Question

A radioactive sample ${{S}_{1}}$ having an activity of 5μCi has twice the number of nuclei as another sample ${{S}_{2}}$ which has an activity of 10μCi. The half-lives of ${{S}_{1}}$and ${{S}_{2}}$can be
(A) 20 years and 5 years respectively
(B) 20 year and 10 years respectively
(C) 10 years each
(D) 5 years each

Answer 1

Activity of the radioactive sample is given microcurie. We convert it into curie and one curie is $3.7\times {{10}^{10}}$ disintegrations per second. To calculate the activity, we need to have the initial number of atoms of the sample and its half-life or decay constant.

Complete step by step answer:
We know activity is given by $\dfrac{dN}{dt}$. Let $\dfrac{d{{N}_{1}}}{dt}$be the activity of the first sample and $\dfrac{d{{N}_{2}}}{dt}$be the activity of the second sample.
Given data in the question says, $2{{N}_{2}}={{N}_{1}}$
$\dfrac{d{{N}_{1}}}{dt}$= 5 μCi
$\dfrac{d{{N}_{2}}}{dt}$=10 μCi
We know activity is given by $\dfrac{dN}{dt}=\lambda N$
Therefore, $\dfrac{d{{N}_{1}}}{dt}={{\lambda }_{1}}{{N}_{1}}$and $\dfrac{d{{N}_{2}}}{dt}={{\lambda }_{2}}{{N}_{2}}$
Substituting the values given in the question

& {{\lambda }_{1}}{{N}_{1}}=5 \\\ & {{\lambda }_{2}}{{N}_{2}}=10 \\\ \end{aligned}$$ Dividing we get $$\dfrac{{{\lambda }_{2}}{{N}_{2}}}{{{\lambda }_{1}}{{N}_{1}}}=2$$ But $$2{{N}_{2}}={{N}_{1}}$$ Therefore, $${{\lambda }_{2}}=4{{\lambda }_{1}}$$---------(1) Also, the relationship between decay constant and the half-life is $${{T}_{\dfrac{1}{2}}}\lambda =.69$$ So eq (1) can be modified into $${{T}_{(\dfrac{1}{2})1}}=4{{T}_{(\dfrac{1}{2})2}}$$ It is clear that the ratio of half-lives of the two samples is 4:1, therefore half-lives can be 20 years and 5 years respectively. Hence, the correct option is (A) **Note:** In this problem the unit of activity of radioactive sample was not given in standard units but we did not bother to convert the same because we have to use the ratio and ultimately, they cancel each other. Half life is the time period in which the number of undecayed nuclei becomes one half of the initial number of atoms.