The activity of a radioactive material is 6.4 | vedclass

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Question

$A_{0}=6.4 	imes 10^{-4}$ Curie

$T _{1 / 2}=5 	ext { days }=\frac{\ln 2}{\lambda}$

$A = A _{0} e ^{-\lambda t }$

$5 	imes 10^{-6}=6.4 	im

Vedclass · Accepted Answer

$35$

Vedclass · Answer

$A_{0}=6.4 	imes 10^{-4}$ Curie

$T _{1 / 2}=5 	ext { days }=\frac{\ln 2}{\lambda}$

$A = A _{0} e ^{-\lambda t }$

$5 	imes 10^{-6}=6.4 	imes 10^{-4} e ^{-\lambda t }$

$\frac{5}{6.4} 	imes 10^{-2}=e^{-\lambda t}$

$7.8 	imes 10^{-3}= e ^{-\lambda t}$

$\log \left(7.8 	imes 10^{-3}ight)=-\lambda t \ln e$

$\ln \left(7.8 	imes 10^{-3}ight)=-\frac{\lambda n 2}{5} \cdot t$

$	herefore \frac{5 	imes 4.853}{0.693}= t =35 	ext { days }$

The activity of a radioactive material is $6.4 \times 10^{-4}$ curie. Its half life is $5\; days$. The activity will become $5 \times 10^{-6}$ curie after $.......day$

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