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Hi Guys, 

Any ideas for this one? Here is what I was able to figure out: since frequency is increased while intensity is constant, current flow must decrease. Hence, it's between A and C. That is because for both of those, the x-intercept is more to the left, meaning that it takes less voltage in order for the current flow to be 0 (makes sense because current flow is weaker when blue light is used). However, I'm stuck when it comes to differentiating between A and C. 

If anyone could tell me if my reasoning so far is correct, and what are the final steps, that would be greatly appreciated,

Good luck tomorrow for everyone!

 

P.S: the answer is C. 

Screen Shot 2016-05-05 at 11.14.17 AM.png

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34 minutes ago, Odelouche said:

Hi Guys, 

Any ideas for this one? Here is what I was able to figure out: since frequency is increased while intensity is constant, current flow must decrease. Hence, it's between A and C. That is because for both of those, the x-intercept is more to the left, meaning that it takes less voltage in order for the current flow to be 0 (makes sense because current flow is weaker when blue light is used). However, I'm stuck when it comes to differentiating between A and C. 

If anyone could tell me if my reasoning so far is correct, and what are the final steps, that would be greatly appreciated,

Good luck tomorrow for everyone!

P.S: the answer is C. 

I would disagree with your analysis that the "current flow is weaker when blue light is used". What happens is that when you replace the red light with the blue light, the energy of each individual photon is increased (due to E = hf), which means that the kinetic energy of each emitted electron also increases (i.e. the emitted electrons are moving faster towards the collecting plate). And in order to stop the flow of these faster-moving electrons, you need a higher stopping voltage. This stopping voltage must be negative in order to resist the flow of electrons in the clockwise direction. Hence, you are correct when saying that the correct answer is either A or C.

Now, to differentiate between A and C, you need to look at the intensity of light. The question states that the blue light emits the same number of photons per second. This means that after you've replaced the light source, you'll still have the same number of electrons being emitted per second. And since current is defined as the amount of charge passing through a cross section of a wire per second, so the current must be the same for both colours of light. The only graph that fits with this analysis is graph C.

Tell me if that makes sense. Good luck with your exam!

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Thanks a lot this makes a lot of sense! Small clarification: the fact that the currents aren't exactly superposed until a certain voltage is applied can be attributed to the fact that, as my teacher often says: "it's not a perfect situation"? 

 

Thanks! 

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3 minutes ago, Odelouche said:

Thanks a lot this makes a lot of sense! Small clarification: the fact that the currents aren't exactly superposed until a certain voltage is applied can be attributed to the fact that, as my teacher often says: "it's not a perfect situation"?

I don't really understand what your teacher means with "it's not a perfect situation". But with my limited knowledge about the photoelectric experiment, I can only speculate that the reason why the 2 graphs are not entirely on top of each other is because there's less supply of electrons to the metal surface when you have a lower voltage supply from the battery.

When you shine the light onto the surface, electrons will be emitted, heading towards the collecting plates. At the same time, electrons in the wire will flow towards the metal surface to replace to electrons that were just emitted. Suppose now that you have a low voltage supply, which implies that the battery is weak and can only pump the electrons in the wire slowly towards the metal surface to replace the emitted electrons. Because this process is slow, so the next time the photons hit the surface, not all the them will get absorbed simply because you don't have enough supply of electrons at the metal surface. Now, if you replace the red light with the blue light, the emitted electrons will move faster to the other side, which consequently increases the pressure forcing the electrons in the wire to flow faster towards the metal surface. In other words, with a blue light, you will have a higher supply of electrons at the surface, which increases the chance for the photons to be absorbed. This explains why the current for blue light is higher compared to the current of the red light when the voltage supply is low. On the other hand, if you have a high voltage supply from the battery, then no matter whether it's blue or red light, you will always have enough supply of electrons at the metal surface waiting to absorb new photons. This explains why the graphs sit on top of each other when the voltage is high.

I'm not sure if my analysis here is correct, but it makes an intuitive sense to me and it also explains why the graph looks the way it looks. Hope this helps!

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