HSC Physics Question Thread

[jamonwindeyer]

Hey!! Thanks so much that definitely helps, way better than the 2 line answer in the book 

Aha totally! Used to hate that when I was doing my HSC  Sorry I couldn't be a little more specific, either X or Y will definitely fall faster, but I'm not sure which one with the info in front of me 

[FallonXay]

Hey FallonXay!!

It depends on the context a little bit. If we are just talking about a metal, totally isolated from anything else, then no! The electrons are not replaced, the metal will gradually build up a positive charge.

If the metal is earthed in any way, either as part of a circuit or just in the circumstances, then electrons from earth will be attracted to the positive charge in the metal (caused by things called holes, which is covered in the next section) and replace the missing electrons. Remember, earth just defines an infinite reservoir of electrons, so if the metal is attached to anything like this, the electrons are replaced I hope this helps!!

oh ok, thanks!

[FallonXay]

Hiiii, I've attached a question from the 2014 Physics HSC exam (Q26b) and I don't quite understand how to solve it. I'm assuming it has something to do with the stopping voltage equation? But i'm not sure how to come to this conclusion.

[jamonwindeyer]

Hiiii, I've attached a question from the 2014 Physics HSC exam (Q26b) and I don't quite understand how to solve it. I'm assuming it has something to do with the stopping voltage equation? But i'm not sure how to come to this conclusion.

I sat this paper, it was nasty let me tell you! I have a nicely typed solution to this, bear with me 

[jamonwindeyer]

Hiiii, I've attached a question from the 2014 Physics HSC exam (Q26b) and I don't quite understand how to solve it. I'm assuming it has something to do with the stopping voltage equation? But i'm not sure how to come to this conclusion.

Okay, this question is actually quite tricky, and requires a bit of thought. Consider the initial experiment; when the frequency of the light is 0 (no light), we have no current flow. For any other frequency, we do have current flow. What this means is that the potential difference of 4.1V exactly takes care of the work function of the metal; any frequency of light (and thus, any amount of energy) will release electrons. The work function is, therefore, 4.1eV.

When the external voltage is removed, this work function comes back into play, and here lies our work. We need to calculate the x-intercept of the new line, by calculating the frequency required for photoelectrons to have precisely 4.1eV of energy. This is the new threshold frequency, the minimum frequency to now overcome this work function. The gradient of the new line is the same (Planck’s Constant), and so doing the math, the new line is as follows (remember to convert between electron-volts and Joules, the conversion is in your data sheet):



So, you would draw a new line with the same gradient (the gradient is always Planck's Constant), with an x-intercept at this frequency. Without the voltage, this is the frequency you need to overcome the work function.

For the second part of the question, you just use Planck's formula and the work function:



I hope this helps a little bit!!  It's a little confusing actually, is there anything here you needed clarified? Happy to help if so 

[FallonXay]

I sat this paper, it was nasty let me tell you! I have a nicely typed solution to this, bear with me 

oh haha. yeah, a lot of tricky questions in the exam 

Okay, this question is actually quite tricky, and requires a bit of thought. Consider the initial experiment; when the frequency of the light is 0 (no light), we have no current flow. For any other frequency, we do have current flow. What this means is that the potential difference of 4.1V exactly takes care of the work function of the metal; any frequency of light (and thus, any amount of energy) will release electrons. The work function is, therefore, 4.1eV.

When the external voltage is removed, this work function comes back into play, and here lies our work. We need to calculate the x-intercept of the new line, by calculating the frequency required for photoelectrons to have precisely 4.1eV of energy. This is the new threshold frequency, the minimum frequency to now overcome this work function. The gradient of the new line is the same (Planck’s Constant), and so doing the math, the new line is as follows (remember to convert between electron-volts and Joules, the conversion is in your data sheet):



So, you would draw a new line with the same gradient (the gradient is always Planck's Constant), with an x-intercept at this frequency. Without the voltage, this is the frequency you need to overcome the work function.

For the second part of the question, you just use Planck's formula and the work function:



I hope this helps a little bit!!  It's a little confusing actually, is there anything here you needed clarified? Happy to help if so 

Ooooh, I see, thanks! Interesting question with an interesting solution   

Only part I don't understand is how you concluded that the work function is 4.1eV. Is there some direct relation between potential difference/volts and electronvolts?
Also, is the SI unit for work function always joules?

[jamonwindeyer]

oh haha. yeah, a lot of tricky questions in the exam 
Ooooh, I see, thanks! Interesting question with an interesting solution   

Only part I don't understand is how you concluded that the work function is 4.1eV. Is there some direct relation between potential difference/volts and electronvolts?
Also, is the SI unit for work function always joules?

Yep! So the definition of an electron volt is: "a unit of energy equal to the work done on an electron accelerating it through a potential difference of one volt."

I suppose a way to think of it is this. Since the voltage source of 4.1 volts exactly cancels the work function, we can conclude that the work function is deaccelerating the electrons (keeping them on/within the metal) with an energy equal to that provided by the 4.1 volts. Accelerating an electron out of a 4.1 volt potential difference, takes 4.1eV of energy, by the definition. So, accelerating the electron OUT of the metal, must take 4.1eV of energy also. And there is the answer.

And yep, SI unit for energy is Joules. If you try to use eV with Planck's Constant, for example, the answer won't be correct. Always convert 

[FallonXay]

Yep! So the definition of an electron volt is: "a unit of energy equal to the work done on an electron accelerating it through a potential difference of one volt."

I suppose a way to think of it is this. Since the voltage source of 4.1 volts exactly cancels the work function, we can conclude that the work function is deaccelerating the electrons (keeping them on/within the metal) with an energy equal to that provided by the 4.1 volts. Accelerating an electron out of a 4.1 volt potential difference, takes 4.1eV of energy, by the definition. So, accelerating the electron OUT of the metal, must take 4.1eV of energy also. And there is the answer.

And yep, SI unit for energy is Joules. If you try to use eV with Planck's Constant, for example, the answer won't be correct. Always convert 

ahh ok, thanks a ton!!! 

[jamonwindeyer]

ahh ok, thanks a ton!!! 

Not a problem at all, happy to help!! 

[Happy Physics Land]

Hiya!

I was wondering (In regards to the photoelectric effect), If a metal surface is hit with an incident EMR above the threshold frequency, electrons are emitted. What happens to the missing electrons in the metal? Are they replaced? How so?

thanks.

Hey FallonXay!

Just reinforcing Jamon's point. When a high frequency light ray hits the surface of the metal, the electron is ejected when the frequency is greater than the threshold frequency (i.e. f > f₀) and this means that the energy given off by each photon is able to overcome the work function of the metal (hf > hf₀).

Ok but this doesn't really answer your question. So when the electrons is ejected (i.e. photoelectron), it can end up in different places depending on the scenario, just like what jamon said. I will just explain some of the following that are related to the hsc course:

1. If a circuit is provided, the photoelectron will travel through the circuit in the form of a photocurrent
2. If no circuits are provides and the metal is just exposed to air, then the photoelectron is ejected into the air and will perhaps ionise air molecules (This is seen in Hertz's experiment)
3. If a voltage is applied to the metal, then the electron ejected from the metal will accelerate through the air (This is seen in einstein's photoelectric experiment, when the photoelectron emitted from the cathode metal accelerates towards the anode metal)
4. If the light ray is shone upon the depletion layer of a p-n junction (or a solar cell), then the ejected electron can travel to the n-type semiconductor layer and this creates a potential difference between n-type and p-type layers. Consequently, when a circuit is connected to the solar cell, electricity can flow.

[FallonXay]

Hey FallonXay!

Just reinforcing Jamon's point. When a high frequency light ray hits the surface of the metal, the electron is ejected when the frequency is greater than the threshold frequency (i.e. f > f₀) and this means that the energy given off by each photon is able to overcome the work function of the metal (hf > hf₀).

Ok but this doesn't really answer your question. So when the electrons is ejected (i.e. photoelectron), it can end up in different places depending on the scenario, just like what jamon said. I will just explain some of the following that are related to the hsc course:

1. If a circuit is provided, the photoelectron will travel through the circuit in the form of a photocurrent
2. If no circuits are provides and the metal is just exposed to air, then the photoelectron is ejected into the air and will perhaps ionise air molecules (This is seen in Hertz's experiment)
3. If a voltage is applied to the metal, then the electron ejected from the metal will accelerate through the air (This is seen in einstein's photoelectric experiment, when the photoelectron emitted from the cathode metal accelerates towards the anode metal)
4. If the light ray is shone upon the depletion layer of a p-n junction (or a solar cell), then the ejected electron can travel to the n-type semiconductor layer and this creates a potential difference between n-type and p-type layers. Consequently, when a circuit is connected to the solar cell, electricity can flow.

ok, makes sense, thanks for the additional info regarding possible scenarios!

However one thing I wanted to clear up (another question ) is in regards to the photon if the frequency is below the threshold frequency. If the photon is below the threshold frequency, is the energy still transferred from the photon into the electron - consequently making the electron 'jump' up a 'shell' however since there is insufficient energy, the electron in the new shell is unstable and returns to its original shell and in the process, emits the energy back off? Or what happens to the photon, is it just reflected off the surface of the metal?

[jamonwindeyer]

ok, makes sense, thanks for the additional info regarding possible scenarios!

However one thing I wanted to clear up (another question ) is in regards to the photon if the frequency is below the threshold frequency. If the photon is below the threshold frequency, is the energy still transferred from the photon into the electron - consequently making the electron 'jump' up a 'shell' however since there is insufficient energy, the electron in the new shell is unstable and returns to its original shell and in the process, emits the energy back off? Or what happens to the photon, is it just reflected off the surface of the metal?

Hey FallonXay! That's a little trickier, I'll give you the simple explanation and the more complex one (which is way beyond the scope of the course):

Simple: The photon will be re-emitted, and it happens pretty much exactly as you say it would. The electron is excited to a new energy level, which is unstable, and when it falls back the photon is then re-emitted. The photon may also just simply not give its energy and, it has to give everything or nothing. This is called the All or Nothing Principle. Either way, the outcome is the same, photons are then re-emitted from the metal as reflected light.

Complex: There is an equation in quantum mechanics called the Schrodinger Equation which dictates the probability certain events occurring in a system, in this case, an electron/photon system. This equation actually says that it is possible for an electron to be excited into a higher energy level, but it can only realistically occur when the energy of a photon almost exactly matches the difference in energy between the electron shells. At all other energies, the probability of it occurring is very low (in Quantum Physics, every outcome has a probability of occurring, even those which seemingly defy classical physical logic). When the frequency is too low, the highest probability lies with the option above, the photon is re-emitted and the electron is unaffected. This is the manifestation of the All or Nothing Principle.

It is important to mention that, if a light has a very high intensity, and there are HEAPS of photons striking the metal, it is actually very slightly possible for a photon to strike an electron, excite it to a higher energy level, and then another photon to induce emission. Teamwork! However, for this to happen, the next photon has to strike the electron AFTER it has been excited, but BEFORE it jumps back down. This is highly unlikely, and so we definitely don't get any substantial amounts of electrons being emitted from the metal 

Reminder: The latter two paragraphs are definitely NOT required knowledge for this course! However, I'm happy to expand on some of it, if you feel like getting a head start on university physics 

[FallonXay]

Hey FallonXay! That's a little trickier, I'll give you the simple explanation and the more complex one (which is way beyond the scope of the course):

Simple: The photon will be re-emitted, and it happens pretty much exactly as you say it would. The electron is excited to a new energy level, which is unstable, and when it falls back the photon is then re-emitted. The photon may also just simply not give its energy and, it has to give everything or nothing. This is called the All or Nothing Principle. Either way, the outcome is the same, photons are then re-emitted from the metal as reflected light.

Complex: There is an equation in quantum mechanics called the Schrodinger Equation which dictates the probability certain events occurring in a system, in this case, an electron/photon system. This equation actually says that it is possible for an electron to be excited into a higher energy level, but it can only realistically occur when the energy of a photon almost exactly matches the difference in energy between the electron shells. At all other energies, the probability of it occurring is very low (in Quantum Physics, every outcome has a probability of occurring, even those which seemingly defy classical physical logic). When the frequency is too low, the highest probability lies with the option above, the photon is re-emitted and the electron is unaffected. This is the manifestation of the All or Nothing Principle.

It is important to mention that, if a light has a very high intensity, and there are HEAPS of photons striking the metal, it is actually very slightly possible for a photon to strike an electron, excite it to a higher energy level, and then another photon to induce emission. Teamwork! However, for this to happen, the next photon has to strike the electron AFTER it has been excited, but BEFORE it jumps back down. This is highly unlikely, and so we definitely don't get any substantial amounts of electrons being emitted from the metal 

Reminder: The latter two paragraphs are definitely NOT required knowledge for this course! However, I'm happy to expand on some of it, if you feel like getting a head start on university physics 

Wouldn't mind hearing some more, if you don't mind. Seems like an interesting phenomenon ^.^

[jamonwindeyer]

Wouldn't mind hearing some more, if you don't mind. Seems like an interesting phenomenon ^.^

You have opened Pandora's Box 

Okay, so in classical physics we learn about Newton's 2nd Law, which can be used (in various forms) to predict the behaviour of a system over time. Classically, this is absolutely accurate. In quantum mechanics, we cannot predict the outcome of an event until it happens, we can only calculate the probabilities of certain events occurring. This links to the whole idea of Schrodingers Cat.

Picture a cat inside a box (we can't see inside), and next to it is a contraption with a radioactive isotope, a detector, and a flask of poison. Two things in this 'system' can occur. The particle can decay, and if this happens, the resultant gamma radiation will trigger the detector, release the poison, and kill the cat. If the particle does not decay, nothing happens and the cat stays alive.

Certain interpretations of quantum mechanics (Copenhagen Interpretations) would suggest that, since we cannot know whether the cat is alive or dead until we check, the cat is simultaneously alive and dead. This is known as a quantum superposition. Only when we open the box, does this superposition collapse into a reality where the cat is alive (yay!) or dead (wahh...). Really, the cat here is just enabling us to extend an atomic state (whether a particle decays or not) to a macroscopic, real living thing.

So, how do we predict the probability of the cat dying or living, or more generally, how do we predict the outcomes of a system? This is where the Schrodinger Equation comes in. This is a partial differential equation which describes how the quantum state of a system evolves over time. The equation looks like this, the subject of the equation being the weird symbol Psi appearing on both sides of the equation (the rest are an essay in itself)



The subject is the wave equation. Now, the probability of certain quantum states is proportional to the square of this wave function. To solve for the wave equation, therefore, is to have complete understanding of a system and the probability of any event occurring. In some interpretations of quantum physics, the wave function can be used to predict the evolution of HUGE systems, perhaps even the entire universe. Solving the Schrodinger Equation for a particle is hard enough though, for a universe is simply impossible 

The Schrodinger Equation has many implications:

- Depending on measurements, the states of a system are quantised, and thus, energy is quantised in all forms (not just for electromagnetic quanta). This is verified already, for example, electron energies in atoms are proven to be quantised.
- Under the Copenhagen interpretation of quantum mechanics, particles do not have set positions, and thus, the result when we measure is drawn from a probability distribution (wave function gives us this). Flowing on from this, we cannot know the precise position of a particle, unless we completely abandon any attempt to measure its momentum. The product of error in momentum and error in position of a particle must be larger than Planck's constant divided by 2 pi, this is Heisenberg's Uncertainty Principle. 
- There are, in certain systems, small probabilities of classical physics being completely broken. This is called Quantum Tunnelling. For example, there is always a slight probability that a particle will pass through a classically insurmountable barrier (the microscopic and basic equivalent of me appearing next to you, from where I am, with no lead up, right as you finish this sentence).

This is a very quick run through of a SUPER fascinating topic, you should definitely do some extra research if you can! And study Physics at uni, it is awesome 

[FallonXay]

You have opened Pandora's Box 

Okay, so in classical physics we learn about Newton's 2nd Law, which can be used (in various forms) to predict the behaviour of a system over time. Classically, this is absolutely accurate. In quantum mechanics, we cannot predict the outcome of an event until it happens, we can only calculate the probabilities of certain events occurring. This links to the whole idea of Schrodingers Cat.

Picture a cat inside a box (we can't see inside), and next to it is a contraption with a radioactive isotope, a detector, and a flask of poison. Two things in this 'system' can occur. The particle can decay, and if this happens, the resultant gamma radiation will trigger the detector, release the poison, and kill the cat. If the particle does not decay, nothing happens and the cat stays alive.

Certain interpretations of quantum mechanics (Copenhagen Interpretations) would suggest that, since we cannot know whether the cat is alive or dead until we check, the cat is simultaneously alive and dead. This is known as a quantum superposition. Only when we open the box, does this superposition collapse into a reality where the cat is alive (yay!) or dead (wahh...). Really, the cat here is just enabling us to extend an atomic state (whether a particle decays or not) to a macroscopic, real living thing.

So, how do we predict the probability of the cat dying or living, or more generally, how do we predict the outcomes of a system? This is where the Schrodinger Equation comes in. This is a partial differential equation which describes how the quantum state of a system evolves over time. The equation looks like this, the subject of the equation being the weird symbol Psi appearing on both sides of the equation (the rest are an essay in itself)



The subject is the wave equation. Now, the probability of certain quantum states is proportional to the square of this wave function. To solve for the wave equation, therefore, is to have complete understanding of a system and the probability of any event occurring. In some interpretations of quantum physics, the wave function can be used to predict the evolution of HUGE systems, perhaps even the entire universe. Solving the Schrodinger Equation for a particle is hard enough though, for a universe is simply impossible 

The Schrodinger Equation has many implications:

- Depending on measurements, the states of a system are quantised, and thus, energy is quantised in all forms (not just for electromagnetic quanta). This is verified already, for example, electron energies in atoms are proven to be quantised.
- Under the Copenhagen interpretation of quantum mechanics, particles do not have set positions, and thus, the result when we measure is drawn from a probability distribution (wave function gives us this). Flowing on from this, we cannot know the precise position of a particle, unless we completely abandon any attempt to measure its momentum. The product of error in momentum and error in position of a particle must be larger than Planck's constant divided by 2 pi, this is Heisenberg's Uncertainty Principle. 
- There are, in certain systems, small probabilities of classical physics being completely broken. This is called Quantum Tunnelling. For example, there is always a slight probability that a particle will pass through a classically insurmountable barrier (the microscopic and basic equivalent of me appearing next to you, from where I am, with no lead up, right as you finish this sentence).

This is a very quick run through of a SUPER fascinating topic, you should definitely do some extra research if you can! And study Physics at uni, it is awesome 

  I think my brain just imploded. In a good way