question_answer

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:     [AIPMT 1998]

A)  1 : 16   

B)                   4 : 1                

C)  1 : 4

D)       1 : 1

Correct Answer: C

Solution :

Key-Idea: Total no. of nuclei remained after n half -lives is  \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\]

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 N0 is initial number of nuclei

\[\therefore \]   \[\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{{{\left( \frac{1}{2} \right)}^{{{n}_{A}}}}}{{{\left( \frac{1}{2} \right)}^{{{n}_{B}}}}}\]

or      \[\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{{{\left( \frac{1}{2} \right)}^{4}}}{{{\left( \frac{1}{2} \right)}^{2}}}=\frac{\left( \frac{1}{16} \right)}{\left( \frac{1}{4} \right)}\]

or      \[\frac{{{N}_{A}}}{{{N}_{B}}}=\frac{1}{4}\]

Note: The graph between number of nuclei decayed with time is shown along side.