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

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    Hi I don't get why E=mc^2 gives total energy in part ii
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    For anyone doing Astrophysics,

    what is the definition of an Airy disc?
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    Can anyone explain question 2 a ii

    I dont get how you can answer it knowing only 1 piece of information?

    http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF
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    (Original post by AR_95)
    Mass of an electron is insignificant compared to the mass of a proton (10^4 bigger)??

    I don't think it would make much difference, Hydrogen atoms are equivalent to protons anyway
    Thanks for the response. On the mark scheme it says you subtract the mass of the positrons produced so this is why I thought it would be significant?
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    (Original post by CD223)
    It is like multiplying by N over N - when you square the root mean square speed to get the mean square speed, you end up with N on the bottom. To cancel this when substituting it into the u squared term on top, multiply by N on top. Does that help?


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    I thought N was due to N number of molecules present because if you don't have N then the equation only applies for a single molecule??
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    (Original post by SuperMushroom)
    Can anyone explain question 2 a ii

    I dont get how you can answer it knowing only 1 piece of information?

    http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF
    Decay constant = ln2/half life
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    Can anyone doing turning points explain to me how the electron tunnelling microscope works? It doesn't make much sense :P
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    (Original post by you-only-live-once)
    i'm doing med phy!! what do you think the 6marker would be on?
    To be honest i have no idea!!! there are so many things to memorise.... and I seriously hope something nice would come out considering today's biology R paper wasn't nice at all...
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    (Original post by AR_95)
    You don't need to know any binding energies. They've given you the mass before the reaction and after the reactions.

    Total mass before - Total mass after , and you'll get a small value of 0. something (u)

    times that value by 931.5 to get the energy released in MeV
    Oh ok, I get it now, thank you
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    I'm reading completely different things in different places about radioactive decay. The Nelson Thornes textbook says that alpha decay occurs in nuclides above the line of stability, but the notes and the guide I'm using both say that is occurs in proton-rich nuclides which would mean below the line, wouldn't it?
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    An Airy disk is the central bright circular region of the pattern produced by light diffracted when passing through a small circular aperture
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    (Original post by miraj3)
    For anyone doing Astrophysics,

    what is the definition of an Airy disc?
    An Airy disk is the central bright circular region of the pattern produced by light diffracted when passing through a small circular aperture
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    (Original post by JordanL_)
    I'm reading completely different things in different places about radioactive decay. The Nelson Thornes textbook says that alpha decay occurs in nuclides above the line of stability, but the notes and the guide I'm using both say that is occurs in proton-rich nuclides which would mean below the line, wouldn't it?
    I believe alpha decay occurs above the line as in past a proton number of around 80Name:  I14-02-pnratio.jpg
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    (Original post by JordanL_)
    I'm reading completely different things in different places about radioactive decay. The Nelson Thornes textbook says that alpha decay occurs in nuclides above the line of stability, but the notes and the guide I'm using both say that is occurs in proton-rich nuclides which would mean below the line, wouldn't it?
    I think alpha emission is just heavy nuclides in general.
    Beta plus/electron capture is for proton rich ones
    Beta - is for neutron rich ones.

    Pretty certain that's right.
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    (Original post by JordanL_)
    I'm reading completely different things in different places about radioactive decay. The Nelson Thornes textbook says that alpha decay occurs in nuclides above the line of stability, but the notes and the guide I'm using both say that is occurs in proton-rich nuclides which would mean below the line, wouldn't it?
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    (Original post by DCMed96)
    To be honest i have no idea!!! there are so many things to memorise.... and I seriously hope something nice would come out considering today's biology R paper wasn't nice at all...
    is R paper meant to be harder..? Yeah i hope something nice would come up too.. I'm thinking more of X-rays or eyes though
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    (Original post by sykik)
    I thought N was due to N number of molecules present because if you don't have N then the equation only applies for a single molecule??
    That's also true, but in the definition of the mean square speed N is on the bottom. In order to sub in the mean square speed for u squared on the top, you must multiply by N on top so the equation for the pressure remains the same.

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    How to do this one?

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    (Original post by JamGrip)
    Ok. I posted basically the same thing back on page 50 but 2b(ii) here: http://filestore.aqa.org.uk/subjects...A-QP-JUN12.PDF (June 2012 astro)
    http://filestore.aqa.org.uk/subjects...W-MS-JUN12.PDF (markscheme)

    It doesn't multiply the answer by two afterwards even though it says it's an orbital radius. Any reason why?

    Thanks
    my guess is you need the diameter of the satellite as your using theta=s/r with s being arc length which is approx the diameter of the satellite and r is the radius from circle which is basically the distance from earth to the satellite. Which is why it multiplied by two in this situation and not in the past paper, I hope this helps
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    (Original post by Sam.B57)
    Can anyone doing turning points explain to me how the electron tunnelling microscope works? It doesn't make much sense :P
    There's two types - constant height and constant current, either is applicable but I find constant height easier to understand.

    Basically, the microscope has a really sharp tip, which is negatively charged, the surface of the sample is positively charged. This means since electrons exhibit wave like properties they propagate between the gap (only from - to +). This transfer of electrons means there's a current between the tip and the surface. The currents magnitude depends on the gap between tip and surface - bigger the gap the less electrons that transfer and hence a smaller current. So as the tip scans across the surface the current increases and decreases if there is a dip or peak, and hence you can build up an image of the surface.

    Probably not very clear sorry
 
 
 
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