(C) Radioactive decay follows first-order kinetics: $N_t = N_0 \times (1/2)^{t/t_{1/2}}$.Given: $N_0 = 1 \, \mu g$,$t_{1/2} = 30 \ years$,$t = 100 \ y

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(C) Radioactive decay follows first-order kinetics: $N_t = N_0 \times (1/2)^{t/t_{1/2}}$.Given: $N_0 = 1 \, \mu g$,$t_{1/2} = 30 \ years$,$t = 100 \ years$.$N_t = 1 \times (1/2)^{100/30} = (1/2)^{10/3}$.Taking $\log$ on both sides: $\log N_t = \frac{10}{3} \log(0.5) = \frac{10}{3} \times (-0.30) = -1$.$N_t = 10^{-1} \, \mu g$.Comparing with $n \times 10^{-1} \, \mu g$,we get $n = 1$.

Class 12 Chemistry Nuclear Chemistry Rate of decay and Half-life

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Assuming $1 \, \mu g$ of trace radioactive element $X$ with a half-life of $30 \ years$ is absorbed by a growing tree. The amount of $X$ remaining in the tree after $100 \ years$ is $n \times 10^{-1} \, \mu g$. Find the value of $n$. $[Given : \ln 10 = 2.303 ; \log 2 = 0.30]$

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Assuming $1 \, \mu g$ of trace radioactive element $X$ with a half-life of $30 \ years$ is absorbed by a growing tree. The amount of $X$ remaining in the tree after $100 \ years$ is $n \times 10^{-1} \, \mu g$. Find the value of $n$. $[Given : \ln 10 = 2.303 ; \log 2 = 0.30]$

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$8 \ g$ of a radioactive substance is reduced to $0.5 \ g$ after $1 \ hour$. The ${t_{1/2}}$ of the radioactive substance is ......... $\min$.

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$A$ piece of wood was found to have the $C^{14}/C^{12}$ ratio $0.7$ times that in a living plant. The time period when the plant died is ....... $years$ (Half-life of $C^{14} = 5760 \, yrs$).

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