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# AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread] watch

1. (Original post by ashwinderk)
If the data was uncorrected, that would mean 2050 count rate given included the background radiation, so surely just using the inverse square equation to find the distance at 0.90m would include the background radiation
No, that would assume that the background radiation also obeyed the inverse square law. In actual fact it doesn't (it's a constant 40 counts/minute), so you need to first take 40 from 2050 to find the corrected count rate, then apply the inverse square law, then re-add the background count rate at the end.
2. (Original post by marioman)
No, that would assume that the background radiation also obeyed the inverse square law. In actual fact it doesn't (it's a constant 40 counts/minute), so you need to first take 40 from 2050 to find the corrected count rate, then apply the inverse square law, then re-add the background count rate at the end.

Ah that makes sense, damn.
3. (Original post by Questioner1234)
Ah that's a good. I got 16700 I think for the indicated power and 48400 for the input

Also on the very last page which box did you tick? I think I did the second one
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Sorry i just saw this
I ticked the second one as well iirc
4. In the turning points paper and the nuclear paper, which boxes did you have to tick?
5. There were boxes in the turning points 😱
6. (Original post by Abc321zxc)
In the turning points paper and the nuclear paper, which boxes did you have to tick?
I didn't do Turning Points but I'm fairly sure that the correct box in Nuclear/Thermal was "some of the bonds are broken" or whatever the exact wording was.
7. (Original post by marioman)
I didn't do Turning Points but I'm fairly sure that the correct box in Nuclear/Thermal was "some of the bonds are broken" or whatever the exact wording was.
Yeah boii. On dat ting bruhhhhhh😎
8. How did you guys find the turning point section compared to the previous years?

Most of the people in the class including myself found it harder than usual
9. (Original post by kelvin1338)
How did you guys find the turning point section compared to the previous years?

Most of the people in the class including myself found it harder than usual
Agreed. What do reckon the a* boundary is ¿
10. (Original post by Leonard Dev)
Agreed. What do reckon the a* boundary is ¿
Most of my classmates saying it's gonna be in the low 20's out of 35 for an A* for the turning points section
11. anyone have a model answer for the 6 marker on turning points, I had literally just revised photocells/photoelectric effect but I thought the wording was a little different.

Was I right in saying that as pd increases KE of electrons decreases as more attractive force/work function higher?
12. (Original post by philo-jitsu)
anyone have a model answer for the 6 marker on turning points, I had literally just revised photocells/photoelectric effect but I thought the wording was a little different.

Was I right in saying that as pd increases KE of electrons decreases as more attractive force/work function higher?
The kinetic energy stays the same. The potential they have to go through increases and because the electrons have a range of kinetic energies some don't have enough energy to overcome the increasing potential so current decreases
13. (Original post by Leonard Dev)
The kinetic energy stays the same. The potential they have to go through increases and because the electrons have a range of kinetic energies some don't have enough energy to overcome the increasing potential so current decreases
Damn....Thats right didn't think of it that deeply, of course the KE transferred stays constant because the frequency of light is constant...Really couldnt wrap my head around this question at the time. Cheers
14. Hello I got an answer wrong and I'm wondering how many marks I lost.

For the count rate one I got the corrected count rate as 2010, then did 2010/36 (6x the distance so 1/36x less power) to get 55.8 counts/min. However I forgot to add the 40 back on at the end. How many marks will I lose?

In addition, with the question with the m - M = log(d/10) I forgot to divide d by 10. Will I lose both marks?

Also, what were the spectral classes for that star question? I remember getting K somewhere but can't remember where.
15. (Original post by PressTabStart)
Hello I got an answer wrong and I'm wondering how many marks I lost.

For the count rate one I got the corrected count rate as 2010, then did 2010/36 (6x the distance so 1/36x less power) to get 55.8 counts/min. However I forgot to add the 40 back on at the end. How many marks will I lose?

In addition, with the question with the m - M = log(d/10) I forgot to divide d by 10. Will I lose both marks?

Also, what were the spectral classes for that star question? I remember getting K somewhere but can't remember where.
Not sure I remember the last two, but you'll drop 1 for the first one

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16. Will I lose a mark for leaving the count rate as 95.8? I hope not since I did all the right things to get the answer!
17. (Original post by Leonard Dev)
The kinetic energy stays the same. The potential they have to go through increases and because the electrons have a range of kinetic energies some don't have enough energy to overcome the increasing potential so current decreases
(Original post by philo-jitsu)
Damn....Thats right didn't think of it that deeply, of course the KE transferred stays constant because the frequency of light is constant...Really couldnt wrap my head around this question at the time. Cheers
This isn't strictly true. The energy supplied is the same, definitely, and they do definitely have the exact same INITIAL kinetic energy, but kinetic energy definitely does decrease when you increase the voltage.

As an example, let's say the work function of a metal is 10J and you supply an electron with 20J. It's maximum kinetic energy is 10J (20J - 10J). No matter what voltage, this will always be the same. However, as you increase the voltage of the metal, it's becoming increasing positive, the electrostatic force of attraction is getting stronger on the electrons. This means that there is a resultant force in the direction of the positive plate, which in turn decelerates the electrons. As KE = 1/2mv^2, as the velocity reduces, the kinetic energy also decreases, so you've no longer got 10J of kinetic energy - this is why they can't reach the negative plate to allow a current to flow

Initial kinetic energy is constant, but kinetic energy is not conserved throughout due to deceleration.
18. (Original post by GrandMasti)
Will I lose a mark for leaving the count rate as 95.8? I hope not since I did all the right things to get the answer!

The question specified an original count rate of 2050, including 40 for the background rate.

2050 - 40 = 2010 from the source at 0.15m

Increasing distance to 0.90m = 6x further away, therefore intensity = 36x less
2010 / 36 = 55.833

Adding the background rate back on gives you 95.83333, or 95.8

You'll get all 3 marks!

Hope it helps
19. (Original post by Jay1421)
This isn't strictly true. The energy supplied is the same, definitely, and they do definitely have the exact same INITIAL kinetic energy, but kinetic energy definitely does decrease when you increase the voltage.

As an example, let's say the work function of a metal is 10J and you supply an electron with 20J. It's maximum kinetic energy is 10J (20J - 10J). No matter what voltage, this will always be the same. However, as you increase the voltage of the metal, it's becoming increasing positive, the electrostatic force of attraction is getting stronger on the electrons. This means that there is a resultant force in the direction of the positive plate, which in turn decelerates the electrons. As KE = 1/2mv^2, as the velocity reduces, the kinetic energy also decreases, so you've no longer got 10J of kinetic energy - this is why they can't reach the negative plate to allow a current to flow

Initial kinetic energy is constant, but kinetic energy is not conserved throughout due to deceleration.
This is correct Ek decreases

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20. (Original post by Jay1421)
This isn't strictly true. The energy supplied is the same, definitely, and they do definitely have the exact same INITIAL kinetic energy, but kinetic energy definitely does decrease when you increase the voltage.

As an example, let's say the work function of a metal is 10J and you supply an electron with 20J. It's maximum kinetic energy is 10J (20J - 10J). No matter what voltage, this will always be the same. However, as you increase the voltage of the metal, it's becoming increasing positive, the electrostatic force of attraction is getting stronger on the electrons. This means that there is a resultant force in the direction of the positive plate, which in turn decelerates the electrons. As KE = 1/2mv^2, as the velocity reduces, the kinetic energy also decreases, so you've no longer got 10J of kinetic energy - this is why they can't reach the negative plate to allow a current to flow

Initial kinetic energy is constant, but kinetic energy is not conserved throughout due to deceleration.
I'm pretty sure the electrons have a range of kinetic energies up to a maximum given by ekmax= hf - phi. Standard unit 2¿🌝

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