Question

A radioactive sample has an initial activity of $28dpm$ 30 minutes later its activity $14dpm$. How many atoms of nuclide were present initially?

Answer 1

Step-$1$:Given data

1. The initial activity of the radioactive sample is $A_1=28dpm.$
2. The final activity of the radioactive sample is $A_2=14dpm.$
3. The half time of the radioactive sample is $T_{\frac{1}{2}}=30min.$

Step-$2$: The activity and half time of radioactive element

1. The activity or decay rate is the number of decays per unit time and it is given by the form, $A=\lambda N$, where, N is the number of radioactive nuclei present at a time, and $\lambda$ is the decay constant.
2. The half-life of a radioactive element is defined as, $T_{\frac{1}{2}}=\frac{0.693}{\lambda }$.

Step-$3$: Finding the initial activity

As we know, the half-life of a radioactive element is$T_{\frac{1}{2}}=\frac{0.693}{\lambda }$.

So,

$$\begin{gathered} \lambda =\frac{0.693}{T_{\frac{1}{2}}}=\frac{0.693}{30}=0.231/min. \\ or\lambda =0.231/min \end{gathered}$$

Again, the initial activity of a radioactive element is $A_1=\lambda N_o$, where $N_o$ is the number of nuclei at the time $t=0$.

So,

$$\begin{gathered} N_o=\frac{A_1}{\lambda }=\frac{28}{0.0231}=1212 \\ or{N}_o=1212 \end{gathered}$$

Therefore, the total number of nuclei present at the initial time is $1212$.