to flow from positive to negative therefore we use the diode to forward bias he circuit

Thanx bro
But um the ms says that there should be a second diode in the circuit, i am confused as to where do i have to place it.

 

[Soldier313]

Aoa wr wb

Can someone please draw for me the diagrams for 7b i and 7b ii ?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_41.pdf

Thank you

 

[Tabi Sheikh]

Thanx bro
But um the ms says that there should be a second diode in the circuit, i am confused as to where do i have to place it.

u can put another diode to protect the relay its always parallel to the relay and is reverse biased always it used for protection only .... but not sure about it whether ms talk abut the same BTW Ur welcome

 

[Tabi Sheikh]

i

Aoa wr wb

Can someone please draw for me the diagrams for 7b i and 7b ii ?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_ms_41.pdf

Thank you

Diode A will act as half wave rectifier as it will on be active while the voltage is positives when voltage is negative it will be zero
second part will be the other way round when negative voltage then positive voltage but when positive i will be zero
in diagram i have just pointed out the points where it will be zero the others on both graphs will be positive Curve .
BTW u can switch the graphs .....