Question

The half-life of ${}_{38}^{90}Sr$ is 28 years. What is the disintegration rate of 15mg of this isotope?

Answer 1

To find the disintegration rate of 15mg ${}_{38}^{90}Sr$ isotope first of all we have to calculate the number of radioactive nuclei present in the 15mg of ${}_{38}^{90}Sr$ and the rate constant for the disintegration. Then, we will go the formula: $\left( { - \dfrac{{dN}}{{dt}}} \right) = \lambda N$, where $\left( { - \dfrac{{dN}}{{dt}}} \right)$ represent the rate of decay or disintegration rate, $\lambda$ is the rate constant and $N$ is the number of radioactive nuclei present at the time $t$. Finally, we will calculate the value of $\lambda N$ in order to get the required disintegration rate. All the calculations that we will use to solve the problems are taken in the SI system.

Complete step-by-step answer:
To find the required value for the disintegration rate of 15mg of ${}_{38}^{90}Sr$:-
The formula used is: $\left( { - \dfrac{{dN}}{{dt}}} \right) = \lambda N$……………….. (i) (Where $\lambda$ is the rate constant and $N$ is the number of radioactive nuclei present at the time $t$.)
Given:-
${t_{\dfrac{1}{2}}} = 28$ Years
$= 28 \times 365 \times 24 \times 3600{\text{ s}}$
$\Rightarrow {t_{\dfrac{1}{2}}} = 8.83 \times {10^8}{\text{ s}}$…………….. (ii)
Mass of ${}_{38}^{90}Sr = 15mg$
$= 15 \times {10^{ - 3}}g$,
Since the atomic mass of ${}_{38}^{90}Sr = 87.62{\text{ }}g/mol$
As we know that $87.62{\text{ }}g/mol$ of ${}_{38}^{90}Sr$ containing $6.022 \times {10^{23}}$ atoms, So
Number of radioactive nuclei (N) present in 15mg of ${}_{38}^{90}Sr = \left( {\dfrac{{6.022 \times {{10}^{23}} \times 15 \times {{10}^{ - 3}}}}{{87.62{\text{ }}}}} \right)$
$= 1.03 \times {10^{20}}$
Thus, $N = 1.03 \times {10^{20}}$
Calculating the rate constant $\lambda$:
$\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$ disintegration/s
$= \dfrac{{0.693}}{{8.83 \times {{10}^8}}}$ disintegration/s
Substitute the value of $\lambda$ and $N$ in equation(i), we get
Rate of disintegration $\left( { - \dfrac{{dN}}{{dt}}} \right) = \left( {\dfrac{{0.693}}{{8.83 \times {{10}^8}}} \times 1.03 \times {{10}^{20}}} \right)$ disintegration/s
$= 8.08 \times {10^{10}}$ disintegration/s
Hence, the required disintegration rate of 15mg of ${}_{38}^{90}Sr = 8.08 \times {10^{10}}$ disintegration/s.

Note: In order to master these kinds of problems we have to keep practicing a lot of conceptual questions on radioactive disintegration law. Students often confuse term activity and rate of disintegration so do not get confused with that they both are identical although at many radioactive reactions we generally use the term activity. One should also care about the data given in the problem that must be used in the SI system while solving the problem.