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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 number of $A$ and $B$ nuclei will be
$1 : 4$
$5 : 4$
$1 : 16$
$4 : 1$
Solution
For $A_{t / 2}=20\, min$, $t=80\,min$, number of half lifes $n=4$
$\therefore $ Nuclei remaining $=\frac{\mathrm{N}_{\mathrm{o}}}{2^{4}} .$ Therefore nuclei decayed $=\mathrm{N}_{0}-\frac{\mathrm{N}_{0}}{2^{4}}$
For $\mathrm{B}_{\mathrm{t} / 2}=40 \mathrm{min} ., \mathrm{t}=80 \mathrm{min},$ number of half lifes $\mathrm{n}=2$
$\therefore $ Nuclei remaining $=\frac{\mathrm{N}_{\mathrm{o}}}{2^{2}} .$ Therefore nuclei decayed $=\mathrm{N}_{0}-\frac{\mathrm{N}_{\mathrm{o}}}{2^{2}}$
$\therefore$ Required ratio$ = \frac{{{\text{No}} – \frac{{{\text{No}}}}{{{2^4}}}}}{{{\text{No}} – \frac{{{\text{No}}}}{{{2^2}}}}} = $ $\frac{{1 – \frac{1}{{16}}}}{{1 – \frac{1}{4}}} = \frac{{15}}{{16}} \times \frac{4}{3} = \frac{5}{4}$
