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    It is predicted that for most nuclei R = RoA^1/3 where Ro is a constant.

    Plot a suitable graph to verify this relationship."
    If you see a power law, it doesn't neccesarily have to be a log log graph... maybe try plotting R against A^1/3 as this should have a constant gradient of Ro with an intercept of 0.

    This is because if y = R and x = A^1/3, we have for y = mx + c that m = Ro and c = 0.

    Let me know if that works out!
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    (Original post by spread_logic_not_hate)
    If you see a power law, it doesn't neccesarily have to be a log log graph... maybe try plotting R against A^1/3 as this should have a constant gradient of Ro with an intercept of 0.

    This is because if y = R and x = A^1/3, we have for y = mx + c that m = Ro and c = 0.

    Let me know if that works out!
    wait wait, wouldn't it be y = c + mx cause it's logr = logro + 1/3logA and 1/3 would be the gradient?
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    wait wait, wouldn't it be y = c + mx cause it's logr = logro + 1/3logA and 1/3 would be the gradient?
    If you used logs then yes, but I reckon that you don't need to here... I think that plotting R against A^1/3 will do (especially as there is only one column to work stuff out, and using this you would only need to work out A^1/3...) So no logs need to be taken at all!
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    (Original post by spread_logic_not_hate)
    If you used logs then yes, but I reckon that you don't need to here... I think that plotting R against A^1/3 will do (especially as there is only one column to work stuff out, and using this you would only need to work out A^1/3...) So no logs need to be taken at all!
    AHHH yess, i need to do A^1/3 only...... and for the graph that would be on the x-axis and for the y-axis i just include the numbers for nuclear radius and not the power, i just put that as the units.
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    AHHH yess, i need to do A^1/3 only...... and for the graph that would be on the x-axis and for the y-axis i just include the numbers for nuclear radius and not the power, i just put that as the units.
    Exactly. Just don't forget the units when finding the gradient, e.g. if the gradient is 6 the value of Ro you want is actually 6 x 10^-15!
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    (Original post by spread_logic_not_hate)
    Exactly. Just don't forget the units when finding the gradient, e.g. if the gradient is 6 the value of Ro you want is actually 6 x 10^-15!
    I just have a question on the gradient now, i know you said how to do it but i just wanna make sure i understand this why Ro is the gradient. So
    y = mx + c

    y = R, x= A^1/3 by that logic you are saying that Ro has to be the gradient isn't it?
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    y = R, x= A^1/3 by that logic you are saying that Ro has to be the gradient isn't it?
    Yeah thats right - im comparing the form of the equation I have to that of y = mx + c and equating parts. So on my graph I have

     y = R

     x = A^{\frac{1}{3}}

    which if I compare to y = mx + c gives

     R(y) = R_0(m)\times A^{\frac{1}{3}}(x) + 0 (c)


    Also as a tip for the future, log graphs are usually only plotted for relationships involving an exponential. For example

     I = I_0e^{(-\alpha x)}

    which describes attentuation would be a good candidate for a log graph. In this case you would plot ln(I) against x, as this gives a gradient of  -\alpha and a y-intercept of ln(Io).
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    (Original post by spread_logic_not_hate)
    Yeah thats right - im comparing the form of the equation I have to that of y = mx + c and equating parts. So on my graph I have

     y = R

     x = A^{\frac{1}{3}}

    which if I compare to y = mx + c gives

     R(y) = R_0(m)\times A^{\frac{1}{3}}(x) + 0 (c)


    Also as a tip for the future, log graphs are usually only plotted for relationships involving an exponential. For example

     I = I_0e^{(-\alpha x)}

    which describes attentuation would be a good candidate for a log graph. In this case you would plot ln(I) against x, as this gives a gradient of  -\alpha and a y-intercept of ln(Io).
    Oh i see!!!. The way i have taught myself and my revision guide is that the following

    Y = Ax^p this is a power law, so you use a log-log graph
    Y = Ae^-kx, this is an exponential law, so you use a log-linear graph.... and now when the exponent is there already (for exam purposes) i don't need to do either cause i just use the relationship and if they only give one column that's another indicator it's not either of them. Would that be right?
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    Y = Ae^-kx, this is an exponential law, so you use a log-linear graph
    This is right - you'll always make the question easier by plotting a log-linear graph, as it is impossible to find the gradient of an exponential curve accuratly by hand! (If you want proof, try finding the gradient of a sketch of e^x for x = 1 - 10!)

    Y = Ax^p this is a power law, so you use a log-log graph
    This is like the example we were just talking about - no logs needed here, a plot of Y against x^p will do just fine! The reason is that the graph is a power graph, not an exponential graph - logs should only be taken when you have an exponential graph.

    It is possible to do a log-log graph for power law relationships, but im my experience in most cases this causes unneccesary headaches! If you were to do it it would be like this:

     y = Ax^p

    take logs of both sides (im taking the natural log, ln, as a force of habit)

     ln(y) = ln(Ax^p) = ln(A) + ln(x^p) = ln(A) + xln(p)

    Now comparing this to y = mx + c we see

     m = ln(p)

     c = ln(A)

    so to get A and p you would take exponentials, i.e.

     A = e^c = e^{y intercept}

     p = e^m = e^{gradient}

    Hope that clears it up for you!
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    (Original post by spread_logic_not_hate)
    This is right - you'll always make the question easier by plotting a log-linear graph, as it is impossible to find the gradient of an exponential curve accuratly by hand! (If you want proof, try finding the gradient of a sketch of e^x for x = 1 - 10!)



    This is like the example we were just talking about - no logs needed here, a plot of Y against x^p will do just fine! The reason is that the graph is a power graph, not an exponential graph - logs should only be taken when you have an exponential graph.

    It is possible to do a log-log graph for power law relationships, but im my experience in most cases this causes unneccesary headaches! If you were to do it it would be like this:

     y = Ax^p

    take logs of both sides (im taking the natural log, ln, as a force of habit)

     ln(y) = ln(Ax^p) = ln(A) + ln(x^p) = ln(A) + xln(p)

    Now comparing this to y = mx + c we see

     m = ln(p)

     c = ln(A)

    so to get A and p you would take exponentials, i.e.

     A = e^c = e^{y intercept}

     p = e^m = e^{gradient}

    Hope that clears it up for you!
    Oh ok i see what you mean!!!!! but for the exams they expect you to use log-log graphs
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    Oh ok i see what you mean!!!!! but for the exams they expect you to use log-log graphs
    Ouch, bit harsh. Thinking about it though, in this form

    ln(y) = ln(A) + xln(p)

    you'd only have to actually work out ln(y) to plot the graph, as x would be given, so you could do this with only 1 column available to you for working...
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    (Original post by spread_logic_not_hate)
    Ouch, bit harsh. Thinking about it though, in this form

    ln(y) = ln(A) + xln(p)

    you'd only have to actually work out ln(y) to plot the graph, as x would be given, so you could do this with only 1 column available to you for working...
    Yeah, but the examples they will give in the exam tomoz afternoon should be simple ones and not hard ones hopefully. Anyways, thanks mate for your help, much appreciated!!!!
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    OkAy this is on "thermoluminescence".... this is mainly for electrical insulators that are heated and they are this gives rise to extra energy levels which damages the material. These extra energy levels are known as the "defect levels" this is where the electrons become "trapped". If they are consequently heated the electrons return to their stable energy and as they do so, it emits out light ( or any other electromagnetic radiation). The amount of light given off depends on the amount of electrons in the defect levels. So, it also depends on the radiation as well. So what is the use of this and what possible drawbacks are of this?
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    someone?
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    The amount of light given off depends on the amount of electrons in the defect levels. So, it also depends on the radiation as well. So what is the use of this and what possible drawbacks are of this?
    Usage - thermoluminescent dating. Bit like radiocarbon dating, where the luminescence of the material when it is heated indicates how long ago the material was subjected to stress and radiation.

    Drawbacks - just gonna think a few up here...

    If the material has undergone multiple periods of heating/cooling, only the most recent will be measurable as every time it is heated the luminescence level is effectivley reset. This means it will be dated more recently than it should be.

    Over a very long time the trapped electrons will free themselves of their own accord, so the signal (i.e. luminescence) will eventually be lost. Related to this is that older materials will have a weak signal, so it may be difficult to get a good reading.

    Since the signal is measured on heating the sample, you only get one shot at measuring it - if you mess up, all the electrons have been freed and you can't do the test again!

    Materials that are heated in regions that happen to be shielded from radiation won't have any trapped electrons, so can't be thermoluminescently dated.

    That's all I can think of for now!
 
 
 
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