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

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    (Original post by jf1994)
    I'm after the answer to this too, so stupid that they basically just say 'diagram drawn correctly' for full marks...

    I drew horizontal line from tip of object to lens line, then projected it back onto the focal point BEFORE the lens, then drew another line from tip of object to the point of intersection of the axes and drew the image where these two lines intersected, between -F and lens axis
    I did it different, my line went to the focal point after the lens. I would think your way was right, because it's a diverging lens, but that would give a virtual image wouldn't it? The question asks for a real image




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    Can someone explain how electron diffraction experiments supported de brogile hypothesis


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    (Original post by RobHunter97)
    I did it different, my line went to the focal point after the lens. I would think your way was right, because it's a diverging lens, but that would give a virtual image wouldn't it? The question asks for a real image




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    Nah it says real object not real image
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    (Original post by Sbarron)
    Attachment 428295For some reason when I do these rearranging ln ones, the mark scheme puts the values seen in my brackets the other way round!? In this instance they had the same answer.... Now I don't know anything about maths I just learn the rearrangements for physics off by heart really... But I would assume that this make a difference!?
    Does this make sense why is it same both ways



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    Boys I'm happy to say I've figured out June 2014 energy released question!

    You can infact do it two ways:

    1. Calculate the change in binding energy (short) not account for neutrons since they have no binding energy

    2. Calculate the mass defect (long) accounting for neutrons - this does work if you do it properly

    both come down to the same exact answer

    Still gobsmacked by the fact that AQA do not account for the product neutrons K.E at the end..
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    (Original post by betbi3etwerrd)
    Boys I'm happy to say I've figured out June 2014 energy released question!

    You can infact do it two ways:

    1. Calculate the change in binding energy (short) not account for neutrons since they have no binding energy

    2. Calculate the mass defect (long) accounting for neutrons - this does work if you do it properly

    both come down to the same exact answer

    Still gobsmacked by the fact that AQA do not account for the product neutrons K.E at the end..
    So essentially it doesn't matter which method you use. Simply use the quicker one based one what you're given in the question?


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    (Original post by betbi3etwerrd)

    1. Calculate the change in binding energy (short) not account for neutrons since they have no binding energy
    Not sure about this

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    (Original post by CD223)
    It's the way log rules work - don't worry if it confuses you, but you can cancel the minus sign by flipping the fraction inside the "ln" brackets.

    If you want I can provide a more detailed explanation as to why this works, but by the looks of things you get the correct answer your way anyway and won't lose marks for dividing by the negative number instead.


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    Oh ok so the minus makes it work... That's cool that's all I need to know.... I would love to understand why but I doubt I would understand an explanation of it lol and at this moment in time I've got other things to learn. Thanks though 😊
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    (Original post by a.a.k)
    Not sure about this

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    I've got multiple sources approving this and the mark scheme does it this way for that question.

    read pg188 Nelson Thornes AQA A2 physics
    paragraph starting with "Energy is released.."

    or just see that the mark scheme worked out the change in binding energy : http://filestore.aqa.org.uk/subjects...W-MS-JUN14.PDF

    By the way I said the binding energy of neutrons are zero but this is obviously only true for the neutrons that are not in a nucleus.
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    (Original post by CD223)
    So essentially it doesn't matter which method you use. Simply use the quicker one based one what you're given in the question?


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    Well it didn't work for most people who used the mass defect because they were calculating it wrong.

    For this question to find the mass of a nucleus , you had to add up the mass of what it'd be if the neutrons and protons were apart then subtract this value from the binding energy of the nucleus (since by definition binding energy of nucleus is the energy needed to pull a part a nucleus into it's constituent protons and neutrons).

    Do that for each nucleus and then find the change in mass, AKA mass defect, (also account for the neutron masses).

    OR you could simply find the change in binding energy (where the binding energy of neutrons are 0 since they're already constituent.
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    Are you ever asked to calculate the rms speed, or is it always given?
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    (Original post by ubisoft)
    Are you ever asked to calculate the rms speed, or is it always given?
    I'm sure they could easily get us to do that
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    http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF

    HELPPP!!! Q.1 (c) ii)

    heres the mark scheme: http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

    Why subtract the two electron mass not add?
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    (Original post by GraceYux)
    http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF

    HELPPP!!! Q.1 (c) ii)

    heres the mark scheme: http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

    Why subtract the two electron mass not add?
    You want to find the difference in mass between reactants and products. By taking the total mass of the products away from the reactant you get the mass defect in atomic mass units, which you can multiply by 931.3MeV to get the energy released from the process.


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    This question is from June 2011 ASTRO

    2 (b) (ii) The power output of Deneb is 70000 times greater than the Sun. Calculate the radius of Deneb.
    surface temperature of the Sun = 5700 K
    black body temperature of Deneb is 8.5x10^3K

    Please help. I don't understand the last step in the mark scheme.
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    (Original post by CD223)
    You want to find the difference in mass between reactants and products. By taking the total mass of the products away from the reactant you get the mass defect in atomic mass units, which you can multiply by 931.3MeV to get the energy released from the process.


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    But surely to find the total mass of the products, you would add the mass of the He nucleus with the two electrons released?
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    (Original post by GraceYux)
    But surely to find the total mass of the products, you would add the mass of the He nucleus with the two electrons released?
    Yes, then subtract that figure from the mass of the 4 hydrogen nuclei (what I termed as the "reactants").


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    (Original post by GraceYux)
    http://filestore.aqa.org.uk/subjects...1-QP-JUN13.PDF

    HELPPP!!! Q.1 (c) ii)

    heres the mark scheme: http://filestore.aqa.org.uk/subjects...W-MS-JUN13.PDF

    Why subtract the two electron mass not add?
    This is simply a mass defect question. From the binding energy per nucleon curve you can probably estimate that the binding energy per nucleon of Hydrogen (1 proton 0 neutron) is 0 MeV.
    Which basically means that you can take the mass of a proton for the Hydrogen.

    Do everything in atomic mass units (u) as in the formula book it gives you the masses of electrons, neutrons and protons in atomic mass units.

    This makes it easier and at the end you simply multiply your answer by 931.5 to get MeV.


    SO to asnwer your question:

    find TOTAL mass of left hand side (hydrogen)
    find TOTAL mass of right hand side (helium and electron)
    find the difference in these masses (hydrogen - (helium + electron)) = (hydrogen - helium - electron)
    give answer in MeV
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    (Original post by betbi3etwerrd)
    This is simply a mass defect question. From the binding energy per nucleon curve you can probably estimate that the binding energy per nucleon of Hydrogen (1 proton 0 neutron) is 0 MeV.
    Which basically means that you can take the mass of a proton for the Hydrogen.

    Do everything in atomic mass units (u) as in the formula book it gives you the masses of electrons, neutrons and protons in atomic mass units.

    This makes it easier and at the end you simply multiply your answer by 931.5 to get MeV.


    SO to asnwer your question:

    find TOTAL mass of left hand side (hydrogen)
    find TOTAL mass of right hand side (helium and electron)
    find the difference in these masses (hydrogen - (helium + electron)) = (hydrogen - helium - electron)
    give answer in MeV
    Name:  Screen Shot 2015-06-14 at 22.04.20.png
Views: 83
Size:  102.0 KB but its not '(hydrogen - (helium + electron))' , its 'H - (He - electron)
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    (Original post by GraceYux)
    Name:  Screen Shot 2015-06-14 at 22.04.20.png
Views: 83
Size:  102.0 KB but its not '(hydrogen - (helium + electron))' , its 'H - (He - electron)
    no it's not


    it is hydrogen - helium - electron

    which is the same as hydrogen - (helium + electron)
 
 
 
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