Question

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

Answer 1

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: ${T^{\dfrac{1}{2}}} = 3.825$ days
Quantity of substance=1mg= ${10^{ - 3}}$g
Where ${T^{\dfrac{1}{2}}}$ is the half life
Now as we know,
The relation between half life and decay constant is
${T^{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda }$..........(where $\lambda$ is decay constant)

Putting the values of ${T^{\dfrac{1}{2}}}$ in seconds, we get
$3.825 \times 24 \times 60 \times 60 = \dfrac{{0.693}}{\lambda }$
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
$\text{Number of atoms (N)}= \dfrac{{{{10}^{ - 3}}}}{{222}} \times 6.022 \times {10^{23}} \\\ \Rightarrow\text{Number of atoms (N)}= 2.71 \times {10^{18}}$
$N$ 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
$A= - \dfrac{{dN}}{{dt}} \\\ \Rightarrow A = \lambda N$
Where $A$ is a symbol representing activity.
So on putting values of $\lambda$ and $N$
We get
$A = 2.096 \times {10^{ - 6}} \times 2.71 \times {10^{18}} \\\ \Rightarrow A = 5.68 \times {10^{12}}Bq$
As we were required to give answer in curie(Ci) so
Remember 1curie $= 3.7 \times {10^{10}}Bq$
$A = \dfrac{{5.68 \times {{10}^{12}}}}{{3.7 \times {{10}^{10}}}} \\\ \therefore A= 153.5\,Ci$

Hence, activity is $153.5\,Ci$.

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.