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Potassium-40 can decay into either calcium-40 or argon-40. All three of these atoms have essentially the same weight. Ninety percent of the potassium-40 will decay into calcium-40, and only ten percent will decay into argon-40. When argon-40 is produced by the radioactive decay of potassium-40 inside a rock, the argon-40 produced by the decay is a gas and is trapped inside the rock. The amount argon-40 trapped in a rock can be measured by grinding up the rock and capturing the liberated argon-40 gas.

Suppose the amount of potassium-40 inside a rock is measured to be 0.81 milligrams, and the amount of argon-40 gas trapped in the rock is measured to be 0.377 milligrams.


1. How much of the potassium-40 that was originally present inside the rock has undergone radioactive decay to produce argon-40?

Respuesta :

Edufirst
Answser:

3.77 mg of K-40 decayed into Ar-40.

Data:

1) K-40, Ca-40, Ar-40: all three have the same atomic mass
2) 90% of the potassium-40 will decay into calcium-40
3) 10% of the potassium-40 will decay into argon-40.
4) K-40 inside the rock = 0.81 mg
5) Ar-40 trapped = 0.377 mg

Soltuion:

1) 0.377 mg of Ar-40 is the 10% of the mass of the K-40 that decayed

=> x * 10% = 0.377 mg => x * 0.1 = 0.377mg

=> x = 0.377 mg / 0.1 = 3.77 mg

That means that 3.77 mg of K-40 decayed into Ar-40. And this is the answer to the question.

Additionaly, you can analyze the content of all K-40 and Ca-40, to understand better the case.

2) The mass of the K-40 that decayed into Ca-40 is 9 times (ratio 9:1) the amount that decayed into Ar-40 =>

mass of K-40 that decayed into Ca-40 = 9 * 0.377 = 3.393 mg

3) Total amount of K-40 that decayed = amount that decayed into Ar-40 + amount that decayed into Ca-40 = 0.377mg + 3.393mg = 3.77 mg

4) Original amount of K-40 = amount of K-40 that decayed + amount of K-40 present in the rock = 3.77mg + 0.81 mg = 4.58 mg

5) amount of K-40 that decayed into Ar-40 as percent

% = [3.77 mg / 4.58mg] * 100 = 82.31 %.



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