The activity of a radioactive material is 6.4 | vedclass

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(D) Given: Initial activity $A_0 = 6.4 	imes 10^{-4} 	ext{ Ci}$,Final activity $A = 5 	imes 10^{-6} 	ext{ Ci}$,Half-life $T_{1/2} = 5 	ext{ days}

Vedclass · Accepted Answer

$35$

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(D) Given: Initial activity $A_0 = 6.4 	imes 10^{-4} 	ext{ Ci}$,Final activity $A = 5 	imes 10^{-6} 	ext{ Ci}$,Half-life $T_{1/2} = 5 	ext{ days}$.Using the law of radioactive decay: $A = A_0 e^{-\lambda t}$.Substituting the values: $5 	imes 10^{-6} = 6.4 	imes 10^{-4} e^{-\lambda t}$.Rearranging: $\frac{5 	imes 10^{-6}}{6.4 	imes 10^{-4}} = e^{-\lambda t} \implies \frac{5}{640} = e^{-\lambda t} \implies \frac{1}{128} = e^{-\lambda t}$.Taking natural logarithm on both sides: $\ln(1/128) = -\lambda t \implies -\ln(2^7) = -\lambda t \implies 7 \ln 2 = \lambda t$.Since $\lambda = \frac{\ln 2}{T_{1/2}}$,we have $7 \ln 2 = \left(\frac{\ln 2}{5}ight) t$.Solving for $t$: $7 = \frac{t}{5} \implies t = 35 	ext{ days}$.

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

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