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    (Original post by Yo12345)
    This is correct Ek decreases


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    Yo. Initial max ke stays the same. final max ke decreases until v=stopping potential. What do you mean by "ek decreases" anyway ¿🤔
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    (Original post by Leonard Dev)
    I'm pretty sure the electrons have a range of kinetic energies up to a maximum given by ekmax= hf - phi. Standard unit 2¿🌝
    They have a range up to a max when there's no potential applied across the photocell. Once you apply a voltage across it, the initial max KE = (electron charge, e) multiplied by the potential difference across the photocell.

    The range still exists, but they only have a maximum kinetic energy very briefly when they're first released from the metal. After that, an electrostatic force of attraction in the opposite direction of travel causes them to slow down. As KE = 1/2mv^2 and v reduces (m is constant), kinetic energy decreases.

    The bottom line is, they receive the same energy from the incident photons regardless of whether there's a potential difference, but if there is a potential difference, that kinetic energy will THEN decrease, after the electrons have received the energy.

    Hope it helps! If you still don't understand, let me know and I'll try to help you
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    (Original post by Jay1421)
    This answer is correct!

    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
    Yes! Thanks a lot!
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    (Original post by GrandMasti)
    Yes! Thanks a lot!
    No problem! If there's any others you're unsure about, let me know and I'll try to help!
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    (Original post by Leonard Dev)
    Yo. Initial max ke stays the same. final max ke decreases until v=stopping potential. What do you mean by "ek decreases" anyway ¿🤔
    We use 'Ek' has kinetic energy


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    (Original post by Jay1421)
    No problem! If there's any others you're unsure about, let me know and I'll try to help!
    Hey Jay , you`ve been really helpful on this thread! what do you think the A boundary will be ? any guesses?
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    For the specific charge of an electron on the turning points paper, what numerical value were we supposed to get. I got 1.9x10 to the something
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    (Original post by Hdnvwizndb)
    For the specific charge of an electron on the turning points paper, what numerical value were we supposed to get. I got 1.9x10 to the something
    I got this too.....I think its correct and 1.6*10^-19/9.11*10^-31 gives a similar answer and this is the actual e/m ratio
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    (Original post by allofthestars)
    Hey Jay , you`ve been really helpful on this thread! what do you think the A boundary will be ? any guesses?
    Hey - for the Turning Points option which I did, based on the last few years and it's difficulty relative to those, I think the A boundary will be at around 49 / 50 marks.
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    Euclidean C0balt

    there was a question on applied about adiabatic gas rising or something in kilometres. Did you get 0.9x km for that?
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    (Original post by Jay1421)
    Hey - for the Turning Points option which I did, based on the last few years and it's difficulty relative to those, I think the A boundary will be at around 49 / 50 marks.
    How difficult did you think it was compared to other papers?


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    (Original post by Yo12345)
    How difficult did you think it was compared to other papers?


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    I think it was definitely harder than 2010, 2011 and 2015 - I think it was pretty on par with 2012 / 2013, not massively difficult but I think there will be reduced grade boundaries compared to the first 3 years I mentioned because there were some 'wordy' questions that could cause some people distress.

    Overall, I don't think it was the hardest or easiest paper, rather just in the middle (as I said, 2012, 2013 as a general difficult comparison).

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    (Original post by Mango Milkshake)
    Euclidean C0balt

    there was a question on applied about adiabatic gas rising or something in kilometres. Did you get 0.9x km for that?
    I can't remember off the top of my head. Gut is telling me I got around 1.7km though.
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    (Original post by Euclidean)
    I can't remember off the top of my head. Gut is telling me I got around 1.7km though.
    Sorry i didnt mean the actual height, I mean the ratio.

    Any ideas?
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    (Original post by Jay1421)
    It was 4 marks because the equation isn't given, so you either have to know the equation or the derivation for it. It asked you to calculate the specific charge given some values of B, r and V - it never asked anything about deriving it. They will probably give marks for derivation, but also the mark scheme will likely say "If 1.9x10^11 Ckg^-1 seen, 4 marks". They can't take marks away because you know the equation MORE than other people, that'd be completely nonsensical.
    1.9? Jesus christ. I knew the equation off the top of my head, correctly wrote it down onto the paper, yet somehow my calculator got 8.6*10^11.

    My examiner is going up think I'm retarded
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    (Original post by Mango Milkshake)
    Sorry i didnt mean the actual height, I mean the ratio.

    Any ideas?
    Sorry I can't remember
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    (Original post by lucabrasi98)
    1.9? Jesus christ. I knew the equation off the top of my head, correctly wrote it down onto the paper, yet somehow my calculator got 8.6*10^11.

    My examiner is going up think I'm retarded
    It sounds to me like you've forgotten to square either B or r, since your answer is around 4x the correct one - just an idea - Either way, if you correctly wrote it, you'll get some credit for that, as well as a mark for getting the unit right (Ckg^-1), so considering the mistake you won't fair too badly, hopefully.

    Fingers crossed for you!
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    for the density question in nuclear/thermal was one of the marks awarded for putting the units of density?
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    (Original post by Mango Milkshake)
    Euclidean C0balt

    there was a question on applied about adiabatic gas rising or something in kilometres. Did you get 0.9x km for that?
    Sounds familiar... That's how much the air actually rose right?
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    (Original post by C0balt)
    Sounds familiar... That's how much the air actually rose right?
    I think it was a ratio of some sort, adiabatic lapse rate. You had to divide your worked out height by the given one to find some ratio. 0.9 something i got
 
 
 
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