HSC Physics Question Thread

[jakesilove]

This is how my physics teacher explained this specific question to us - not 100% sure though ._.

Aha yeah it's totally fair enough, talking about forces is complicated and teachers tend to over simplify to the point of not really making sense. Which is why you can basically write whatever and not get penalised for it, because people teach in so many different ways! I just personally hate pseudo-Physics words that make Physics even more complicated than it needs to be (ie. slowing down is accelerating, speeding up is accelerating, we don't need two different words aha)

[teapancakes08]

Hey! If an object DOES WORK, it's going to have to work AGAINST SOMETHING. If an object has work DONE ON IT, that work has to be produced by SOMETHING. So, if a satellite has work DONE ON IT, how could that happen? Well, the gravitational field will do the work, thus PUSHING the satellite downwards! So, it moves to a lower orbit.

It sounds like you actually do have a good understanding of the topic! I think it's easiest to use a diagram in situations like this; hopefully the below gives you all the information you need! Let me know if I can clarify anything.

(Image removed from quote.)

Oh, I see now. So if the object is moved away from the other body/object then it is working against the gravitational field, hence it moves to a high orbit. And then if it falls back down the field is doing the work on the object, thus the GPE is being converted into kinetic energy, hence it moves to a lower orbit.

Thanks so much for the explanation ^^

[Cindy2k16]

Hi could someone explain how to 2014 HSC Q26 b)? I completely don't understand the working out provided in the answer especially since I've never done calculations where there's a non-zero voltage across the electrodes for the photoelectric effect.
Thanks in advance! (also how do you upload images on posts? do you have to upload them to imgur first?)

[Happy Physics Land]

Hi could someone explain how to 2014 HSC Q26 b)? I completely don't understand the working out provided in the answer especially since I've never done calculations where there's a non-zero voltage across the electrodes for the photoelectric effect.
Thanks in advance! (also how do you upload images on posts? do you have to upload them to imgur first?)

Hey Cindy!

No problems! A conceptually difficult question on the stopping voltage experiment here, requires a bit of calculations as well! What's so difficult about it are two things:

1. Understanding the significance of stopping voltage and maximum kinetic energy
2. Applying and understanding the formula E=qV

Here at Atarnotes we want the best for our students, so I will hold your hand through this.

The diagram of the experiment indicates 4.1V as the stopping voltage, which is the point at which the voltage source provides enough energy to repel the ejected electron so that it doesnt reach the anode (or collector) and hence no current will be detected. The point of having a stopping voltage is just so that we can work out the work function, which is the minimum energy required to eject electrons from a metal.

I will prove to you why this is the case:

Maximum kinetic energy = hf - work function _____ equation 1

This is the equation for calculating the kinetic energy of ejected electrons as a light beam shines upon the metal.

But since stopping voltage works against the motion of electrons:

Maximum kinetic energy = hf - qV ________ equation 2

Hence by balancing equation 2 and equation 1, we can see that work function = qV
So work function of the metal = 4.1eV

Good good, now the difficult part is over, congratulations.

We need to do two things for this question

1. Draw the line on the graph for the experiment done with 0.0V

To draw the line, we must know its x-intercept, which is the threshold frequency. We can calculate threshold frequency through the formula:

Work function = planck's constant x threshold frequency

We know work function = 4.1eV = 4.1 x (1.602 x 10^-19)J
We know planck's constant = 6.632 x 10^-34
Substitute in the values:

4.1 x (1.602 x 10^-19)J = 6.632 x 10^-34 x threshold frequency

Hence threshold frequency of the metal = 9.9 x 10¹⁴ Hz. So your line for the repeated experiment should have an x-intercept at 9.9. Now how can I draw this line without knowing another point? Easy, because we know that the gradient is always constant for kinetic energy versus frequency graphs because it is Planck's constant. So just draw your line with roughly the same gradient as the original line they have provided you with!

Moving onto the next part of the question:

2. Determine the radiation frequency which produces photoelectrons with maximum kinetic energy of 1.2eV

Recall our formula:

Maximum kinetic energy = hf - work function

We know that maximum kinetic energy as provided = 1.2eV = 1.2 x (1.602 x 10^-19)J
We know that work function = 4.1eV = 4.1 x (1.602 x 10^-19)J
And finally we know the value for h, which is Planck's constant

So now substitute all your values into the formula, and you obtain an answer of 1.3 x 10¹⁵Hz as the radiation frequency

Very tough question, definitely a band 6 range response (only if there's a band 7). I hope my explanation helped!

Best Regards
Happy Physics Land

[jakesilove]

Hey Cindy!

No problems! A conceptually difficult question on the stopping voltage experiment here, requires a bit of calculations as well! What's so difficult about it are two things:

1. Understanding the significance of stopping voltage and maximum kinetic energy
2. Applying and understanding the formula E=qV

Here at Atarnotes we want the best for our students, so I will hold your hand through this.

The diagram of the experiment indicates 4.1V as the stopping voltage, which is the point at which the voltage source provides enough energy to repel the ejected electron so that it doesnt reach the anode (or collector) and hence no current will be detected. The point of having a stopping voltage is just so that we can work out the work function, which is the minimum energy required to eject electrons from a metal.

I will prove to you why this is the case:

Maximum kinetic energy = hf - work function _____ equation 1

This is the equation for calculating the kinetic energy of ejected electrons as a light beam shines upon the metal.

But since stopping voltage works against the motion of electrons:

Maximum kinetic energy = hf - qV ________ equation 2

Hence by balancing equation 2 and equation 1, we can see that work function = qV
So work function of the metal = 4.1eV

Good good, now the difficult part is over, congratulations.

We need to do two things for this question

1. Draw the line on the graph for the experiment done with 0.0V

To draw the line, we must know its x-intercept, which is the threshold frequency. We can calculate threshold frequency through the formula:

Work function = planck's constant x threshold frequency

We know work function = 4.1eV = 4.1 x (1.602 x 10^-19)J
We know planck's constant = 6.632 x 10^-34
Substitute in the values:

4.1 x (1.602 x 10^-19)J = 6.632 x 10^-34 x threshold frequency

Hence threshold frequency of the metal = 9.9 x 10¹⁴ Hz. So your line for the repeated experiment should have an x-intercept at 9.9. Now how can I draw this line without knowing another point? Easy, because we know that the gradient is always constant for kinetic energy versus frequency graphs because it is Planck's constant. So just draw your line with roughly the same gradient as the original line they have provided you with!

Moving onto the next part of the question:

2. Determine the radiation frequency which produces photoelectrons with maximum kinetic energy of 1.2eV

Recall our formula:

Maximum kinetic energy = hf - work function

We know that maximum kinetic energy as provided = 1.2eV = 1.2 x (1.602 x 10^-19)J
We know that work function = 4.1eV = 4.1 x (1.602 x 10^-19)J
And finally we know the value for h, which is Planck's constant

So now substitute all your values into the formula, and you obtain an answer of 1.3 x 10¹⁵Hz as the radiation frequency

Very tough question, definitely a band 6 range response (only if there's a band 7). I hope my explanation helped!

Best Regards
Happy Physics Land

THE PRODIGAL SON RETURNS!

[Happy Physics Land]

THE PRODIGAL SON RETURNS!

Who said you cant write an essay in physics?

[Cindy2k16]

Hey Cindy!

No problems! A conceptually difficult question on the stopping voltage experiment here, requires a bit of calculations as well! What's so difficult about it are two things:

1. Understanding the significance of stopping voltage and maximum kinetic energy
2. Applying and understanding the formula E=qV

Here at Atarnotes we want the best for our students, so I will hold your hand through this.

The diagram of the experiment indicates 4.1V as the stopping voltage, which is the point at which the voltage source provides enough energy to repel the ejected electron so that it doesnt reach the anode (or collector) and hence no current will be detected. The point of having a stopping voltage is just so that we can work out the work function, which is the minimum energy required to eject electrons from a metal.

I will prove to you why this is the case:

Maximum kinetic energy = hf - work function _____ equation 1

This is the equation for calculating the kinetic energy of ejected electrons as a light beam shines upon the metal.

But since stopping voltage works against the motion of electrons:

Maximum kinetic energy = hf - qV ________ equation 2

Hence by balancing equation 2 and equation 1, we can see that work function = qV
So work function of the metal = 4.1eV

Good good, now the difficult part is over, congratulations.

We need to do two things for this question

1. Draw the line on the graph for the experiment done with 0.0V

To draw the line, we must know its x-intercept, which is the threshold frequency. We can calculate threshold frequency through the formula:

Work function = planck's constant x threshold frequency

We know work function = 4.1eV = 4.1 x (1.602 x 10^-19)J
We know planck's constant = 6.632 x 10^-34
Substitute in the values:

4.1 x (1.602 x 10^-19)J = 6.632 x 10^-34 x threshold frequency

Hence threshold frequency of the metal = 9.9 x 10¹⁴ Hz. So your line for the repeated experiment should have an x-intercept at 9.9. Now how can I draw this line without knowing another point? Easy, because we know that the gradient is always constant for kinetic energy versus frequency graphs because it is Planck's constant. So just draw your line with roughly the same gradient as the original line they have provided you with!

Moving onto the next part of the question:

2. Determine the radiation frequency which produces photoelectrons with maximum kinetic energy of 1.2eV

Recall our formula:

Maximum kinetic energy = hf - work function

We know that maximum kinetic energy as provided = 1.2eV = 1.2 x (1.602 x 10^-19)J
We know that work function = 4.1eV = 4.1 x (1.602 x 10^-19)J
And finally we know the value for h, which is Planck's constant

So now substitute all your values into the formula, and you obtain an answer of 1.3 x 10¹⁵Hz as the radiation frequency

Very tough question, definitely a band 6 range response (only if there's a band 7). I hope my explanation helped!

Best Regards
Happy Physics Land

Thank you for your explanation it did help!!

I have a few more questions if thats okay. How do you know that 4.1V is the stopping voltage? Do you interpret it from the graph because I'm having trouble understanding the graph (not the one to draw, but the one provided)- I get that its graphing E=hf but how does it relate to the experiment with 4.1V (or does it not relate at all?). I thought that the x-intercept of max KE vs frequency graphs was only dependent on the work function of the metal but does it change when there is a voltage applied across the electrodes? I hope my questions make sense?
It seems today's a bad day for studying physics I've never been so confused about it before 
Thanks in advance!

[jamonwindeyer]

Thank you for your explanation it did help!!

I have a few more questions if thats okay. How do you know that 4.1V is the stopping voltage? Do you interpret it from the graph because I'm having trouble understanding the graph (not the one to draw, but the one provided)- I get that its graphing E=hf but how does it relate to the experiment with 4.1V (or does it not relate at all?). I thought that the x-intercept of max KE vs frequency graphs was only dependent on the work function of the metal but does it change when there is a voltage applied across the electrodes? I hope my questions make sense?
It seems today's a bad day for studying physics I've never been so confused about it before 
Thanks in advance!

Hey your question definitely makes sense! Don't worry, we all have those days

This is a really tough question, but think of it this way. Those electrons in the metal are held in place with a specific amount of energy, and it is the light that has to provide enough energy to overcome all of this and thus release the electron. However, what if we loosened the grip, that would effecitvely cancel out part of the work function of the metal. Some of the work would be done for us. That is done by applying a voltage. The electrons are pulled away from their bonds ever so slightly by the voltage, thus reducing the effects of the work function.

In our situation, notice that all frequencies cause emitted photoelectrons. This would imply that the work function is non-existent. Close. It has been completely overcome by our applied voltage, which has pulled the electrons away just enough so that they are floating in mind air (sort of), waiting for any incoming photon to excite them and cause emission.

So that must mean that our 4.1V cancels out our work function exactly. That means that our work function is 4.1eV! Why? Well, one electron volt of energy is the energy given to an electron when exposed to a 1V potential difference. We have the same thing with 4.2

[Happy Physics Land]

Hey your question definitely makes sense! Don't worry, we all have those days

This is a really tough question, but think of it this way. Those electrons in the metal are held in place with a specific amount of energy, and it is the light that has to provide enough energy to overcome all of this and thus release the electron. However, what if we loosened the grip, that would effecitvely cancel out part of the work function of the metal. Some of the work would be done for us. That is done by applying a voltage. The electrons are pulled away from their bonds ever so slightly by the voltage, thus reducing the effects of the work function.

In our situation, notice that all frequencies cause emitted photoelectrons. This would imply that the work function is non-existent. Close. It has been completely overcome by our applied voltage, which has pulled the electrons away just enough so that they are floating in mind air (sort of), waiting for any incoming photon to excite them and cause emission.

So that must mean that our 4.1V cancels out our work function exactly. That means that our work function is 4.1eV! Why? Well, one electron volt of energy is the energy given to an electron when exposed to a 1V potential difference. We have the same thing with 4.2

HAPPY BIRTHDAY JAMON!!! <3 [/color]

[Cindy2k16]

Hey your question definitely makes sense! Don't worry, we all have those days

This is a really tough question, but think of it this way. Those electrons in the metal are held in place with a specific amount of energy, and it is the light that has to provide enough energy to overcome all of this and thus release the electron. However, what if we loosened the grip, that would effecitvely cancel out part of the work function of the metal. Some of the work would be done for us. That is done by applying a voltage. The electrons are pulled away from their bonds ever so slightly by the voltage, thus reducing the effects of the work function.

In our situation, notice that all frequencies cause emitted photoelectrons. This would imply that the work function is non-existent. Close. It has been completely overcome by our applied voltage, which has pulled the electrons away just enough so that they are floating in mind air (sort of), waiting for any incoming photon to excite them and cause emission.

So that must mean that our 4.1V cancels out our work function exactly. That means that our work function is 4.1eV! Why? Well, one electron volt of energy is the energy given to an electron when exposed to a 1V potential difference. We have the same thing with 4.2

Hi thanks for your help! I understand this idea of pulling the electrons away but i still have a few questions (sorry)
1. Just to clarify, are the electrons being pulled away because the emitter electrode is negative and the opposite electrode is positive? So this pulling away won't happen if the voltage switched to the emitter being positive etc?
2. I forgot to ask this earlier, but if the plate opposite the emitter is positive as shown in the diagram, wouldn't it be attracting the photoelectrons? If so, then how can the 4.1V be the stopping voltage if its not repelling the electrons at all and rather, is pulling them towards the opposite electrode? (I looked at a few websites and I think they all showed the opposite electrode to be negative when the stopping voltage is applied- unless i misunderstood something?)

Thank you! (sorry I'm taking so long to get this..)

[Happy Physics Land]

Hi thanks for your help! I understand this idea of pulling the electrons away but i still have a few questions (sorry)
1. Just to clarify, are the electrons being pulled away because the emitter electrode is negative and the opposite electrode is positive? So this pulling away won't happen if the voltage switched to the emitter being positive etc?
2. I forgot to ask this earlier, but if the plate opposite the emitter is positive as shown in the diagram, wouldn't it be attracting the photoelectrons? If so, then how can the 4.1V be the stopping voltage if its not repelling the electrons at all and rather, is pulling them towards the opposite electrode? (I looked at a few websites and I think they all showed the opposite electrode to be negative when the stopping voltage is applied- unless i misunderstood something?)

Thank you! (sorry I'm taking so long to get this..)

Hey Cindy!

Its alright! It is a very tough question to understand. Your first dotpoint is right on the money. If the collector is positive, then electrons would travel towards it spontaneously due to electrostatic attraction (i.e. negative is attracted to positive). If the collector is negative then electrons are repelled and the only way the electrons can ever reach the collector is if they get given enough kinetic energy from the incident light ray.

So yeah with regards to your second question, the only reason why we supply a voltage to the emitter and collector is so that we can repel the electrons up to a stopping voltage value where electrons cannot reach the collector anymore and so we can work out work function. Theoretically if the collector is just positive then yes we do not need any voltage input because electrons will simply be attracted to collector. But experimentally, we still need some voltage supply purely just to bring charge to the collector (i.e. make it positive), but this voltage wouldnt determine anything valuable for us.

Best Regards
Happy Physics Land

[student123456]

Hola guys and gals!
Hoping someone can help me out with this multiple choice question please ::

[RuiAce]

Hola guys and gals!
Hoping someone can help me out with this multiple choice question please ::

Addressed by Jake in post #987

[jakesilove]

Addressed by Jake in post #987

Rui you are a machine, efficient af

[jamonwindeyer]

Rui you are a machine, efficient af

I feel like he has a bloody spreadsheet in his head that he just closes his eyes and navigates through