A rock has 12.5 percent of its original amount of potassium-40 remaining in it; potassium-40 has a half-life of 1.25 billion years. How long ago was the rock formed?


A1.25 billion years ago

B2.5 billion years ago

C3.75 billion years ago

D5 billion years ago

Verified Answer :

If the rock has 12.5 % of its original amount of potassium-40 remaining in it, this has a half-life of 1.25 billion years, the rock was formed approximately 3.75 billion years ago. The correct answer is C.

Answer : The correct option is, (C) 3.75 billion years ago.

Explanation :

Half-life = 1.25 billion years

First we have to calculate the rate constant, we use the formula :

\(k=\frac{0.693}{t_{1/2}}\)

\(k=\frac{0.693}{1.25\text{ years}}\)

\(k=0.554\text{ billion years}^{-1}\)

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

\(t=\frac{2.303}{k}\log\frac{a}{a-x}\)

where,

k = rate constant  = \(0.554\text{ billion years}^{-1}\)

t = time passed by the sample  = ?

a = let initial amount of the reactant  = X g

a - x = amount left after decay process = \(12.5\% \times (x)=\frac{12.5}{100}\times (X)=0.125Xg\)

Now put all the given values in above equation, we get

\(t=\frac{2.303}{0.554}\log\frac{X}{0.125X}\)

\(t=3.75\text{ billion years}\)

Therefore, the time ago the rock formed was \(3.75\text{ billion years}\)