The Governing equation is shown below. Additionaly part (i) I got 3.6 Gyrs I don

Question

The Governing equation is shown below. Additionaly part (i) I got 3.6 Gyrs I don't know if this is correct.

2. Surprises in 40 K decay and 40Ar production. (i) A rock sample from the Canadian shield is found to have the following elemental concentrations: l"Ar 37.63 x 10 7 and Kl 10 6. If the half-life of 40K is 1.2Gyr, how old is this rock sample? (Hint: you may use the equation on page 21 of Lectures 8+9. Also, remember that logo r Inr/ In 2. (ii) When my grand-grand-gran was young, some 3.6 Gyr ago, he placed a rock eontaining 3g of 40 K with no (0g) 40Ar on his bookshelf in a sealed container, for my students to analyze. I just found it, and so finally it's your turn: How much 40 K and 40Ar should you find? Remember that the latter grows like 0.109 x (1 -2 t/tan) for 1g of initial 40K; see again page 21 of Lectures 8+9. (iii) What are the ratios of l'oArl/I40 K] that you found in parts (i) and (ii)? Do you notice anything peculiar about these values?

Explanation / Answer

i) t / 1.2 = ln [1 + [7.63*10-7] / 0.109*10-6 ]
  t = 3.6 Gyr
ii) 40K= 3 / 40 = 0.075
   40Ar = 3 * 0.109*[1-2-3.6/1.2] = 0.286125
iii) [40Ar] / [40K] = 0.286125 / 0.075 = 3.815
   The ratio [40Ar] / [40K] signifies the time elapsed.
  

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