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13.Nuclei
medium
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 $.......day$
A
$7$
B
$15$
C
$25$
D
$35$
(JEE MAIN-2022)
Solution
$A_{0}=6.4 \times 10^{-4}$ Curie
$T _{1 / 2}=5 \text { days }=\frac{\ln 2}{\lambda}$
$A = A _{0} e ^{-\lambda t }$
$5 \times 10^{-6}=6.4 \times 10^{-4} e ^{-\lambda t }$
$\frac{5}{6.4} \times 10^{-2}=e^{-\lambda t}$
$7.8 \times 10^{-3}= e ^{-\lambda t}$
$\log \left(7.8 \times 10^{-3}\right)=-\lambda t \ln e$
$\ln \left(7.8 \times 10^{-3}\right)=-\frac{\lambda n 2}{5} \cdot t$
$\therefore \frac{5 \times 4.853}{0.693}= t =35 \text { days }$
Standard 12
Physics
