Half-lives of two radioactive elements A and | vedclass

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Question

For $A_{t / 2}=20\, min$, $t=80\,min$, number of half lifes $n=4$

$	herefore $ Nuclei remaining $=\frac{\mathrm{N}_{\mathrm{o}}}{2^{4}} .$ Therefo

Vedclass · Accepted Answer

$5 : 4$

Vedclass · Answer

For $A_{t / 2}=20\, min$, $t=80\,min$, number of half lifes $n=4$

$	herefore $ 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 $\mathrm{n}=2$

$	herefore $ Nuclei remaining $=\frac{\mathrm{N}_{\mathrm{o}}}{2^{2}} .$ Therefore nuclei decayed $=\mathrm{N}_{0}-\frac{\mathrm{N}_{\mathrm{o}}}{2^{2}}$

$	herefore$ Required ratio$ = \frac{{{	ext{No}} - \frac{{{	ext{No}}}}{{{2^4}}}}}{{{	ext{No}} - \frac{{{	ext{No}}}}{{{2^2}}}}} = $ $\frac{{1 - \frac{1}{{16}}}}{{1 - \frac{1}{4}}} = \frac{{15}}{{16}} 	imes \frac{4}{3} = \frac{5}{4}$

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

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