Half lives of two radioactive substances A an | vedclass

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Question

(c) For $80$ minutes, number of half lives of sample $A = {n_A} = \frac{{80}}{{20}} = 4$ and number of half lives of sample

$B = {n_B} = \frac{{80}

Vedclass · Accepted Answer

$1 : 4$

Vedclass · Answer

(c) For $80$ minutes, number of half lives of sample $A = {n_A} = \frac{{80}}{{20}} = 4$ and number of half lives of sample

$B = {n_B} = \frac{{80}}{{40}} = 2.$

Also by using $N = {N_0}{\left( {\frac{1}{2}} \right)^n}$

==> $N \propto \frac{1}{{{2^n}}}$

==> $\frac{{{N_A}}}{{{N_B}}} = \frac{{{2^{{n_B}}}}}{{{2^{{n_A}}}}} = \frac{{{2^2}}}{{{2^4}}} = \frac{1}{4}$

Half lives of two radioactive substances $A$ and $B$ are respectively $20$ minutes and $40$ minutes. Initially the sample of $A$ and $B$ have equal number of nuclei. After $80$ minutes, the ratio of remaining number of $A$ and $B$ nuclei is

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