Question

: How to calculate the decay constant, half-life, and the mean life for a radioisotope which activity is found to decrease by 25% in one week?

Answer 1

Hint: Decay constant $(\text{ }\!\!\lambda\!\!\text{ )}$ is the fraction of total nuclei that will decay per unit time $(\text{t)}$. It can be calculated using the following formula:
\[{{\text{N}}_{\text{t}}}={{\text{N}}_{0}}{{e}^{-\text{ }\!\!\lambda\!\!\text{ t}}}\]
The half-life$\left( {{\text{T}}_{{1}/{2}\;}} \right)$ for a radioisotope is the time taken to decay nuclei of a radioactive sample to half of its initial number of nuclei. It is inversely proportional to the decay constant.
\[{{\text{T}}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\text{ }\!\!\lambda\!\!\text{ }}\]
Mean life $(\text{ }\!\!\tau\!\!\text{ })$ is the average life of all the nuclei before decay.
\[\text{ }\!\!\tau\!\!\text{ }=\dfrac{1}{\text{ }\!\!\lambda\!\!\text{ }}\]

Complete step-by-step answer: The radioisotope is a radioactive isotope of an element that is very unstable in nature due to the presence of an unnatural combination of protons and neutrons in its nucleus and it tends to decay into a smaller stable nucleus by emitting some elementary particles such as an electron, positron, etc.
The total activity of a radioisotope is the number of nuclei that decays per unit of time. It is directly proportional to the number of nuclei present in the sample.
\[
   \text{A}\propto \text{N} \\
  \text{A}=\text{ }\!\!\lambda\!\!\text{ N} \\
\]
Where $\text{ }\!\!\lambda\!\!\text{ }$ is called decay constant.
It is defined as the fraction of total nuclei that will decay per unit of time. If ${{\text{N}}_{0}}$ represents the initial number of nuclei before decay and $\text{N}$ represents the number of nuclei left after time $\text{t}$ then, the decay constant can be given as:
\[\text{ }\!\!\lambda\!\!\text{ N}=-\dfrac{\text{dN}}{\text{dt}}\]
or \[-\text{ }\!\!\lambda\!\!\text{ dt}=\dfrac{\text{dN}}{\text{N}}\]
On integrating both sides, we get\[\]
\[-\text{ }\!\!\lambda\!\!\text{ }\int\limits_{0}^{\text{t}}{\text{dt}}=\int\limits_{{{\text{N}}_{0}}}^{\text{N}}{\dfrac{\text{dN}}{\text{N}}}\]
\[
   \Rightarrow -\text{ }\!\!\lambda\!\!\text{ t}+\ln {{\text{N}}_{0}}=\ln \text{N} \\
  \Rightarrow \text{N}={{\text{N}}_{0}}{{e}^{-\text{ }\!\!\lambda\!\!\text{ t}}} \\
\]
Since $\text{N}\propto \text{A}$ so we can write this equation in terms of activity also.
\[\text{A}={{\text{A}}_{0}}{{e}^{-\text{ }\!\!\lambda\!\!\text{ t}}}.........(1)\]
Now, according to the question, in 1 week activity is decreased by 25% i.e.,
\[
   \text{A}=(100-25)%\text{ of }{{\text{A}}_{\text{0}}} \\
  \Rightarrow \text{A}=75%\text{ of }{{\text{A}}_{\text{0}}} \\
  \Rightarrow \text{A}=0.75{{\text{A}}_{0}} \\
\]
So, putting all the values in the equation $(1)$, we get
\[
   0.75{{\text{A}}_{0}}={{\text{A}}_{0}}{{e}^{-\text{ }\!\!\lambda\!\!\text{ }(1\text{week)}}} \\
  \Rightarrow 0.75={{e}^{-\text{ }\!\!\lambda\!\!\text{ }(1\text{week)}}} \\
  \Rightarrow \ln 0.75=-\text{ }\!\!\lambda\!\!\text{ (1week)} \\
  \Rightarrow -\text{ }\!\!\lambda\!\!\text{ }=-0.288 \\
  \Rightarrow \text{ }\!\!\lambda\!\!\text{ }=0.288\text{ wee}{{\text{k}}^{-1}} \\
\]
Now, the half-life $\left( {{\text{T}}_{{}^{1}/{}_{2}}} \right)$ is the time taken to decay nuclei of a radioactive sample to half of its initial number of nuclei. So, putting $\text{N}=\dfrac{{{\text{N}}_{0}}}{2}$ and $\text{t}={{\text{T}}_{{}^{1}/{}_{2}}}$ in the equation$(1)$ and on solving we get the following equation for half-life:
\[{{\text{T}}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\text{ }\!\!\lambda\!\!\text{ }}\]
Putting $\text{ }\!\!\lambda\!\!\text{ }=0.288\text{ wee}{{\text{k}}^{-1}}$ in the above equation,
\[
   {{\text{T}}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{\text{0}\text{.288}}\text{ week} \\
  \Rightarrow {{\text{T}}_{{}^{1}/{}_{2}}}=2.41\text{ weeks} \\
\]
Lastly, mean life $(\text{ }\!\!\tau\!\!\text{ })$ is the average life of all the nuclei before decay and it is given as:
\[\text{ }\!\!\tau\!\!\text{ }=\dfrac{1}{\text{ }\!\!\lambda\!\!\text{ }}\]
Putting the value of $\text{ }\!\!\lambda\!\!\text{ }=0.288\text{ wee}{{\text{k}}^{-1}}$ in the above equation, we get
\[\text{ }\!\!\tau\!\!\text{ }=\dfrac{1}{\text{0}\text{.288}}\text{ weeks}=3.47\text{ weeks}\]
Hence, the decay constant is $\text{ }\!\!\lambda\!\!\text{ }=0.288\text{ wee}{{\text{k}}^{-1}}$ , the half-life is ${{\text{T}}_{{1}/{2}\;}}=2.41\text{ weeks}$ and the mean life is $\text{ }\!\!\tau\!\!\text{ }=3.47\text{ weeks}$.

Additional information: The radioactive decay law and the half-life formula are used to find the age of organic material, which is known as radioactive dating. One of the popular forms of radioactive dating is carbon dating. By knowing the half-life of carbon-14 isotope the rate of disintegration of the nuclei within the organism or substance can be calculated and thereby its age can be determined.

Note: The activity is measured in the unit of Becquerel and Curie.
\[
   1\text{ Becquerel (Bq)}=1\text{ disintegration per second} \\
  1\text{ Curie (Ci)}=3.7\times {{10}^{10}}\text{Bq} \\
\]
Also, The mean life is about 1.44 times longer than half-life and can be calculated directly from the value of half-life by using the relation: