Question:
Half-lives of two radioactive elements $A$ and $B$ are $20\, $ minutes and $40\,$ minutes, respectively. Initially, the samples have equal number of nuclei. After $80\,$ minutes, the ratio of decayed numbers of $A$ and $B$ nuclei will be :
Half-lives of two radioactive elements $A$ and $B$ are $20\, $ minutes and $40\,$ minutes, respectively. Initially, the samples have equal number of nuclei. After $80\,$ minutes, the ratio of decayed numbers of $A$ and $B$ nuclei will be :
Updated On: Feb 2, 2026
- $1 : 16$
- $4 : 1$
- $1 : 4$
- $5 : 4$
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The Correct Option is D
Solution and Explanation
$80$ minutes $= 4$ half-lives of $A = 2$ half-lives of $B$
Let the initial number of nuclei in each sample be $N$
$N_A$ after $80$ minutes $ = \frac{N}{2^4} $
$\Rightarrow $ Number of $A$ nuclides decayed $= \frac{15}{16} N $
$N_B$ after $80$ minutes $ = \frac{N}{2^2} $
$ \Rightarrow $ Number of $B$ nuclides decayed $= \frac{3}{4} N $
Required ratio $ = \frac{15/16}{3/4} = \frac{5}{4}$
Let the initial number of nuclei in each sample be $N$
$N_A$ after $80$ minutes $ = \frac{N}{2^4} $
$\Rightarrow $ Number of $A$ nuclides decayed $= \frac{15}{16} N $
$N_B$ after $80$ minutes $ = \frac{N}{2^2} $
$ \Rightarrow $ Number of $B$ nuclides decayed $= \frac{3}{4} N $
Required ratio $ = \frac{15/16}{3/4} = \frac{5}{4}$
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Following are the terms related to nucleus:
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