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AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] Watch

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    Anyone got a link for the old spec past papers?


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    (Original post by BoigaDendrophila)
    Anyone got a link for the old spec past papers?


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    They're in the first post


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    What does it mean by certain radioactive isotopes are easier to screen? Where can I find information on this in the nelson thornes book
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    (Original post by AR_95)
    What does it mean by certain radioactive isotopes are easier to screen? Where can I find information on this in the nelson thornes book
    I assume it's referring to the various penetrating powers (of alpha, beta +-, gamma) of the radioactive emissions. For example, alpha would be easier to screen than gamma because alpha has a lower penetrating power.
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    (Original post by AR_95)
    What does it mean by certain radioactive isotopes are easier to screen? Where can I find information on this in the nelson thornes book
    (Original post by PotterPhysics)
    I assume it's referring to the various penetrating powers (of alpha, beta +-, gamma) of the radioactive emissions. For example, alpha would be easier to screen than gamma because alpha has a lower penetrating power.
    Yeah screening is an alternative to "shielding"

    Hence why lead aprons are often worn etc.


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    (Original post by CD223)
    Yeah screening is an alternative to "shielding"

    Hence why lead aprons are often worn etc.


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    That makes sense.

    In the cgp book it says that a nuclear reactor is surrounded by a thick concrete casing which acts as a shield. Nuclear fission produces gamma radiation. On another page, it says that gamma radiation can be blocked by several meters of concrete. Does that mean nuclear reactors are cased in several meters of concrete?! (Edit: What I was thinking is it would be more practical (due to size considerations, etc.) to use a lead casing, as then it would only be a few cm thick.)
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    (Original post by CD223)
    Does anyone know when you do and don't you take into account the mass of a particle/atom when considering binding energy? Is it when they're on their own? Ie: 1 electron/neutron/proton?


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    You can't have a binding energy between 1 thing. I considered a hadron comprised of quarks but they can't be classed as binding energy due to gluons and whatnot, anyway, tangent.
    The binding energy is the energy required to break up a nucleus into its consistent protons and neutrons. Hence, the mass difference times c² is the binding energy. We always account for the mass for a particle. It's important to note for these questions you use the different masses for the proton and neutron unless otherwise stated (as the 3rd decimal place is important). If a particle is solitary binding energy=0 as it cannot be broken down further


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    Funny enough I was talking about the topic yesterday but...


    How am I supposed to work out the mass of the left hand side ffs

    EDIT :
    Nevermind I didn't realise it was in the data sheet
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    (Original post by AR_95)
    Funny enough I was talking about the topic yesterday but...


    How am I supposed to work out the mass of the left hand side ffs

    EDIT :
    Nevermind I didn't realise it was in the data sheet
    Energy is released from the nucleus so we only need to consider nuclear mass. Nuclear mass of left hand side is just 4 times the mass of a hydrogen nucleus, that is, 4 times the mass of a proton, i.e. 4u. This is because a lone proton doesn't have any mass defect.

    Regarding edit: Is my answer incorrect?

    Oh wait, yes, sorry. The proton rest mass isn't exactly 1u, we have to take it up to 5 decimal places since that is what the He one is given to. Silly oversight
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    (Original post by PotterPhysics)
    Energy is released from the nucleus so we only need to consider nuclear mass. Nuclear mass of left hand side is just 4 times the mass of a hydrogen nucleus, that is, 4 times the mass of a proton, i.e. 4u. This is because a lone proton doesn't have any mass defect.

    Regarding edit: Is my answer incorrect?
    4u yes but you're not supposed to consider it as 4u. I realised it was 4 protons beforehand, but I couldn't find anywhere that the mass of proton = 1.00728U which is usually given in these types of questions. After a while and constant staring at the formula sheet, I found that the mass (1.00728u) for proton is actually in the list of all the constants (it's just under proton rest mass). The same is given for neutron and electron which is quite frankly wonderful
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    (Original post by AR_95)
    Funny enough I was talking about the topic yesterday but...


    How am I supposed to work out the mass of the left hand side ffs
    Assume that the electron neutrino has zero mass.
    u is a constant. It would be wise to convert to kg for simplicity.
    u=1.661x10^-27
    Mass after=(4.00150u=6.646915x10-27kg)+2(9.11x10-31)
    Mass before=4(1.673x10-27)
    change in mass=4.3263x10-29
    E=mc^2
    E=3.89367x10-12J
    Converting to electron Volts is to divide by the magnitude charge of the electron (1.60x10-19)
    In MeV=24.34MeV
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    It took so long for me to type I didn't see you'd solved
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    Btw I saw and have done most of the past papers from the "OLD PPQ's" link, however I did see some people mentioning 2002/2003 questions. where are these from?
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    (Original post by Amanzz)
    Assume that the electron neutrino has zero mass.
    u is a constant. It would be wise to convert to kg for simplicity.
    u=1.661x10^-27
    Mass after=(4.00150u=6.646915x10-27kg)+2(9.11x10-31)
    Mass before=4(1.673x10-27)
    change in mass=4.3263x10-29
    E=mc^2
    E=3.89367x10-12J
    Converting to electron Volts is to divide by the magnitude charge of the electron (1.60x10-19)
    In MeV=24.34MeV
    I always knew how to do the question. My only problem (a small one) was finding the value of Mp (mass of proton in U) which I found was given on the formula sheet

    Your way of doing it brings you close to the final answer but it wouldn't of been accepted in the mark scheme, which specifies 24.7 MeV

    The way I did it/mark scheme suggested was

    (4x1.00728u) - [ (4.00150u) + ( 2 x 5.4x10^-4u)

    Which gave a remainder of 0.02652 u
    Simply multiplying it by 931.3 gives 24.69 IE 24.7

    I think you'd get 1 or 2 marks for the right method though
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    Do we use 931.3MeVu^-1 (as given in teh cgp book) or 931.5MeVu^-1 (as given in the aqa data booklet)? I'm thinking the latter but everyone uses 931.3...
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    Also guys I just want reassurance. Heard a dreadful rumour from my teacher last year that they weren't giving you an equations sheet in the exam this year? Is that even remotely true?
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    (Original post by PotterPhysics)
    Do we use 931.3MeVu^-1 (as given in teh cgp book) or 931.5MeVu^-1 (as given in the aqa data booklet)? I'm thinking the latter but everyone uses 931.3...
    Oh wow, I didn't even realise this haha. The mark scheme for that question included 931.5 initially, but in brackets next to it said allow 931.3. In this question using both values you still end up rounding to 24.7 so it wouldn't have mattered. I think both are just as acceptable but I'll keep an eye out from now on, on what they prefer.
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    (Original post by AR_95)
    Also guys I just want reassurance. Heard a dreadful rumour from my teacher last year that they weren't giving you an equations sheet in the exam this year? Is that even remotely true?
    I highly doubt it they would suddenly change the data booklet when the specification is the same. Perhaps he was referring to next year's new spec?
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    (Original post by AR_95)
    Also guys I just want reassurance. Heard a dreadful rumour from my teacher last year that they weren't giving you an equations sheet in the exam this year? Is that even remotely true?
    Nope, completely wrong. They may be removing them or decreasing how much is on them for the new syllabus (I've heard rumours), but nothing to affect us.
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    (Original post by PotterPhysics)
    That makes sense.

    In the cgp book it says that a nuclear reactor is surrounded by a thick concrete casing which acts as a shield. Nuclear fission produces gamma radiation. On another page, it says that gamma radiation can be blocked by several meters of concrete. Does that mean nuclear reactors are cased in several meters of concrete?! (Edit: What I was thinking is it would be more practical (due to size considerations, etc.) to use a lead casing, as then it would only be a few cm thick.)
    Yeah they use concrete.

    Purely due to structural engineering it's less practical to use lead due to container fatigue etc.


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