Rate Of Decay Formula

The decay of a particular nucleus cannot be predicted and is not affected by physical influences like temperature, unlike chemical reactions. The rate of isotope decay depends on two factors. The total number of undecayed nuclei present in the system on doubling the average and undecayed nuclei must double the rate of decay. The stability of the isotope since some isotope decay rapidly. The rate of decay gives the number of nuclei that decay per second.

Rate of Decay Formula is given as

\(\begin{array}{l}\frac{-d[A]}{dt} = k[A]\end{array} \)

On integrating the above equation, we have

\(\begin{array}{l}ln\frac{[A_0]}{[A]}=kt\end{array} \)

The half-life (t1/2) is given by

\(\begin{array}{l}t_{1/2}=\frac{0.693}{k}\end{array} \)

Example 1

The half-life of a 226-radium is 1622 years. Determine the time it will take for a sample of 226-radium to decay to 10% of its original radioactivity.


Use t1/2 equation to find the rate constant.

k = 0.693 / 1622

= 4.27 × 10-4 year

\(\begin{array}{l}ln\frac{[A_0]}{[A]}=kt\end{array} \)

Insert the value for k into the integrated form of the rate equation

4.27 × 10−4 × t = ln(100/10)

4.27 × 10−4 × t = 2.303

t = 5392 years

Example 2

A piece of wood is found to give 10 counts per minute per gram of carbon when subjected to 14C analysis. New wood counts of 15 cpmg-1. The half-life of 14C is 5570 years. Determine the age of the old wood.


Use t1/2 equation to find the rate constant

k = 0.693 / 5570

= 1.24 × 10-4 years

Substitute the value of k into the integrated form of the rate equation.

1.24 × 10-4 × t = ln(14C content in new wood /  content in old wood)

1.24 × 10-4 × t = ln(15/10)

1.24 × 10-4 × t = ln1.5

t = 3270 years old


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