Question
Half-lives of two radioactive substances $$A$$ and $$B$$ are respectively $$20\,min$$ and $$40\,min.$$ Initially, the samples of $$A$$ and $$B$$ have equal number of nuclei. After $$80\,min$$ the ratio of remaining number of $$A$$ and $$B$$ nuclei is
A.
$$1:16$$
B.
$$4:1$$
C.
$$1:4$$
D.
$$1:1$$
Answer :
$$1:4$$
Solution :
Total time given $$= {80\,\min }$$
Number of half-lives of $$A,$$ $${n_A} = \frac{{80\,\min }}{{20\;\,\min }} = 4$$
Number of half-lives of $$B,$$ $${n_B} = \frac{{80\,\min }}{{40\;\,\min }} = 2$$
Number of nuclei remained undecayed $$N = {N_0}{\left( {\frac{1}{2}} \right)^n}$$
where $${N_0}$$ is initial number of nuclei and $$N$$ is final number of nuclei
So for two different cases $$\left( A \right)$$ and $$\left( B \right),$$
$$\eqalign{
& \frac{{{N_A}}}{{{N_B}}} = \frac{{{{\left( {\frac{1}{2}} \right)}^{{n_A}}}}}{{{{\left( {\frac{1}{2}} \right)}^{{n_B}}}}}\,\,{\text{or}}\,\,\frac{{{N_A}}}{{{N_B}}} = \frac{{{{\left( {\frac{1}{2}} \right)}^4}}}{{{{\left( {\frac{1}{2}} \right)}^2}}} = \frac{{\left( {\frac{1}{{16}}} \right)}}{{\left( {\frac{1}{4}} \right)}} \cr
& {\text{or}}\,\,\frac{{{N_A}}}{{{N_B}}} = \frac{1}{4} \cr} $$