HSC Physics Question Thread

[DrDusk]

Hey Everyone,

I am in Year 11 and am going to do my Physics prelim exams soon. I was wondering whether anyone has any good tips and what the most important things I should study that I can use for the exam, to achieve a high mark?

Also, does anyone have any past prelim papers for Physics?... (The new syllabus is making it hard to find them).

Thanks in advance,
Coolmate

The most important thing is to make sure you have an understanding of the concepts. Do NOT rote learn the Math type questions for Physics, rather train your mind to think critically. Other than that practice your essay type questions, ask your teacher to mark them.

Good luck

[Coolmate]

The most important thing is to make sure you have an understanding of the concepts. Do NOT rote learn the Math type questions for Physics, rather train your mind to think critically. Other than that practice your essay type questions, ask your teacher to mark them.

Good luck!

Thankyou so much for your response and the help, I will do this!

[classof2019]

Hi, could someone please help with this question (multiple choice)?

Light with a frequency above the threshold frequency is incident on a metal surface. If light of the same intensity but double the frequency was used, how would the photocurrent (number of electrons emitted per second) and the kinetic energy of photoelectrons be affected?

Answer: The photoelectrons would have a higher kinetic energy and photocurrent would decrease.


I'm a bit confused because doesn't intensity, by definition, refer to the number of photons incident on a given area per second? And since intensity is the same, by the All or Nothing principle, shouldn't this mean the same number of electrons are liberated per second?

Any help would be greatly appreciated. Cheers!

[DrDusk]

Sorry, did you mean the photoelectric current would increase?
Because it should increase and not decrease..

You are right in saying that the same amount of electrons will be ejected but the Photo current will still increase. The reason for this is we have I=Q/t. Suppose in a little thought experiment we take a stopwatch and measure a time T, and measure that a total of Q' electrons flow in time T. Let the frequency of the light be f and the frequency of the metal be f' Now I'll perform some derivations:

Now the remaining energy Ef becomes the Kinetic energy of the electron so we have:


We can assume 'd' is constant. Now if you look at the equation, INCREASING f increases the velocity factor, which in turn increases the measured current I

Now for a time interval T we defined above, we can use the formula T = distance/speed which gives us:


Clearly you can see increasing v increases the Photo current. In your mind you can picture it as an olympic running race. Suppose you take a time interval of 10 seconds for a 100 meter race. The slower  the runners are, the less of them will pass through the finish line in 10 seconds. However the faster they are, more and more will start to pass through the finish line.

Just like that running race, more electrons will pass a given point in a time 't'. This increase the factor Q/t, which increases the current I = Q/t

Also just an interesting thing you can note from that equation. If the incoming frequency of the light " f " equals the frequency of the metal " f' ", then we have Current = 0. This is expected, Why?, well because when you shine a light with a frequency equal to the min frequency, the electrons exit with ZERO VELOCITY. Current is only formed when Electrons are MOVING.

[classof2019]

Sorry, did you mean the photoelectric current would increase?
Because it should increase and not decrease..

You are right in saying that the same amount of electrons will be ejected but the Photo current will still increase. The reason for this is we have I=Q/t. Suppose in a little thought experiment we take a stopwatch and measure a time T, and measure that a total of Q' electrons flow in time T. Let the frequency of the light be f and the frequency of the metal be f' Now I'll perform some derivations:

Now the remaining energy Ef becomes the Kinetic energy of the electron so we have:


We can assume 'd' is constant. Now if you look at the equation, INCREASING f increases the velocity factor, which in turn increases the measured current I

Now for a time interval T we defined above, we can use the formula T = distance/speed which gives us:


Clearly you can see increasing v increases the Photo current. In your mind you can picture it as an olympic running race. Suppose you take a time interval of 10 seconds for a 100 meter race. The slower  the runners are, the less of them will pass through the finish line in 10 seconds. However the faster they are, more and more will start to pass through the finish line.

Just like that running race, more electrons will pass a given point in a time 't'. This increase the factor Q/t, which increases the current I = Q/t

Also just an interesting thing you can note from that equation. If the incoming frequency of the light " f " equals the frequency of the metal " f' ", then we have Current = 0. This is expected, Why?, well because when you shine a light with a frequency equal to the min frequency, the electrons exit with ZERO VELOCITY. Current is only formed when Electrons are MOVING.

This makes complete sense to me, however the back-of-the-book answer does say that photocurrent decreases. Here is their justification:

"If the frequency was increased, each photon would have more energy. As the intensity remains constant, less photons will be incident on the surface per second and hence photocurrent will be reduced. Because energy is conserved and because it takes a specific amount of energy to remove an electron, the increased photon energy will cause the kinetic energy of the ejected electrons to increase".

What do you think?

[DrDusk]

" As the intensity remains constant, less photons will be incident on the surface per second and hence photocurrent will be reduced"

See this is not wrong, it IS a valid statement, and I did think of this as well.
I did a bit of research and half the sources say that current should increase and the other half say it should decrease.

It appears there is no 'straight' answer to this question which is really weird. See I believe both arguments are technically correct, it just depends on which one outweighs the other in terms of its affect on the photo-current.

Maybe its just because I'm currently on 2 hours sleep, so my brain is just fatigued

Still this is indeed really interesting. I'll consult my Professor for Quantum Physics and get back to you on this...

[classof2019]

See this is not wrong, it IS a valid statement, and I did think of this as well.
I did a bit of research and half the sources say that current should increase and the other half say it should decrease.

It appears there is no 'straight' answer to this question which is really weird. See I believe both arguments are technically correct, it just depends on which one outweighs the other in terms of its affect on the photo-current.

Maybe its just because I'm currently on 2 hours sleep, so my brain is just fatigued

Still this is indeed really interesting. I'll consult my Professor for Quantum Physics and get back to you on this...

Thank you for looking into it.

Just an idea ... could it perhaps have something to do with differing definitions of intensity? I.e. they seem to be referring to intensity as the amount of 'energy' passing through a given area per second (power per metre squared - consider the units of intensity, Wm^-2), whereas I, and other textbooks, have interpreted intensity to be the number of photons per second?

[DrDusk]

Thank you for looking into it.

Just an idea ... could it perhaps have something to do with differing definitions of intensity? I.e. they seem to be referring to intensity as the amount of 'energy' passing through a given area per second (power per metre squared - consider the units of intensity, Wm^-2), whereas I, and other textbooks, have interpreted intensity to be the number of photons per second?

My mind is actually functioning properly now that I'm fully awake and not sleep deprived.
Firstly that answer is most definitely wrong. Your are right, its due to their definition of intensity. I'll explain to you what their answer means and why its wrong. Even though its wrong, developing an understanding is crucial.

Okay so this is the justification of their answer:
We have E = hf for each photon. If the light contains 'n' photons, it must carry energy E = nhf. Now this is where  they make their crucial mistake. There are two types of Particle physics, Classical and Quantum. In classical Physics the intensity of a wave as you said is defined by W/m^2. Using this definition we can clearly see Intensity is proportional to the Energy of a wave which is E = nhf. So if Intensity stays constant, E must stay constant. However h is a constant, so the only way to keep E constant with an increasing f is by decreasing n.

So now we have that the number of photons decrease. I hope you can see why this spells trouble, because in Quantum Physics decreasing the number of Photons decreases the intensity of the wave, but lets carry on. So photons decrease, meaning the rate of arrival for the photons will decrease. Which means the photo-current will decrease.

However as I mentioned above this is problematic.
The real definition of intensity in this case is proportional to the amount of photons, so 'n' must stay constant in the equation E = nhf. This means the energy of the light wave and subsequently each photon increases. This means each photon carries greater kinetic energy as I mentioned. However now that I'm not half asleep there is something more important to consider.

Greater kinetic energy per photon means each ejected electron will travel a greater distance before the next one is ejected. This effect should more or less cancel out the fact that they are travelling faster. This will lead to basically a constant current, or one that increases just slightly... It wont increase by a lot as I said in my first post.

[not a mystery mark]

Hey squad! I have no idea what's happening - I can only figure this out to be anticlockwise, but the answer is clockwise??
Pls help. My trial is tomorrow and the end is neigh.

[DrDusk]

Hey squad! I have no idea what's happening - I can only figure this out to be anticlockwise, but the answer is clockwise??
Pls help. My trial is tomorrow and the end is neigh.

The loop must generate a current that opposes the change in flux. This is a very confusing statement so I'll make it easier for you.
Basically it means the loop must create a current that fights to return the system back to its original setting. In the original setting the Magnet is stationary at some distance below the loop. The current in this loop will aim to get the magnet back into this position. The only way of doing this is to generate a magnetic field that doesn't attract the one of the magnet, i.e. it must repel the magnet. For this to happen a north pole must be generated by the loop at the bottom.

Now using the right hand coil rule, your thumb points downwards and your fingers will curl in the direction of the current, which will be clockwise.

Good luck for Physics tomorrow. Remember to think critically and outside the box, and if your stuck on a question stay calm and brainstorm any Laws that you can apply. The most common ones will be Conservation of Energy & Momentum.

[not a mystery mark]

The loop must generate a current that opposes the change in flux. This is a very confusing statement so I'll make it easier for you.
Basically it means the loop must create a current that fights to return the system back to its original setting. In the original setting the Magnet is stationary at some distance below the loop. The current in this loop will aim to get the magnet back into this position. The only way of doing this is to generate a magnetic field that doesn't attract the one of the magnet, i.e. it must repel the magnet. For this to happen a north pole must be generated by the loop at the bottom.

Now using the right hand coil rule, your thumb points downwards and your fingers will curl in the direction of the current, which will be clockwise.

Good luck for Physics tomorrow. Remember to think critically and outside the box, and if your stuck on a question stay calm and brainstorm any Laws that you can apply. The most common ones will be Conservation of Energy & Momentum.

OH MY GOD SORRY, THE ANSWER SAYS ANTICLOCKWISE. THIS WAS MY SAME TRAIN OF THOUGHT.
I'm guessing the answer on the book is wrong.
Also thank you heaps!!! I will try my best!

[DrDusk]

OH MY GOD SORRY, THE ANSWER SAYS ANTICLOCKWISE. THIS WAS MY SAME TRAIN OF THOUGHT.
I'm guessing the answer on the book is wrong.
Also thank you heaps!!! I will try my best!

Books are always infamous for getting wrong answers its not even funny.

Remember though, as it begins to leave the loop, the current will flow Anti-clockwise

[not a mystery mark]

Books are always infamous for getting wrong answers its not even funny.

Remember though, as it begins to leave the loop, the current will flow Anti-clockwise

I'm becoming sadly familiar with that by the day ahah.
Also yess, because of the system will induce a downwards South Pole to oppose it's motion away from the loop. #induction4dayzzz

[DrDusk]

How was the trial @Mystery mark?

Easy, hard, regardless how did you go? Any questions that required you to think outside the box or were most of them essay based?

I have still yet to see a real paper from the new syllabus.

[not a mystery mark]

How was the trial @Mystery mark?

Easy, hard, regardless how did you go? Any questions that required you to think outside the box or were most of them essay based?

I have still yet to see a real paper from the new syllabus.

Pretty well actually - felt a lot of it was somewhat formulaic though. I was a bit panicky during the beginning (as one does during the first few minutes) but all the information started surfacing when it needed too which was good. Overall I'm hoping for an 85 maybe?

The questions didn't look too essay focused compared to what I saw in past-old-syllabus HSC exams, although there is an added emphasis on reliability, validity, and accuracy of data. One question asked to refer to my depth study and analyse its reliability - wasn't too bad. Also a question like "How did the discovery of the Motor Effect impact society and the environment" which felt a bit strange haha

Currently working for the Extension 2 Math exam tomorrow and I am screaming ahaha. It's the final exam and after that I'm going to have a week's sleep. I'll be happy to send the paper whenever we get it back if you're interested. It was a nice test.