Question

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?

A. 1.25 billion years ago

B. 2.5 billion years ago

C. 3.75 billion years ago

D. 5 billion years ago

Answer 1

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=(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=(2.303)/(k)\log(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\% * (x)=(12.5)/(100)* (X)=0.125Xg`

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

`t=(2.303)/(0.554)\log(X)/(0.125X)`

`t=3.75\text{ billion years}`

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

Answer 2

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.