
A piece of wood from an archaeological sample has $5.0counts{\min ^{ - 1}}$ per gram of $C - 14$, while a fresh sample of wood has a count of $15.0{\min ^{ - 1}}{g^{ - 1}}$. If half-life of $C - 14$ is $5770yr$, the age of the archaeological sample is:
A. $8500yr$
B. $9200yr$
C. $10,000yr$
D. $11,000yr$
Answer
575.4k+ views
Hint:Radioactive decay of carbon follows first order kinetics. Therefore, we can first find the rate constant from the half-life, and then substitute this in the equation for first order kinetics to get the time required.
Formulas used: $t = \dfrac{{2.303}}{k}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Where $t$ is the time taken, $k$ is the rate constant, ${N_0}$ is the initial rate of decay and $N$ is the present rate of decay.
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$
Where $k$ is the rate constant and ${t_{1/2}}$ is the half-life.
Complete step by step answer:
As the radioactive decay of carbon follows first order kinetics, we have the equation for rate constant as:
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$
Where $k$ is the rate constant and ${t_{1/2}}$ is the half-life.
Half-life is given as $5770yr$. Substituting this, we get:
$k = \dfrac{{0.693}}{{5770}} = 1.201 \times {10^{ - 4}}y{r^{ - 1}}$
The time required for the completion of a first order decay is given by the formula:
$t = \dfrac{{2.303}}{k}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Where $t$ is the time taken, $k$ is the rate constant, ${N_0}$ is the initial rate of decay and $N$ is the present rate of decay.
The decay rate in the fresh piece of wood is used as the value for initial rate of decay, as the properties and rate of decay remains the same for a particular material, irrespective of time. Hence, we have ${N_0} = 15.0{\min ^{ - 1}}{g^{ - 1}}$ and given that the present rate of decay, $N = 5.0counts{\min ^{ - 1}}{g^{ - 1}}$. Substituting this into our equation, we get:
$t = \dfrac{{2.303}}{{1.201 \times {{10}^{ - 4}}}}\log \left( {\dfrac{{15}}{5}} \right)$
On solving, we get:
$t = 1.91757 \times {10^4} \times \log 3$
Computing this, we get the time taken or the age of the archaeological sample as:
$t = 9149yr$, which is approximately equal to $9200yr$.
Hence, the correct option to be marked is option B.
Note:
Radioactive decay is based on the fact that after a plant’s death, it stops the intake of carbon dioxide having radioactive carbon. While the plant is alive, the levels of radioactive carbon present in it is fairly constant and is continuously taken up and released by the plant. After its death, there is no more intake and the residual radioactive carbon starts to decay. When this sample is analysed years later, we are able to deduce the age of the plant as we know radioactive decay follows first order kinetics.
Formulas used: $t = \dfrac{{2.303}}{k}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Where $t$ is the time taken, $k$ is the rate constant, ${N_0}$ is the initial rate of decay and $N$ is the present rate of decay.
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$
Where $k$ is the rate constant and ${t_{1/2}}$ is the half-life.
Complete step by step answer:
As the radioactive decay of carbon follows first order kinetics, we have the equation for rate constant as:
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$
Where $k$ is the rate constant and ${t_{1/2}}$ is the half-life.
Half-life is given as $5770yr$. Substituting this, we get:
$k = \dfrac{{0.693}}{{5770}} = 1.201 \times {10^{ - 4}}y{r^{ - 1}}$
The time required for the completion of a first order decay is given by the formula:
$t = \dfrac{{2.303}}{k}\log \left( {\dfrac{{{N_0}}}{N}} \right)$
Where $t$ is the time taken, $k$ is the rate constant, ${N_0}$ is the initial rate of decay and $N$ is the present rate of decay.
The decay rate in the fresh piece of wood is used as the value for initial rate of decay, as the properties and rate of decay remains the same for a particular material, irrespective of time. Hence, we have ${N_0} = 15.0{\min ^{ - 1}}{g^{ - 1}}$ and given that the present rate of decay, $N = 5.0counts{\min ^{ - 1}}{g^{ - 1}}$. Substituting this into our equation, we get:
$t = \dfrac{{2.303}}{{1.201 \times {{10}^{ - 4}}}}\log \left( {\dfrac{{15}}{5}} \right)$
On solving, we get:
$t = 1.91757 \times {10^4} \times \log 3$
Computing this, we get the time taken or the age of the archaeological sample as:
$t = 9149yr$, which is approximately equal to $9200yr$.
Hence, the correct option to be marked is option B.
Note:
Radioactive decay is based on the fact that after a plant’s death, it stops the intake of carbon dioxide having radioactive carbon. While the plant is alive, the levels of radioactive carbon present in it is fairly constant and is continuously taken up and released by the plant. After its death, there is no more intake and the residual radioactive carbon starts to decay. When this sample is analysed years later, we are able to deduce the age of the plant as we know radioactive decay follows first order kinetics.
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