
The half-life of ${C^{14}}$ is 5760 years. For a “200” mg sample of ${C^{14}}$, the time taken to change to 25 mg is ____.
(A) 11520 years
(B) 23040 years
(C) 5760 years
(D) 17280 years
Answer
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Hint: The half-life of the isotope is defined as the time taken by a substance to reduce to its half value. The expression used to calculate the half-life is applied which relates the initial quantity and the remaining quantity of substance.
Complete step by step answer:
It is given that the half-life of ${C^{14}}$ isotope is 5760 years.
We need to calculate the time taken for 200 mg of sample to reduce to 25 mg.
The half-life is defined as the time required for the original quantity to reduce to its half quantity. The half-life is denoted by ${t^{1/2}}$.
The term half-life is applied in nuclear physics to describe how frequently the unstable atom undergoes radioactive decay. The half-life of the radioactive compound is a constant which measures the time required by the substance to reduce by half of its quantity by the consequence of decay and as a result emits radiation.
The formula used to calculate the half-life is shown below.
$N(t) = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{t_{1/2}}}}}}$
Where,
N(t) is the remaining quantity which has not been decayed.
${N_0}$ is the initial quantity of the substance.
t is the elapsed
${t^{1/2}}$ is the half-life
Here, the initial amount of substance is 200 mg and the remaining amount is 25 mg.
Substitute the values in the above expression.
$ \Rightarrow 25mg = 200{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{5760}}}}$
$ \Rightarrow \dfrac{{25}}{{200}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{5760}}}}$
Substitute log on both sides.
$ \Rightarrow \log \dfrac{1}{4} = \dfrac{t}{{5760}}\log \dfrac{1}{2}$
$ \Rightarrow t = 11520\;years$
Therefore, the time taken to change to 200 mg into 25 mg is 11520 years.
So, the correct answer is Option A.
Note: Don’t get confused as the quantity of sample reduced to its 1/8th value but by the definition of half-life it is the time required by the sample to reduce to its half. The radioactive decay is a first-order reaction, so the time required by the nuclei half of the radioactive isotope to decay is constant.
Complete step by step answer:
It is given that the half-life of ${C^{14}}$ isotope is 5760 years.
We need to calculate the time taken for 200 mg of sample to reduce to 25 mg.
The half-life is defined as the time required for the original quantity to reduce to its half quantity. The half-life is denoted by ${t^{1/2}}$.
The term half-life is applied in nuclear physics to describe how frequently the unstable atom undergoes radioactive decay. The half-life of the radioactive compound is a constant which measures the time required by the substance to reduce by half of its quantity by the consequence of decay and as a result emits radiation.
The formula used to calculate the half-life is shown below.
$N(t) = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{t_{1/2}}}}}}$
Where,
N(t) is the remaining quantity which has not been decayed.
${N_0}$ is the initial quantity of the substance.
t is the elapsed
${t^{1/2}}$ is the half-life
Here, the initial amount of substance is 200 mg and the remaining amount is 25 mg.
Substitute the values in the above expression.
$ \Rightarrow 25mg = 200{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{5760}}}}$
$ \Rightarrow \dfrac{{25}}{{200}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{5760}}}}$
Substitute log on both sides.
$ \Rightarrow \log \dfrac{1}{4} = \dfrac{t}{{5760}}\log \dfrac{1}{2}$
$ \Rightarrow t = 11520\;years$
Therefore, the time taken to change to 200 mg into 25 mg is 11520 years.
So, the correct answer is Option A.
Note: Don’t get confused as the quantity of sample reduced to its 1/8th value but by the definition of half-life it is the time required by the sample to reduce to its half. The radioactive decay is a first-order reaction, so the time required by the nuclei half of the radioactive isotope to decay is constant.
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