Question

Potassium-40 is radioactive and decays into Argon-40. The half-life of Potassium-40 is 1.3 billion years. You measure the # of atom of each isotope and find 75 atoms of Potassium-40 and 525 atoms of Argon-40. How many half-lives have passed since the rock formed?

Answer 1

Answer: 3 half-lives have passed since the rock formed.

Explanation:

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

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{1.3\times 10^9}=0.53\times 10^{-9}years^{-1}$

b) $t=\frac{2.303}{0.53\times 10^{-9}}\log\frac{75+525}{75}$

$t=3.9\times 10^{9}years$

number of half lives = $\frac{\text {total time}}{\text {half life}}=\frac{3.9\times 10^{9}}{1.3\times 10^9}=3$

Thus 3 half-lives have passed since the rock formed.

Answer 2

The number of lives that have passed since the rock should be formed is 3.

Calculation of the number of lives:

Since

The half-life of Potassium-40 is 1.3 billion years.

Now

The expression for the rate law should be

t = 2.303 k log a/a - x

Here,

k = rate constant

t = age of sample

a = let us assume the initial amount of the reactant

a - x = amount left after decay process  

Now the t0.5 = 0.693/k

= 0.639 / 1.3*10^9

= 0.53*10^-9 years ^-1

Now

The time taken should be 3.9*10^9 years

Finally, the no of lives should be

= Total time / half-life

= 3.9*10^9 / 1.3*10^9

= 3

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