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    What does the lnN against time graph luk like..and its anonlagy with capacitors..Sfteee
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    Straight line through origin.

    Intercept is [ln]No, and gradient = (-)[landa] i think

    analogous with

    capacitors
    [ln]I against t
    Were intercept equals Io, and gradient equals 1/RC
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    it obeys y=mx+c which is a stright line but it doesn't have to be through the origin depends if c is zero when x is zero or not so
    the equation is N=N(initial)e^-λt
    so by lenning both sides to get rid of e the equation becomes
    ln N=lnN°-λt
    y= c-mx
    so lnN° is the y intercept the graph would be a stright decreasing line with a negative slop equals to -λ
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    The analogy with capacitors would be that the chrage left on a capacitor (capacitance C) discharging through a resistor (resistance R) would also lead to a straight line graph when you plot ln(Q) against time. Again this is because:

    Q = Q0e-t/RC
    lnQ = ln(Q0e-t/RC)
    lnQ = lnQ0 - t/RC

    y = mx + c

    with y=lnQ, x=t, m= -1/RC and c = lnQ0

    I think..
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    So The number of nuclei decayed is analogous with the charge in capacitor. that makes sence.

    So that must mean that Current(dq/dt) is analogous with activity(dn/dt)??
    What about voltage?
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    (Original post by SinghFello)
    So The number of nuclei decayed is analogous with the charge in capacitor. that makes sence.

    So that must mean that Current(dq/dt) is analogous with activity(dn/dt)??
    What about voltage?
    according to Q=Q°e^-t/RC
    then λ≡1/RC
    dQ/dt=Q/RC
    same like dN/dt=Nλ
    so dQ/dt=I
    and dN/dt=A
    now the voltage can be applied in the same equation as it's directly proportional to Q
    but nothing indicates that it's anologous to any of the decay equation things
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