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Homework Help: Help!:Calculation of decay/sec using t1/2 & No

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data
    The unstable isotope 40K is used to know the age of old rocks formed from molten lava. The half-life of 40K is 1.28 x 10^9 years, and it decays to 40Ar, which is a gas. In the molten rock, all gas escapes, so there is no Ar. However, when the rock is solidified, the gas 40Ar formed by decomposition of 40K is trapped inside the rock.

    a)How many decays per second occur in a sample containing 1.63 x 10^-6 g of 40K?

    b) What is the activity of the sample in Curies?

    c)If the number of 40Ar atoms in the piece of rock is the same as the number of 40K atoms, what is the age of the rock?

    Solution: a)0.421 Bq;b) 1.14 x 10-11 Ci;c) 1.28 x 109 years

    2. Relevant equations
    A= (decay constant) * No * e ^ -(decay constant) * t1/2

    3. The attempt at a solution

    A= (0.693/1.28 * 10^9) * 1.63 * 10^-6 * 2.718 ^ -(decay constant) * t1/2
    A= 882.492

    I attempted to solve part A but was unsuccessful. From some web resources e is approx 2.718. Is this correct? What is e?

    I know to convert Bq to Ci, it's Bq * 2.70*10^−11, is this correct? My answer to Part A is wrong hence i couldn't solve for part B.

    I'm really bad at physics and it would be good if the explanation of the formula could be provided.

    I know A is the activity, t1/2 is the half life, decay constant= 0.693/t1/2, No is the initial quantity but what is e?

    Greatly appreciate all inputs!
  2. jcsd
  3. Jan 16, 2010 #2
    Huh? [tex]A = \lambda N[/tex], where N is the number of atoms of K-40. From there, its just an issue of chucking the numbers in. Though, please watch your units carefully...1.28 x 10^9 is in years and 1.63 x 10^-6 is in grams
  4. Jan 17, 2010 #3
    Ok, so i use [tex]A = \lambda N[/tex] instead of the one i've provided? Is there any difference between the two formula?

    So i convert years to seconds? And let grams remain as grams?
  5. Jan 17, 2010 #4
    From what I think you provided,
    [tex]A = \lambda N = \lambda N_{0} e^{-\lambda t}[/tex]
    This describes how the activity of the sample varies with time. In this case, time does not factor in, for we are already given N at that particular time.

    Yes, you convert years to seconds. Its up to you whether to leave it as grams or not; as long as you correctly obtain the number of atoms of K-40. Do you know how to get that?
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