(C) The radioactive decay law is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.Given that the ratio of $C^{14}$ to $C^{12}$ is one-fourth

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(C) The radioactive decay law is given by $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.Given that the ratio of $C^{14}$ to $C^{12}$ is one-fourth of the original amount,we have $\frac{N(t)}{N_0} = \frac{1}{4}$.Substituting the values,we get $\frac{1}{4} = \left(\frac{1}{2}\right)^{t/5700}$.Since $\frac{1}{4} = \left(\frac{1}{2}\right)^2$,we can equate the exponents: $2 = \frac{t}{5700}$.Solving for $t$,we get $t = 2 \times 5700 = 11400 \, years$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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An archaeologist analyses the wood in a prehistoric structure and finds that the ratio of $C^{14}$ (half-life $= 5700 \, years$) to $C^{12}$ is only one-fourth of that found in the cells of buried plants. The age of the wood is about .......... $years$.

AIIMS 2013 All AIIMS

A
$5700$
B
$2850$
C
$11400$
D
$22800$

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Nuclei

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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An archaeologist analyses the wood in a prehistoric structure and finds that the ratio of $C^{14}$ (half-life $= 5700 \, years$) to $C^{12}$ is only one-fourth of that found in the cells of buried plants. The age of the wood is about .......... $years$.

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$10 \, g$ of radioactive material of half-life $15 \, \text{years}$ is kept in store for $20 \, \text{years}$. The disintegrated material is ............ $g$.

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