Answer
410.1k+ views
Hint: This question is based on the concept of radioactivity so to attempt this question one should have prior knowledge about the radioactivity and also remember to use ${t_{1/2}} = 5,580$ years, using this information will help you to approach the solution of the question.
Complete answer:
According to the given information we have two nuclei ${C^{14}}$ and ${C^{12}}$ where half-life of ${C^{14}}$ is 5,580 years i.e. ${t_{1/2}} = 5,580$ years also given that ratio from ${C^{14}}$to ${C^{12}}$ in a wooden piece is 25% of that in atmosphere
We know that ${C^{14}}$ is a radioactive nuclei and ${C^{12}}$ is a stable nuclei
Also, we know that when the live nuclei become dead it starts disintegrating due to which the ratio stars decreasing
Since, we know that rate of disintegration of a radioactive material is named as activity (A) which is directly proportional to the numbers of atoms left decayed in the sample
Thus, $\dfrac{A}{{{A_o}}} = {\left( {\dfrac{1}{2}} \right)^n}$here ${A_0}$is the initial decay rate or activity and n is the number of half-life which is given as \[n = \dfrac{t}{{{t_{1/2}}}}\]
As it is given that ratio from ${C^{14}}$to ${C^{12}}$is equal to 25% i.e. $\dfrac{A}{{{A_o}}} = \dfrac{{25}}{{100}}$
Substituting the value in the above equation we get
$\dfrac{{25}}{{100}} = {\left( {\dfrac{1}{2}} \right)^n}$
$ \Rightarrow $${\left( {\dfrac{1}{2}} \right)^n} = \dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^2}$
Therefore, n = 2
Now we know that \[n = \dfrac{t}{{{t_{1/2}}}}\] substituting the value of n and ${t_{1/2}}$ in this equation we get
\[2 = \dfrac{t}{{5580}}\]
$ \Rightarrow $t = 11, 160 years
Therefore, the age of wood is 11,160 years
So, the correct answer is “Option D”.
Note:
In the above solution we used the term “radioactive disintegration” which can be explained as the process in which an unstable nuclei or nucleus energy decreases due to the radiation this process is also named as radioactive decay and the material which consist unstable nuclei is named as the radioactive there are 3 types of radioactive decay such as Alpha decay, Beta decay and gamma decay.
Complete answer:
According to the given information we have two nuclei ${C^{14}}$ and ${C^{12}}$ where half-life of ${C^{14}}$ is 5,580 years i.e. ${t_{1/2}} = 5,580$ years also given that ratio from ${C^{14}}$to ${C^{12}}$ in a wooden piece is 25% of that in atmosphere
We know that ${C^{14}}$ is a radioactive nuclei and ${C^{12}}$ is a stable nuclei
Also, we know that when the live nuclei become dead it starts disintegrating due to which the ratio stars decreasing
Since, we know that rate of disintegration of a radioactive material is named as activity (A) which is directly proportional to the numbers of atoms left decayed in the sample
Thus, $\dfrac{A}{{{A_o}}} = {\left( {\dfrac{1}{2}} \right)^n}$here ${A_0}$is the initial decay rate or activity and n is the number of half-life which is given as \[n = \dfrac{t}{{{t_{1/2}}}}\]
As it is given that ratio from ${C^{14}}$to ${C^{12}}$is equal to 25% i.e. $\dfrac{A}{{{A_o}}} = \dfrac{{25}}{{100}}$
Substituting the value in the above equation we get
$\dfrac{{25}}{{100}} = {\left( {\dfrac{1}{2}} \right)^n}$
$ \Rightarrow $${\left( {\dfrac{1}{2}} \right)^n} = \dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^2}$
Therefore, n = 2
Now we know that \[n = \dfrac{t}{{{t_{1/2}}}}\] substituting the value of n and ${t_{1/2}}$ in this equation we get
\[2 = \dfrac{t}{{5580}}\]
$ \Rightarrow $t = 11, 160 years
Therefore, the age of wood is 11,160 years
So, the correct answer is “Option D”.
Note:
In the above solution we used the term “radioactive disintegration” which can be explained as the process in which an unstable nuclei or nucleus energy decreases due to the radiation this process is also named as radioactive decay and the material which consist unstable nuclei is named as the radioactive there are 3 types of radioactive decay such as Alpha decay, Beta decay and gamma decay.
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