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Radioactivity 2

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi!
    I'm stuck on this one too.
    "A rock contains the radioactive isotope Rb-87. A piece of this rock contained 2,05*10^20 Rb-87 atoms and 8,25*10^20 decay products per kg. The half-life of Rb-87 is 47*10^9 years. How old is the rock?"
    2. Relevant equations
    [tex]N(t)=N_0(\frac{1}{2})^{t/T_{1/2}} [/tex]


    3. The attempt at a solution
    No nuclei (or atoms) disappear, they only decay. Therefore, the initial number of Rb-87 atoms must have been 2,05*10^20+8,25*10^20 per kg. Let's assume that we have a sample of 1 kg (the mass will change with decay, but not the number of atoms, so it doesn't matter).
    Now, let x be its age. Then we can write
    2,05*10^20=(2,05*10^20+8,25*10^20)(1/2)^(t/T_{1/2}) which gives us that x is approximately 22,9*10^9 years. However, in the key it says 1,7*10^9 years. What's wrong here?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 16, 2010 #2
    Is your question is fully written?
    Please don't use comma or dot in numbers..? it is confusing..(Germany it is common, using comma for dot)
    in your case 't' should be age..but why you introduced 'x'..
    is this problem is solved?
    [tex]
    47\times10^9 {\rm or} 4.7\times10^9
    [/tex]?
     
    Last edited: Apr 16, 2010
  4. Apr 16, 2010 #3
    I dont think you have all the data written down correctly, if 1.7*10^9 is the answer. If you start with 10.3 atoms after one half life there would be 5.15 atoms left and 47*10^9 years would have past but you have fewer atoms left and an answer of less time which is clearly wrong. Unless i haven't understood the question
     
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