May 2015- Physics Paper 1 and Paper 2

[Sofia.]

Guys I have a question, for the binding energy per nuclei graph in q2 p2, the be per nucleon axis had values, will marks be deducted if I ignored them because I don't know where the peak is on the y axis, on the x it's 56 but on the y I had no idea,

 

My peak on the x-axis was way off haha! The peak is somewhere around 8-9, but as long as it's higher than uranium you should be fine I think. 

[Megamind]

 

But how Ossih, isn't because they are in parallel, they will have the same voltage, otherwise I am stupid because I could have write 6 even without understanding, and I wish they Ecf my answer to the next question

Concerning the damping question Ossih, do you think that the liquid will be damped in this significant way ? So consider water oscillating in a bottle ?

 

If that's the paper 1 question, I think it would be damped. I mean, just thinking about it, if you shake a bottle of water, the water is going to stop sooner or later!

 

 

Couldn't the utube be frictionless? :/ 

[Hassan76]

I think they have to tell you

First when I read "u tube" I thought it is something in the syllabus that I didn't review

Guys I have a question, for the binding energy per nuclei graph in q2 p2, the be per nucleon axis had values, will marks be deducted if I ignored them because I don't know where the peak is on the y axis, on the x it's 56 but on the y I had no idea,

 

My peak on the x-axis was way off haha! The peak is somewhere around 8-9, but as long as it's higher than uranium you should be fine I think.

Where ,?

[Ossih]

 

 

Ossih what about the 2nd part of the 9th question about atoms, wave functions..... How did you find it?

 

Find what? The uncertainty in momentum, you use ∆x∆p is approximately h/4π, and ∆x is the length of the box.

 

Then the momentum = h/wavelength

 

And for the graph, its a standing wave in its 6th harmonic.

No, I mean how do you find the question

And how should I now that it is at the 6 th harmonic, i didn't know the length, they gave me the position?

 

But how Ossih, isn't because they are in parallel, they will have the same voltage, otherwise I am stupid because I could have write 6 even without understanding, and I wish they Ecf my answer to the next question

Concerning the damping question Ossih, do you think that the liquid will be damped in this significant way ? So consider water oscillating in a bottle ?

 

If that's the paper 1 question, I think it would be damped. I mean, just thinking about it, if you shake a bottle of water, the water is going to stop sooner or later!

Yes, but also we need to consider that the graph shows light damping

 

 

Oh the question was interesting And oh you do have the wavelength! That's how you got p = h/lambda isnt it? (or was it the other way around, regardless, you know wavelength). If you do the total length of the box divided by wavelength, that'll tell you how many waves to fit It was 6 so I drew the 6th harmonic. 

[Sofia.]

Where on the x-axis? Mine was definitely before 50, couldn't remember the atomic no of iron. 

[Hassan76]

There will be a range I think Sofia as in m14 tz2 hp2 q3, do not worry

[Vioh]

 

No not at all. Intensity is not a measure of the number of the photons but rather a measure of the total energy of photons.

 

Thanks for correcting me! I was definitely very wrong when I said intensity is the number of photons. So to avoid this, let's completely ignore the "intensity" all together (as the question doesn't even mention it anyway). However, I still stand my ground that the current does increase.

 

But even if this does increase their speed, this does not mean that the current increases! The current is the charge over time. The total charge is determined by the total number of electrons, NOT on their speeds or energies. The total number of charge is the same.

 

BUT kinetic energy of the electron doesn't have anything to do with the current! Rather, it affects the voltage because eV = 1/2mv^2 for an electron remember from Topic 5? You will increase the potential difference but NOT the actual charge/current. The current is solely the charge divided by time, and the charge hasn't changed.

 

I think you have this picture in your head that they're moving faster, and maybe they are, but the effect of that, even though its contrary to intuition, is to increase the potential difference.

 

Let me be a little bit clearer with my previous explanation:

According to many sources (wiki, book, etc.), the current is defined as: "the amount of electric charge passing across a surface in one unit of time (e.g. second)"

 

Imagine if you have a surface where all the emitted electrons are positioned on the left-hand-side of this surface. The electrons will have to flow across this surface to reach the other side. Let's suppose that in this situation, there are 'n' electrons passing through the surface per second. Thus, the current is 'n' by definition.

 

Now, we'll increase each electron's velocity by 2 (this can be done by increasing KE), then obviously those 'n' electrons can now pass through the surface within the period of only 0.5 seconds. This means that '2n' electrons will pass through the surface in 1 second. Thus the current has been increased by 2 to become '2n'.

 

From this conceptual picture, then increasing the kinetic energy of the electrons will increase the velocity, which will consequently increase the current as well. This is also in agreement with the Tsokos physics book. Here is an excerpt from page 390 (sorry for the bad quality):

 

 

Also, remember that all I'm saying is that current must increase. But regarding the claim that the stopping voltage becomes more negative, we are perfectly in agreement.

Edited May 10, 2015 by Vioh

[Ossih]

 

 

But how Ossih, isn't because they are in parallel, they will have the same voltage, otherwise I am stupid because I could have write 6 even without understanding, and I wish they Ecf my answer to the next question

Concerning the damping question Ossih, do you think that the liquid will be damped in this significant way ? So consider water oscillating in a bottle ?

 

If that's the paper 1 question, I think it would be damped. I mean, just thinking about it, if you shake a bottle of water, the water is going to stop sooner or later!

 

 

Couldn't the utube be frictionless? :/ 

 

I mean I guess it could but, I'm sure they would state it if it were frictionless. And I don't think it damps cause of friction, you know the fact that if you have a thing like with many places where water can go, water will be at the same level due to the air pressure right? The air pressure on both sides balances it out. Like this image http://www.uq.edu.au/_School_Science_Lessons/12.2.5a.GIF thats it's natural position cause of air pressure and pascal's law (although Pascal's law isn't on the IB syllabus).

[Hassan76]

Ossih what about the 2nd part of the 9th question about atoms, wave functions..... How did you find it?

 

Find what? The uncertainty in momentum, you use ∆x∆p is approximately h/4π, and ∆x is the length of the box.

 

Then the momentum = h/wavelength

 

And for the graph, its a standing wave in its 6th harmonic.

No, I mean how do you find the question

And how should I now that it is at the 6 th harmonic, i didn't know the length, they gave me the position?

But how Ossih, isn't because they are in parallel, they will have the same voltage, otherwise I am stupid because I could have write 6 even without understanding, and I wish they Ecf my answer to the next question

Concerning the damping question Ossih, do you think that the liquid will be damped in this significant way ? So consider water oscillating in a bottle ?

 

If that's the paper 1 question, I think it would be damped. I mean, just thinking about it, if you shake a bottle of water, the water is going to stop sooner or later!

Yes, but also we need to consider that the graph shows light damping

 

Oh the question was interesting And oh you do have the wavelength! That's how you got p = h/lambda isnt it? (or was it the other way around, regardless, you know wavelength). If you do the total length of the box divided by wavelength, that'll tell you how many waves to fit It was 6 so I drew the 6th harmonic.

But I didn't have ether length this was the position from the nucleus, and the graph was wave function against position, the more wider the curve the less the uncertainty in momentum, and this question was in HpM N10, you can check it

What about the feature ?

[Ossih]

 

 

No not at all. Intensity is not a measure of the number of the photons but rather a measure of the total energy of photons.

 

Thanks for correcting me! I was definitely very wrong when I said intensity is the number of photons. So to avoid this, let's completely ignore the "intensity" all together (as the question doesn't even mention it anyway). However, I still stand my ground that the current does increase.

 

But even if this does increase their speed, this does not mean that the current increases! The current is the charge over time. The total charge is determined by the total number of electrons, NOT on their speeds or energies. The total number of charge is the same.

 

BUT kinetic energy of the electron doesn't have anything to do with the current! Rather, it affects the voltage because eV = 1/2mv^2 for an electron remember from Topic 5? You will increase the potential difference but NOT the actual charge/current. The current is solely the charge divided by time, and the charge hasn't changed.

 

I think you have this picture in your head that they're moving faster, and maybe they are, but the effect of that, even though its contrary to intuition, is to increase the potential difference.

 

Let me be a little bit clearer with my previous explanation:

According to many sources (wiki, book, etc.), the current is defined as: "the amount of electric charge passing across a surface in one unit of time (e.g. second)"

 

Imagine if you have a surface where all the emitted electrons are positioned on the left-hand-side of this surface. The electrons will have to flow across this surface to reach the other side. Let's suppose that in this situations, there are 'n' electrons passing through the surface per second (i.e. the current is 'n').

 

Now, we'll increase each electron's velocity by 2 (this can be done by increasing KE), then obviously those 'n' electrons can now pass through the surface within the period of only 0.5 seconds. This means that '2n' electrons will pass through the surface in 1 second. Thus the current has been increased by 2 to become '2n'.

 

From this conceptual picture, then increasing the kinetic energy of the electrons will increase the velocity, which will consequently increase the current as well. This is also in agreement with the Tsokos physics book. Here is an excerpt from page 390 (sorry for the bad quality):

 

Untitled.png

 

Also, remember that all I'm saying is that current must increase. But regarding the claim that the stopping voltage becomes more negative, we are perfectly in agreement.

 

 

Hm, maybe you're right :$ I'm just not convinced that there is a higher speed, or even if there is, I don't believe that it leads to a higher current because think about it like this

 

Say in the before situation, I had 10 electrons that whizzed off 1 s at a time

in this new situation, you say that the same 10 electrons whizz off faster. But I find it hard to believe that the current depends on the speed of the electron. I agree that current is the charge per unit time at a specific place. But those 10 electrons, are still being fired 1s at a time, they just reach the other side faster. Each cross-section of wire isn't really experiencing more charge per unit time, is it? The way I see it is that the electrons have more kinetic energy, so yes, they may go faster but this extra energy is used to create a potential difference

 

Or is there a fallacy in the way I'm thinking?

[Hassan76]

Guys could you simplify what are you trying to answer so I could try to help please ?

[Ossih]

 

 

 

 

Ossih what about the 2nd part of the 9th question about atoms, wave functions..... How did you find it?

 

Find what? The uncertainty in momentum, you use ∆x∆p is approximately h/4π, and ∆x is the length of the box.

 

Then the momentum = h/wavelength

 

And for the graph, its a standing wave in its 6th harmonic.

No, I mean how do you find the question

And how should I now that it is at the 6 th harmonic, i didn't know the length, they gave me the position?

 

But how Ossih, isn't because they are in parallel, they will have the same voltage, otherwise I am stupid because I could have write 6 even without understanding, and I wish they Ecf my answer to the next question

Concerning the damping question Ossih, do you think that the liquid will be damped in this significant way ? So consider water oscillating in a bottle ?

 

If that's the paper 1 question, I think it would be damped. I mean, just thinking about it, if you shake a bottle of water, the water is going to stop sooner or later!

Yes, but also we need to consider that the graph shows light damping

 

Oh the question was interesting And oh you do have the wavelength! That's how you got p = h/lambda isnt it? (or was it the other way around, regardless, you know wavelength). If you do the total length of the box divided by wavelength, that'll tell you how many waves to fit It was 6 so I drew the 6th harmonic.

But I didn't have ether length this was the position from the nucleus, and the graph was wave function against position, the more wider the curve the less the uncertainty in momentum, and this question was in HpM N10, you can check it

What about the feature ?

 

 

Hm I think it was the wave length of the wave, I can't remember, but I don't remember reading position from the nucleus. If it were, then you would we right, but then there would be no specific harmonic for the question, and so I doubt it would merit 3 marks (I think it was 3). And the feature was the square of the amplitude. I understand what you mean by the HPM N10 question, but the wave is spread out all over the length of the box (cause the harmonic fills the box, doesn't it?) so really , the uncertainty in position is still that whole box.

[Sofia.]

They're trying to answer the question on paper 1 about the graph of the new photocurrent when the light was switched from red to blue

[Hassan76]

I think the 2 marks for the feature were want about saying that the electron fits a standing wave model, and that the square of the amplitude is proportional to the probability of finding it at a certain distance from the nucleus

They're trying to answer the question on paper 1 about the graph of the new photocurrent when the light was switched from red to blue

Yes I know, Sofia, but I want to know what are they debating/ discussing, I mean what question are they asking

[Sofia.]

Ah sorry! Whether the photocurrent is higher because the electrons move faster

[Hassan76]

Ah sorry! Whether the photocurrent is higher because the electrons move faster[/quote

Thank you Sofia, I don't mean to bother you

[Vioh]

 

Hm, maybe you're right :$ I'm just not convinced that there is a higher speed, or even if there is, I don't believe that it leads to a higher current because think about it like this

 

Say in the before situation, I had 10 electrons that whizzed off 1 s at a time

in this new situation, you say that the same 10 electrons whizz off faster. But I find it hard to believe that the current depends on the speed of the electron. I agree that current is the charge per unit time at a specific place. But those 10 electrons, are still being fired 1s at a time, they just reach the other side faster. Each cross-section of wire isn't really experiencing more charge per unit time, is it? The way I see it is that the electrons have more kinetic energy, so yes, they may go faster but this extra energy is used to create a potential difference

 

Or is there a fallacy in the way I'm thinking?

 

 

Well, the kinetic energy IS the measure of the velocity of a particle (because KE = 1/2mv^2). So if you increase the kinetic energy, it's equivalent to saying that the velocity has increased.

 

It's true that the total number electrons getting emitted is exactly the same (as indicated by the total number of photons), but in the later case, all of these electrons move faster (i.e. takes less time). Mathematically, current = total charge divided by time. Total charge stays the same, but time decreases. This means that current increases.

 

Let's now consider the case that you set up there with those 10 electrons. Suppose that these 10 electrons pass through an ammeter in 1 second, then the current is 10e Amperes. Now, increasing the velocity by 2, we will have the 10 electrons passing through that ammeter in 0.5 second, then the current now is 10e/0.5 = 20e Amperes. So current does increase.

 

EDIT: Regarding your last question ("this extra energy is used to create a potential difference"), the answer is that the increase in potential energy is to stop those electrons from moving further. In other words, all the kinetic energy of the electrons will get converted to the potential energy of the electrons

Edited May 10, 2015 by Vioh

[Hassan76]

Look guys, this is simply as in Oxford study guide: the kinetic energy is the energy that remains ,Maths extra

For example consider you enter the cinema and you buy a ticket for 10 pounds, but your father gave you 11 pounds, 10 pounds is the work function ( h f.node), 11 pounds is the hf, and the remaining which is 1 pound is the maximum ke, the additional money, if it is 1 pound he cannot do a lot with it,mouth for example if it's 10 pounds he could buy a lot of things What do you think, and in the Oxford study guide there is a graph that shows that for 2 uv lights of same frequency but different intensity, the photo current is different but Vs is the same, the maximum ke is related to Vs by eVs. Charge is not changed, so vs must be changed

What do you think guys?

[Ossih]

 

 

Hm, maybe you're right :$ I'm just not convinced that there is a higher speed, or even if there is, I don't believe that it leads to a higher current because think about it like this

 

Say in the before situation, I had 10 electrons that whizzed off 1 s at a time

in this new situation, you say that the same 10 electrons whizz off faster. But I find it hard to believe that the current depends on the speed of the electron. I agree that current is the charge per unit time at a specific place. But those 10 electrons, are still being fired 1s at a time, they just reach the other side faster. Each cross-section of wire isn't really experiencing more charge per unit time, is it? The way I see it is that the electrons have more kinetic energy, so yes, they may go faster but this extra energy is used to create a potential difference

 

Or is there a fallacy in the way I'm thinking?

 

 

Well, the kinetic energy IS the measure of the velocity of a particle (because KE = 1/2mv^2). So if you increase the kinetic energy, it's equivalent to saying that the velocity has increased.

 

It's true that the total number electrons getting emitted is exactly the same (as indicated by the total number of photons), but in the later case, all of these electrons move faster (i.e. takes less time). Mathematically, current = total charge divided by time. Total charge stays the same, but time decreases. This means that current increases.

 

Let's now consider the case that you set up there with those 10 electrons. Suppose that these 10 electrons pass through an ammeter in 1 second, then the current is 10e Amperes. Now, increasing the velocity by 2, we will have the 10 electrons passing through that ammeter in 0.5 second, then the current now is 10e/0.5 = 20e Amperes. So current does increase.

 

EDIT: Regarding your last question ("this extra energy is used to create a potential difference"), the answer is that the increase in potential energy is to stop those electrons from moving further. In other words, all the kinetic energy of the electrons will get converted to the potential energy of the electrons

 

 

no but that's what im saying, its not the case of 10 electrons passing through that ammeter in 0.5s. Each individual electron is faster BUT the time after which each electron whizzes off is the same, because there's a one-to-one instantaneous correspondence between a photon hitting and an electron hitting.

 

What I'm saying is that initially, each of those 10 electrons travelled at say 1m/s and travelled say 12m of wire. Each electron was shot with a delay of 1s in between them. Now, those same electrons are travelling at 2m/s, still travelling 12m of wire, still a 1s delay between them. Just the shots are faster, but the overall movement of charge isn't faster itself! Do you get what I'm trying to say haha.

 

I can't think of a good analogy. It's a matter of firing faster bullets and slower bullets, but the time interval between bullets is the same, so overall, the number of bullets per unit time is the same.

[Hassan76]

Ossih what temperature did you get for the thermodynamics question in q7, and the efficiency was 47% right, and what did you answer for the degraded energy?