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Dating ratio of a rock

  1. Oct 18, 2015 #1
    Suppose a rock sample at the time of its formation 3.6 × 10^9 years ago contained N0[87Rb] rubidium atoms 87 (87Rb), N0[87Sr] strontium atoms 87 (87Sr) and N0[86Sr] strontium atoms 86 (86Sr). From the relative abundance of the three species at the height of the rock formation yielded the following ratios: N0 [87Rb] / N0 [86Sr] = 0.2 and N0 [87Sr] / N0 [86Sr] = 0.7. The 87Rb decays into 87Sr with a half-life of 48.8 × 109 years. Turn the 86Sr and 87Sr are stable.

    Determine the percentage of 87Rb atoms that have survived to this day and Determine the ratio N [87Rb] / N [86Sr] that currently exists in this sample of rock ..


    Can someone explain me how this thing of the ratio works? I don't understand what they want in the question?
     
  2. jcsd
  3. Oct 18, 2015 #2

    BvU

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    Hi Granger, welcome to PF :smile: !

    You'll get used to the culture around here. read the guidelines to see why we impose the use of the template. It's really useful.
    Good thing you separate your questions in one exercise per thread.

    You already posted the governing relevant equations. Once you have the right set of equations, all exercises aren't really that much physics any more, but mathematics. Solving m equations with m unknowns mostly.

    This exercise expands on the governing equations you already posted: now it's about ratios. Simply write down the equations for the time development of each species. Ratios gve you expressions of the type ## A_1 \, e^{t/\tau_1} / A_2 \, e^{t/\tau_2} ## and that can be written as ## B \, e^{t/\tau_3} ##, and then you're back on familiar terrain. Good luck !
     
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