An old chair of wood shows, $_{6}{{C}^{14}}$ activity which is 80 percent of the activity found today, calculate the age of the old chair of wood. (${{t}_{1/2}}\text{ of}{{\text{ }}_{6}}{{C}^{14}}=5700$ yr)
A. $t=\dfrac{2.303}{5770}\log \dfrac{100}{80}$
B. $t=\dfrac{5770}{0.301}\log \dfrac{100}{80}$
C. $t=\dfrac{5770}{0.693}\log \dfrac{100}{80}$
D. $t=\dfrac{0.693}{5770}\log \dfrac{100}{80}$
Answer
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Hint: You can start solving this question by writing the general equation of half-life and its relation with the activity. Half-life is the time in which the amount of a radioactive substance is reduced to half of its initial amount. Half life is different for different substances.
Complete step by step solution:
Given in the question:
An old chair of wood shows, $_{6}{{C}^{14}}$ activity which is 80 percent of the activity which means that activity of the present sample is equal to 80 percent the activity of the initial sample
\[{{\dfrac{-dN}{dt}}_{present}}=80\%{{\dfrac{-dN}{dt}}_{\operatorname{int}ial}}\]
This can also be written as : $\lambda ({{N}_{p}})=\dfrac{80}{100}\lambda ({{N}_{0}})$
Hence we can write $\dfrac{{{N}_{o}}}{{{N}_{p}}}=\dfrac{8}{10}$
And half life is given in the question:
So $t=\dfrac{5770}{\ln 2}\ln \dfrac{10}{8}$
$t=\dfrac{5770}{0.693}\log \dfrac{100}{80}$
Hence the correct answer is option (C) i.e. The age of the old chair of wood is $t=\dfrac{5770}{0.693}\log \dfrac{100}{80}$. You can solve this equation to get the exact value of the age of the old chair of wood.
Additional information:
Radioactive carbon dating method is a method which is used for determining the age of an object which contains an organic material by using the properties of radiocarbon which is a radioactive isotope of carbon.
Note: Do not get confused between the half life and mean life as the mean life of a species is always 1.443 times longer than the half life of the substance. For example lead-209 compound decays to form bismuth-209. The half life of this reaction is 3.25 hours and the mean life of the reaction is 4.69 hours. Radioactive elements have a shorter half life in comparison to the stable element and are generally considered more unstable.
Complete step by step solution:
Given in the question:
An old chair of wood shows, $_{6}{{C}^{14}}$ activity which is 80 percent of the activity which means that activity of the present sample is equal to 80 percent the activity of the initial sample
\[{{\dfrac{-dN}{dt}}_{present}}=80\%{{\dfrac{-dN}{dt}}_{\operatorname{int}ial}}\]
This can also be written as : $\lambda ({{N}_{p}})=\dfrac{80}{100}\lambda ({{N}_{0}})$
Hence we can write $\dfrac{{{N}_{o}}}{{{N}_{p}}}=\dfrac{8}{10}$
And half life is given in the question:
So $t=\dfrac{5770}{\ln 2}\ln \dfrac{10}{8}$
$t=\dfrac{5770}{0.693}\log \dfrac{100}{80}$
Hence the correct answer is option (C) i.e. The age of the old chair of wood is $t=\dfrac{5770}{0.693}\log \dfrac{100}{80}$. You can solve this equation to get the exact value of the age of the old chair of wood.
Additional information:
Radioactive carbon dating method is a method which is used for determining the age of an object which contains an organic material by using the properties of radiocarbon which is a radioactive isotope of carbon.
Note: Do not get confused between the half life and mean life as the mean life of a species is always 1.443 times longer than the half life of the substance. For example lead-209 compound decays to form bismuth-209. The half life of this reaction is 3.25 hours and the mean life of the reaction is 4.69 hours. Radioactive elements have a shorter half life in comparison to the stable element and are generally considered more unstable.
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