VCE Physics Question Thread!

[knightrider]

How would you do this question?

What happens to the speed of the waves in a
ripple tank if the frequency of the wave source is
halved?

[Rishi97]

How would you do this question?

What happens to the speed of the waves in a
ripple tank if the frequency of the wave source is
halved?

Hey Knightrider
These sort of questions are best answered by finding a relationship between the variables asked. In this case, since the question is asking for speed and frequency, we can use the formula speed = frequency x wavelength. So, if the frequency is halved (decreased), the speed must also decrease since they are proportional to each other.
Hope that makes sense :-)

[knightrider]

Hey Knightrider
These sort of questions are best answered by finding a relationship between the variables asked. In this case, since the question is asking for speed and frequency, we can use the formula speed = frequency x wavelength. So, if the frequency is halved (decreased), the speed must also decrease since they are proportional to each other.
Hope that makes sense :-)

Thanks Rishi97 

but the answer says there is no change  in speed?

[Rishi97]

Thanks Rishi97 

but the answer says there is no change  in speed?

Oh really?
I'm not sure then sorry

[Floatzel98]

I'm having a bit of trouble with VCAA 2014 Exam Question 22 (If anyone has time to help). With part a) is it's first excited state at 4.9 eV? It can absorb a 1.8eV photon because that is the exact difference between the 1st and 2nd excited states but it can't emit a photon from the first excited state because there is no energy level at (4.9eV - 1.8eV) 3.1 eV? I got really confused on this when i did it today. I read the diagram from top to bottom instead of bottom to top. Just wanted to clear it all up.

For part b I'm really lost with what to do. 0.9, 1.5 and 2.2 eV are all between the first excited state. I don't know how to use them at all to find out the unknown excited state.

Thanks

[Orbit]

How would you do this question?

What happens to the speed of the waves in a
ripple tank if the frequency of the wave source is
halved?

One of the most important things to remember about waves is their speed depends only on the medium they travel though. If you have red and green light passing through a vacuum, for example, they still have the same speed: . A consequence of this is that when the frequency of the wave source is doubled the wavelength must halve to keep the speed constant in accordance with . So to answer your question, the speed does not change, because the medium (water) does not change.

I'm having a bit of trouble with VCAA 2014 Exam Question 22 (If anyone has time to help). With part a) is it's first excited state at 4.9 eV? It can absorb a 1.8eV photon because that is the exact difference between the 1st and 2nd excited states but it can't emit a photon from the first excited state because there is no energy level at (4.9eV - 1.8eV) 3.1 eV? I got really confused on this when i did it today. I read the diagram from top to bottom instead of bottom to top. Just wanted to clear it all up.

For part b I'm really lost with what to do. 0.9, 1.5 and 2.2 eV are all between the first excited state. I don't know how to use them at all to find out the unknown excited state.

Thanks

You are essentially correct for part a. Remember that in general atoms can only take on specific discrete energy levels. So absorbing a photon would give it an energy of which is the second excited state but emitting a photon would bring its energy down to which is not one of its discrete energy states.

For part b, we just need to use some simple observation. Since a photon being emitted does not correspond to a drop in any of the known energy states, we have two possibilities for , one where the energy of the atom drops to and one where it drops from : namely, or . Now we substitute another given value, an emission of a photon, to find that cannot be as if there would be no drop emitting a photon. So must be , and we can verify this by seeing that .

[Floatzel98]

A searchlight is beaming out green light. The wavelength of the light is 520nm  and the power of the search light is 5.0kW.

a) Calculate how many photons leave the searchlight each second.
b) What is the momentum of a single 520nm photon.
c) All these photons are focused onto a perfectly reflecting mirror of area 0.1 m^2. Calculate the average force on the mirror.
d) If the mirror in the previous question was replaced by a perfectly absorbing surface, then the answer would:
A - increase by a factor 2
B - decrease by a factor 2
C - remain the same
D - change, but not by a factor of 2

I'm having a bit of trouble on parts c and d. I don't understand how to incorporate the area into the answer. I mean there is no pressure given so that we can. I tried using F = p/ t and multiplying it by the number of photons per second but that isnt giving me the correct answer either. And I don't have any reasoning for any of the answers in part d either.

Any help would be great. Thanks

[Floatzel98]

If you were asked to find a value for the gradient of a stopping voltage against frequency graph, and just say you didn't know how to do it, would you get marks for just writing down the actual value of Planck's Constant or not?

[odeaa]

If you were asked to find a value for the gradient of a stopping voltage against frequency graph, and just say you didn't know how to do it, would you get marks for just writing down the actual value of Planck's Constant or not?

I dont think so unfortunately. I havent seen a question that explicitly asks for the gradient, but I have seen several (VCAA) ones that ask to calculate h from the graph and they all say students were given 0 marks for just saying the value from the data sheet.

If it was worded as asking specifically for the gradient and not plancks constant, they might give you a mark because that shows you know the gradient is h, but i doubt it

[Floatzel98]

I dont think so unfortunately. I havent seen a question that explicitly asks for the gradient, but I have seen several (VCAA) ones that ask to calculate h from the graph and they all say students were given 0 marks for just saying the value from the data sheet.

If it was worded as asking specifically for the gradient and not plancks constant, they might give you a mark because that shows you know the gradient is h, but i doubt it

Yeah, didn't think you would get a mark.

This is a bit of a specific question again, but if you were finding the value of h from a graph/table with multiple points and you chose a set of points that gave you a value 'outside' the preferable range, would you still get full marks for that? A couple of VCAA questions I did said that a preferable range would be between 4.7 to 5.3 x 10^-15 eV but I ended up using points that gave me a value closer to 4.14x10^-15 eV

[odeaa]

Yeah, didn't think you would get a mark.

This is a bit of a specific question again, but if you were finding the value of h from a graph/table with multiple points and you chose a set of points that gave you a value 'outside' the preferable range, would you still get full marks for that? A couple of VCAA questions I did said that a preferable range would be between 4.7 to 5.3 x 10^-15 eV but I ended up using points that gave me a value closer to 4.14x10^-15 eV

I've had that exact problem, and I didn't give myself any marks. Usually, they don't want you to use the values from the table, but the graph itself (that's what the question asks for after all) because the line of best fit will be slightly different to the gradient of two random points. You're usually safe with the x and y intercepts (threshold freq and work function respectively) because they are actually on the graph and not just values from the table

[Kel9901]

A searchlight is beaming out green light. The wavelength of the light is 520nm  and the power of the search light is 5.0kW.

a) Calculate how many photons leave the searchlight each second.
b) What is the momentum of a single 520nm photon.
c) All these photons are focused onto a perfectly reflecting mirror of area 0.1 m^2. Calculate the average force on the mirror.
d) If the mirror in the previous question was replaced by a perfectly absorbing surface, then the answer would:
A - increase by a factor 2
B - decrease by a factor 2
C - remain the same
D - change, but not by a factor of 2

I'm having a bit of trouble on parts c and d. I don't understand how to incorporate the area into the answer. I mean there is no pressure given so that we can. I tried using F = p/ t and multiplying it by the number of photons per second but that isnt giving me the correct answer either. And I don't have any reasoning for any of the answers in part d either.

Any help would be great. Thanks

I think that the answer for c) would be multiplying the change in momentum of one photon by the number of photons per second (which has a unit of s^-1 anyway). The change in momentum that one photon experiences would be double the answer in b), since the photon REFLECTS off the surface (the mirror provides momentum for it to 'stop' AND 'turn around'). So probably 2*what you did

For d), it's probably B, because the photons only 'stop' and don't 'turn around', hence experiencing only half the change in momentum.

[Floatzel98]

For the attached question, how could an electron return to the ground state with 13.6 eV if that is equal to the ionization energy? I know now that n = 3 obviously isn't the highest excited state, but I thought that 0 eV or n = infinity was the ionisation energy, not actually an excitation state.

[silverpixeli]

For the attached question, how could an electron return to the ground state with 13.6 eV if that is equal to the ionization energy? I know now that n = 3 obviously isn't the highest excited state, but I thought that 0 eV or n = infinity was the ionisation energy, not actually an excitation state.

I guess it's a technicality, i think there are lots of excited states up there at not-quite-ionized, and 13.6 is like an asymptote, in that it actually takes you until n=infinity before you get there. also, if an electron was hanging around just outside the electron with no energy i suppose it could 'fall in' emitting a photon on the way down. (waves hands)

[odeaa]

How come when finding the work done using a F/d graph, you use the area under the graph and not simply the force*distance (from the graph)?