Question

Find the activity in curie of 1 mg of radon-222 whose half life is 3.825 days.

Answer 1

Hint: Activity of the radioactivity material is defined as number of disintegrations per second, or the number of unstable atomic nuclei that decay per second in a given sample.So simply by finding decay constant, we can find activity of material.

Complete step by step answer:
Half life: it is the interval of time required for decaying of one-half of the atomic nuclei of a radioactive sample.
Activity: Activity of the radioactivity material is defined as number of disintegrations per second, or the number of unstable atomic nuclei that decay per second in a given sample
Given: $T^{{\displaystyle \frac{1}{2}}}=3.825$ days
Quantity of substance=1mg= ${10}^{-3}$g
Where $T^{{\displaystyle \frac{1}{2}}}$ is the half life
Now as we know,
The relation between half life and decay constant is
$T^{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{\ln 2}{\lambda }}$..........(where $\lambda$ is decay constant)

Putting the values of $T^{{\displaystyle \frac{1}{2}}}$ in seconds, we get
$3.825\times 24\times 60\times 60={\displaystyle \frac{0.693}{\lambda }}$
As $\ln 2=0.693$
So finding the value of $\lambda$ (decay constant) from this equation we get
$\lambda =2.096\times {10}^{-6}$ $s^{-1}$
Now converting quantity of substance into number of atoms
$$\begin{gathered} \text{Number of atoms (N)}={\displaystyle \frac{{10}^{-3}}{222}}\times 6.022\times {10}^{23} \\ \Rightarrow \text{Number of atoms (N)}=2.71\times {10}^{18} \end{gathered}$$
$N$ is the number of atoms present.

Remember radon has molecular mass 222 as already given.Now as we know activity of substance is given by
$$\begin{gathered} A=-{\displaystyle \frac{dN}{dt}} \\ \Rightarrow A=\lambda N \end{gathered}$$
Where $A$ is a symbol representing activity.
So on putting values of $\lambda$ and $N$
We get
$$\begin{gathered} A=2.096\times {10}^{-6}\times 2.71\times {10}^{18} \\ \Rightarrow A=5.68\times {10}^{12}Bq \end{gathered}$$
As we were required to give answer in curie(Ci) so
Remember 1curie $=3.7\times {10}^{10}Bq$
$$\begin{gathered} A={\displaystyle \frac{5.68\times {10}^{12}}{3.7\times {10}^{10}}} \\ \therefore A=153.5Ci \end{gathered}$$

Hence, activity is $153.5Ci$.

Note: Half-life is the time for half the radioactive nuclei in any sample to undergo radioactive decay. For example, after 2 half-lives, there will be one fourth the original material remains, after three half-lives one eight the original material remains, and so on. Half-life is a convenient way to assess the rapidity of a decay. While solving such problems we have to keep in mind that while using the formula $$N=N_0{e}^{-\lambda t}$$, we have to take undecayed nuclei at that time.