(D) The number of half-lives $n$ is calculated as: $n = \frac{t}{t_{1/2}} = \frac{22920}{5730} = 4$. The amount of ${}^{14}C$ remaining after $n$ half

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(D) The number of half-lives $n$ is calculated as: $n = \frac{t}{t_{1/2}} = \frac{22920}{5730} = 4$. The amount of ${}^{14}C$ remaining after $n$ half-lives is given by: $N = N_0 (\frac{1}{2})^n = N_0 (\frac{1}{2})^4 = \frac{N_0}{16}$. The fraction of ${}^{14}C$ that has decayed is: $\text{Decayed fraction} = \frac{N_0 - N}{N_0} = 1 - \frac{N}{N_0} = 1 - \frac{1}{16} = \frac{15}{16}$.

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

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The half-life for decay of ${}^{14}C$ by $\beta$-emission is $5730 \ yr$. The fraction of ${}^{14}C$ that decays in a sample that is $22920 \ yr$ old would be

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A
$\frac{1}{8}$
B
$\frac{1}{16}$
C
$\frac{7}{8}$
D
$\frac{15}{16}$

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The half-life for decay of ${}^{14}C$ by $\beta$-emission is $5730 \ yr$. The fraction of ${}^{14}C$ that decays in a sample that is $22920 \ yr$ old would be

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