The half-life of {}^{65}Zn is 245 days. Afte | vedclass

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(A) The decay constant $K$ is given by $K = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{245} \ days^{-1}$.For a first-order reaction, the time $t$ required f

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$102$

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(A) The decay constant $K$ is given by $K = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{245} \ days^{-1}$.For a first-order reaction, the time $t$ required for the activity to reach $A_t$ from $A_0$ is $t = \frac{1}{K} \ln \frac{A_0}{A_t}$.Given that $75\%$ of the original activity remains, $A_t = 0.75 A_0 = \frac{3}{4} A_0$, so $\frac{A_0}{A_t} = \frac{4}{3}$.Substituting the values: $x = \frac{245}{\ln 2} \ln \frac{4}{3} = 245 	imes \frac{\log(4/3)}{\log 2}$.$x = 245 	imes \frac{2 \log 2 - \log 3}{\log 2} = 245 	imes \frac{2(0.3010) - 0.4771}{0.3010}$.$x = 245 	imes \frac{0.6020 - 0.4771}{0.3010} = 245 	imes \frac{0.1249}{0.3010} \approx 101.66 \ days$.Rounding to the nearest integer, $x = 102$.

The half-life of ${}^{65}Zn$ is $245 \ days$. After $x$ days, $75\%$ of original activity remained. The value of $x$ in days is . . . . . . . (Nearest integer)
(Given $: \log 3=0.4771$ and $\log 2=0.3010$ )

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The half-life of ${}^{65}Zn$ is $245 \ days$. After $x$ days, $75\%$ of original activity remained. The value of $x$ in days is . . . . . . . (Nearest integer)(Given $: \log 3=0.4771$ and $\log 2=0.3010$ )