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    (Original post by cowie)
    Think back to diffraction grating at AS Physics. So: d sin (theta) = n(lambda). It's the first minima you're talking about so discard n as it = 1. This gives you d sin(theta) = lambda. Rearrange to find d: d = lambda/sin(theta) but they seem to convert it into d = (1.22 x lambda)/ sin(Theta) for diameter of an nucleus and then d = (0.61 x lambda)/ sin(Theta) for radius.
    correct me if i'm wrong..

    https://www.youtube.com/watch?v=L0q8u0N5K_Y
    this guy explains it very well
    is it not: d = 0.61sin (theta)/lambda ?

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    (Original post by arrow_h)
    Guys do we need to know how to derive PV=nRT and PV=nKT ???? If so, how do we derive it
    This covers it: https://www.youtube.com/watch?v=SwdccM91SMM
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    Anyone who does turning points, do we need to know Hertz's experiment ???
    It's come up in a few older papers but not in spec?
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    (Original post by fisher96)
    Anyone who does turning points, do we need to know Hertz's experiment ???
    It's come up in a few older papers but not in spec?
    Yeah because it has come up as a 6 marker where you were asked why stationary waves were formed and how the wavelength could be determined using a detector. Probably won't come up as a 6 marker again but they could just ask a question or two about it.
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    (Original post by Franckenstar)
    What year is that ?


    June 2011
    Sorry forgot to say what year it is !
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    Can someone explain the reason behind this:

    Nuclear power stations in the UK are fuelled by fission of uranium-235. Uranium occurs naturally in the Earth's crust. Unfortunately it contains only 0.7% uranium-235. Natural uranium contains mostly uranium-238. This absorbs neutrons without undergoing fission.

    Is the part in bold mainly to do with the N-Z curve? I looked at the N-Z curve and I can't link it to it . Does it simply absorb neutrons because it's proton rich?
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    (Original post by particlestudent)
    Can someone explain the reason behind this:

    Nuclear power stations in the UK are fuelled by fission of uranium-235. Uranium occurs naturally in the Earth's crust. Unfortunately it contains only 0.7% uranium-235. Natural uranium contains mostly uranium-238. This absorbs neutrons without undergoing fission.

    Is the part in bold mainly to do with the N-Z curve? I looked at the N-Z curve and I can't link it to it . Does it simply absorb neutrons because it's proton rich?
    I can only think of two things

    1- The neutron converts into a proton ( using feynman's diagram ) but this would trigger fission.
    2- The neutron isn't fast enough to trigger fission.
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    (Original post by particlestudent)
    Can someone explain the reason behind this:

    Nuclear power stations in the UK are fuelled by fission of uranium-235. Uranium occurs naturally in the Earth's crust. Unfortunately it contains only 0.7% uranium-235. Natural uranium contains mostly uranium-238. This absorbs neutrons without undergoing fission.

    Is the part in bold mainly to do with the N-Z curve? I looked at the N-Z curve and I can't link it to it . Does it simply absorb neutrons because it's proton rich?
    Fission occurs when a nucleus absorbs a neutron and subsequently is unstable (this is for reasons that include an average increase in size of the nucleus, but with no added attractive force i.e some nucleons are separated further from each other and so the strong force is weaker under this larger average separation), that means there exists two nuclei that it can fission into that would reduce the overall energy (an increase in binding energy per nucleon). U238 upon absorbing a neutron doesn't become unstable with respect to fission, but rather becomes unstable with respect to beta and so it actually undergoes beta decay (I think it undergoes double beta to plutonium which is subsequently fissile).

    U238 is fissionable but has a really huge critical energy and thus you need extremely high energy neutrons to trigger any form of fission.

    I'm not quite sure what you mean by the N-Z curve, do you perhaps mean the binding energy curve?

    There are a range of explanations as to why U238 is not fissionable, but by the definition of fissionable it is, however, it cannot sustain a chain reaction which is the most important thing as not many neutrons are given off at reasonable energies to cause subsequent fission.

    (I took this exam last year so I'm not quite sure what AQA would want as an answer, but these are just general comments on the topic)
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    (Original post by kother2015)
    June 2011
    Sorry forgot to say what year it is !
    Qin = Qout + W ,according to the second law.
    Hence, Qin > Qout and so a heat pump delivers more energy than is supplied to it.
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    Astrophysics question:

    Can anyone help me with this question? Its the astro part of June2010 question 2bii. The question is in the spoiler. Thanks
    Spoiler:
    Show

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    (Original post by Protoxylic)
    Fission occurs when a nucleus absorbs a neutron and subsequently is unstable (this is for reasons that include an average increase in size of the nucleus, but with no added attractive force i.e some nucleons are separated further from each other and so the strong force is weaker under this larger average separation), that means there exists two nuclei that it can fission into that would reduce the overall energy (an increase in binding energy per nucleon). U238 upon absorbing a neutron doesn't become unstable with respect to fission, but rather becomes unstable with respect to beta and so it actually undergoes beta decay (I think it undergoes double beta to plutonium which is subsequently fissile).

    U238 is fissionable but has a really huge critical energy and thus you need extremely high energy neutrons to trigger any form of fission.

    I'm not quite sure what you mean by the N-Z curve, do you perhaps mean the binding energy curve?

    There are a range of explanations as to why U238 is not fissionable, but by the definition of fissionable it is, however, it cannot sustain a chain reaction which is the most important thing as not many neutrons are given off at reasonable energies to cause subsequent fission.

    (I took this exam last year so I'm not quite sure what AQA would want as an answer, but these are just general comments on the topic)
    Thanks that makes sense!

    By the N-Z curve I meant the graph of neutron number against proton number that shows where stable nuclei lie (and also shows where beta +/- and alpha emitters lie).
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    Astrophysics

    Question 1bi.) I don't understand where the 51 comes from? If someone could explain to me how they get the 51 that would be great thankyou!

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
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    (Original post by particlestudent)
    Thanks that makes sense!

    By the N-Z curve I meant the graph of neutron number against proton number that shows where stable nuclei lie (and also shows where beta +/- and alpha emitters lie).
    I'm aware of the N-Z curve I was just confused as to how it fits in the argument of fission
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    (Original post by Hattie28)
    Astrophysics

    Question 1bi.) I don't understand where the 51 comes from? If someone could explain to me how they get the 51 that would be great thankyou!

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
    M=50 and M= fo/fe, 50=fo/fe rearrange to fo = 50fe

    Total length of telescope= fo + fe
    fo + fe = 3.7

    sub in fo=50fe

    50fe + fe= 3.7 therefore 51fe=3.7
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    (Original post by Hattie28)
    Astrophysics

    Question 1bi.) I don't understand where the 51 comes from? If someone could explain to me how they get the 51 that would be great thankyou!

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
    You make two equations:

    Fo + Fe = 3.7
    so Fo = 3.7 - Fe

    Then substitute that into the second equation which is:

    Fo/Fe = 50 so Fo = 50Fe
    3.7 - Fe = 50Fe

    and so 51Fe = 3.7

    That's how they got 51
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    (Original post by Protoxylic)
    I'm aware of the N-Z curve I was just confused as to how it fits in the argument of fission
    Oh right. I just quickly assumed if it absorbs neutrons, it must lie at the bottom of the curve, but then I took a look at the graph and uranium is at the complete other end of the graph .

    Also could you help me with question 2 on here:
    https://e4cf8bb391554b7c9d8e0fc42269...r%20Energy.pdf

    The answers 0.117kg, I've looked at the mark scheme but it's not helpful at all, just numbers every
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    (Original post by particlestudent)
    Oh right. I just quickly assumed if it absorbs neutrons, it must lie at the bottom of the curve, but then I took a look at the graph and uranium is at the complete other end of the graph .Also could you help me with question 2 on here:https://e4cf8bb391554b7c9d8e0fc42269...r%20Energy.pdfThe answers 0.117kg, I've looked at the mark scheme but it's not helpful at all, just numbers every

    Power*time elapsed = Total useful energy released
    Total energy released per fission = 10 * total useful energy released = number of plutonium atoms*energy released per plutonium atom (since efficiency is 10%)
    --> Find number of Pu atoms
    --> Find mass by 239*[Number of Pu atoms]/[Avagadro's Number] (this is in grams)
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    (Original post by Protoxylic)
    Power*time elapsed = Total useful energy released
    Total energy released per fission = 10 * total useful energy released = number of plutonium atoms*energy released per plutonium atom (since efficiency is 10%)
    --> Find number of Pu atoms
    --> Find mass by 239*[Number of Pu atoms]/[Avagadro's Number] (this is in grams)
    Thanks! It's the 0.239 instead of 239 that through me off (the grams part). I'm sure the radioactive and energy chapter can't get any worse than this .
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    (Original post by particlestudent)
    Thanks! It's the 0.239 instead of 239 that through me off (the grams part). I'm sure the radioactive and energy chapter can't get any worse than this .
    No problem. The molar mass (239 for Pu) is in grams per mole
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    (Original post by C0balt)
    Does anybody have the original 2015 paper before it got replaced due to theft?
    We did the Nuclear/Thermal and Astrophysics original papers as a mock. You aren't missing much at all, all they changed were a few numbers in the calculations.
 
 
 
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