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G485 June 2012 Paper and unofficial mark scheme

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    some knows how to do 4.a and 4 .d ?
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    for 4 if nobody has this
    not sure about a but i said something about conservation of momentum and how speed of gold particle = (speed of alpha*mass of alpha)/(mass of gold nucleus) but not sure if its right
    b should be lines curving away from the opposite particle with arrows aiming away from each other i think?
    c use F=Qq/4(pi)(E0)r^2 with Q as 79*1.6*10^-19 and q as 2*1.6*10^-19, substitute r and get about 10N
    d 1/r^2 graph, for safety precautions it should pass through (12,2.5) as well as the X i think, but i didnt extend it beyond the X the other way, not sure if you were meant to since X is the min distance the alpha particle comes to the nucleus, so not sure if it would get both marks

    6a the neutrons released in the decay can be absorbed by more uranium nuclei causing them to undergo fission etc
    b - not sure how well i did on this one so im not sure if this is all marks available

    fuel rods contain the radioactive uranium which undergoes fission for energy
    control rods are made of C-12 which absorb slow neutrons and hence controls the chain reaction as less neutrons can be absorbed by uranium

    i took a guess at the moderators because i had no idea

    moderators control depth of C rods so allows control of energy output of whole nuclear reactor or something?

    PS I also had 5 but Hamer got there first XD
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    (Original post by Yosemite80)
    for 4 if nobody has this
    not sure about a but i said something about conservation of momentum

    I said as the alpha particle approaches the gold atom it decelerates losing kinetic energy , and gold nuclei gains kinetic energy due to the loss of kinetic energy of the alpha particle, therefore it accelerates. I think i'm missing a point or too.

    b should be lines curving away from the opposite particle with arrows aiming away from each other i think?

    I drew a graph on this forum see if it's similar

    c use F=Qq/4(pi)E0r^2 with Q as 79*1.6*10^-19 and q as 2*10^-19, substitute r and get about 10N

    d 1/r^2 graph, for safety precautions it should pass through (12,2.5) as well as the X i think, but i didnt extend it beyond the X, so not sure if it would get both marks

    Did inverse square law but not through that point. How did you calculate that point?

    and 5 if nobody has beaten me to it
    a - neutrons = 234-92=142

    same.

    b- KE=eV=5.6*10^6*1.6*10^-19

    same

    bii - speed=sqr(2KE/m) with mass substituted and KE from prev q

    same.

    c - each nuclei generated 5.6MeV or 9*10^-13J so divide power by that to get initial activity

    same.
    cii - half life * lambda = ln2 so rearrange for lambda, half life = 88*3.16*10^7, lambda should be 2.49*10^-10

    same.

    ciii - rearrange A=N(lambda) and N is 2.81*10^23

    same.

    ciii2 - divide N by avogadro constant and multiply by molar mass (0.24*(2.81*10^23/6.02*10^23)) to get 0.111-0.112kg

    I got different i thought the molar mass was 0.24kg then I multiplied by the relative atomic mass of 238 to get 57.12kg

    6a the neutrons released in the decay can be absorbed by more uranium nuclei

    same.

    b - not sure how well i did on this one so i wont risk putting it on here
    lol
    Well i Put something like this.

    Moderator such as carbon dioxide can be used to slow down neutrons to increase the probability of colliding with a uranium atom.
    Control rods control the rate of reaction lowering them slows down the reaction as neutrons have to pass through a medium of graphite/carbon . Lifting the control rods out of the reaction increases the rate of reaction.
    Fuel rods contain the radioactive material that is used for induced fission such as plutonium-238 uranium-235.
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    that point was 2 times the min distance so force would be a quarter of the force at min distance
    i hope the marks are for starting at 6,10 and correct shape though

    ive put my answer for 6b on there but not sure if it would get all marks :3 going by your answer ive probs lost some marks on the moderator lol XD
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    (Original post by Z REFAN Z)
    Well i Put something like this.

    Moderator such as carbon dioxide can be used to slow down neutrons to increase the probability of colliding with a uranium atom.
    Control rods control the rate of reaction lowering them slows down the reaction as neutrons have to pass through a medium of graphite/carbon . Lifting the control rods out of the reaction increases the rate of reaction.
    Fuel rods contain the radioactive material that is used for induced fission such as plutonium-238 uranium-235.
    yeah your 4b was similar to mine
    and 4c for q i meant 2*1.6*10^-19 XD
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    I made so many stupid little mistakes
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    The control rods are made from boron as it does absorb Neutrons. Moderator is Carbon as it doesn't absorb neutrons
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    (Original post by bobska)
    The control rods are made from boron as it does absorb Neutrons. Moderator is Carbon as it doesn't absorb neutrons
    Water (heavy and light varieties) is probably a more common moderator, and it can be useful if the moderator absorb neutrons especially as their temperature increases as this reduces the chances of a meltdown, or like water just slows down the neutrons less at higher temperatures.
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    oh yeah, forgot the rest of 6 :P

    6c - 3GW/0.22=13.6GW oe (1.3636363636363636*10^10)
    ii - Since it's asking for total energy I did (ci)*8.64*10^4 to get 1.178181818181818*10^15J, hope its right :3
    iii - total energy divided by 3.2*10^-11 then multiplied by mass of one uranium nucleus to get about 14.4kg? I completely forgot what i got for this one lol
    d- long half life so dangerous for a long time i think, not sure if it gets both marks
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    That's what the control rods are for. The moderator is only meant to slow the neutrons down, water can also be used.
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    (Original post by joefoxon)
    I think they deliberately tried to catch people out with it. Seems to have worked
    I really don't...
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    (Original post by Picture~Perfect)
    I think I probably got an A, *fingers crossed* for 90%
    I checked the june 2011 boundaries and 90% was 71. I think the grade boundaries will be similar as not may people will have bothered to memorise the MRI section, as if has already come up. Plus not many people will have gotten full marks on the universe question. I'm definitely hopeful for low boundaries
    However, now the exam is 2hrs the boundaries are likely to go back up...
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    Would you lose all 3 marks for not dividing by 2 in the gamma ray photon wavelength equation if everything else was correct?...
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    (Original post by Dorkus Maximus XII)
    However, now the exam is 2hrs the boundaries are likely to go back up...
    The one in January was also 2 hours though.
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    (Original post by SS*)
    Would you lose all 3 marks for not dividing by 2 in the gamma ray photon wavelength equation if everything else was correct?...
    Probably lose only one mark
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    (Original post by moorbre)
    Probably lose only one mark
    I really really hope so Also, what if you didn't divide by 1000 in the last Mpc question?...

    (Sorry for asking so many questions but i'm so worried!)
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    Power of ten errors only lose you one mark.
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    I think the answer for the first question may well be wrong. on the question, they say it has a steady current which means it's not an alternating current.
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    I kinda liked the paper!
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    (Original post by Z REFAN Z)
    anyone get something like this? I think the lines have to be more curvy and going backwards, but i don't think it matters that much.

    http://imgur.com/7OZCV
    For that question I just drew an arrow pointing left from the gold nucleus and an arrow pointing right from the He nucleus.
    I'm sure you will get the marks for that though
    Seems correct

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Updated: June 15, 2012
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