The activity of a sample is 64 times 10^{-5} | vedclass

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Question

$A=A_{0}\left(\frac{1}{2}ight)^{t / T_{1 / 2}} \Rightarrow 5 	imes 10^{-6}=64 	imes 10^{-5}\left(\frac{1}{2}ight)^{t / 3}$

$\Rightarrow \frac

Vedclass · Accepted Answer

$21$

Vedclass · Answer

$A=A_{0}\left(\frac{1}{2}ight)^{t / T_{1 / 2}} \Rightarrow 5 	imes 10^{-6}=64 	imes 10^{-5}\left(\frac{1}{2}ight)^{t / 3}$

$\Rightarrow \frac{1}{128}=\left(\frac{1}{2}ight)^{t / 3} \Rightarrow t=21 \mathrm{\,days}$

The activity of a sample is $ 64 \times 10^{-5}\ Ci$ . Its half-life $3$ days. The activity will become$5 \times {10^{ - 6}}\ Ci$ after ........... $days$

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