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    On this image there is a graph of LnN=LnN0-yt

    I want to know if the graph touches the X axis, and if any exponential decay equations would touch the x axis if arranged in this form. Many thanks!
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    It would touch the x-axis. When put in this form, with these axes it is simply a straight line graph like any other.
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    (Original post by notastampcollector)
    It would touch the x-axis. When put in this form, with these axes it is simply a straight line graph like any other.
    Would it go through it? And in the exponential form it tends towards the X axis but never touches it, isn't that weird if what you said is true??
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    (Original post by kashim91)
    Would it go through it? And in the exponential form it tends towards the X axis but never touches it, isn't that weird if what you said is true??
    No, because in the exponential form not touching the x-axis means "The number of nuclei (N) never becomes zero". When the ln(N) graph crosses the x-axis, this simply means "ln(N) is zero at some point". Which is perfectly possible, as you can see if you set N=1, and doesn't imply that N becomes zero.
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    (Original post by notastampcollector)
    No, because in the exponential form not touching the x-axis means "The number of nuclei (N) never becomes zero". When the ln(N) graph crosses the x-axis, this simply means "ln(N) is zero at some point". Which is perfectly possible, as you can see if you set N=1, and doesn't imply that N becomes zero.
    This makes sense, but what about other exponential decay equations? Should i touch the graph on the x-axis?
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    If you're using the logarithm technique to put the equation in a linear form then you should draw the graph such that it touches the x-axis (literally just treat it as a normal straight line graph, with ln(N) as y and t as x). Obviously if you're drawing the exponential form of the graph it won't touch the x-axis.
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    For radioactive decay:

    The graph would touch X-axis for \ln N=0 \quad \Longleftrightarrow \quad N=\mathrm{e}^0=1. Theoretically, after some time all particles will have decayed, so the graph could pass below X-axis. There are two problems with that, however - first is that \ln 0 is undefined. Second is that - at least I believe so - it usually does not make sense to draw the graph for small values of N. Radioactive decay is a spontaneous process, and there is no way of determining the time between each single decays. The formula N=N_0\, \exp \left( -\lambda t\right) has just a statistical meaning for large enough values of N.
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    (Original post by jaroc)
    For radioactive decay:

    The graph would touch X-axis for \ln N=0 \quad \Longleftrightarrow \quad N=\mathrm{e}^0=1. Theoretically, after some time all particles will have decayed, so the graph could pass below X-axis. There are two problems with that, however - first is that \ln 0 is undefined. Second is that - at least I believe so - it usually does not make sense to draw the graph for small values of N. Radioactive decay is a spontaneous process, and there is no way of determining the time between each single decays. The formula N=N_0\, \exp \left( -\lambda t\right) has just a statistical meaning for large enough values of N.
    THANKS, but should I draw it below or above, I have an experiment similar to this tomorrow, but with pendulum swings instead
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    At the point Ln N = 0 then N=1
    It's just the point on the graph that corresponds mathematically to their being just one atom left.
    If you take the graph beyond that where Ln N<0, it corresponds to N being a fraction of an atom!
    Mathematically this is no problem.
    The problem arises when you try to fit that result to the real world.
    You can't really analyse what goes on when there are just one or two atoms left using the exponential function. It's now just the probability of a random event.
    And you can't, of course, have a fraction of an atom. (At least not in this context.)
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    (Original post by kashim91)
    THANKS, but should I draw it below or above, I have an experiment similar to this tomorrow, but with pendulum swings instead
    The amplitude will decay exponentially and a graph of Ln A against time will be a straight line with negative slope. The slope will be related to the amount of damping due to friction.
    It is quite possible for this graph to touch the t axis. That would correspond to the value of the amplitude being equal to unity. In other words, if you are measuring in cm it would be the point where it is 1cm.
    If you measure in meters the graph would start below the axis because the values will be less than 1.
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    Many thanks to everyone who replied, I think I've come to a conclusion now
 
 
 
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