
The half-life of a radioactive isotope is 3 hours. If the initial mass of isotope were 256g, the mass of remaining undecayed after 18 hours would be:
(A)- 12g
(B)- 16g
(C)- 4g
(D)- 8g
Answer
594.6k+ views
Hint: The half-life of a substance is defined as the time required to reduce the initial mass of the substance to an amount which is half of its initial mass. To answer this question we will use the formula used to calculate the amount if substance remains undecayed.
\[N\,=\,{{N}_{o}}{{\left( \dfrac{1}{2} \right)}^{n}}\]
Where, N is the amount of substance left undecayed
${{N}_{o}}$is the initial amount of the substance
\[n\,=\,\dfrac{T}{{{t}_{{}^{1}/{}_{2}}}}\] where, T is the time taken in disintegration and ${{T}_{{}^{1}/{}_{2}}}$ is the half- life of the substance
Complete answer:
Let’s look at the solution to the given question:
The disintegration of radioactive substances is a first order reaction. This means that the rate of disintegration is dependent on the concentration of one of the reactants only.
Radioactive substances are highly unstable and keep on emitting energy continuously. Uranium, thorium, etc. are some examples of radioactive substances.
It is given in the question that,${{N}_{o}}\,=\,256g$, T= 18hours and ${{t}_{{}^{1}/{}_{2}}}$= 3hours.
First we will calculate ‘n’
\[n\,=\,\dfrac{T}{{{t}_{{}^{1}/{}_{2}}}}\]
On putting the values given in the question, we get,
n = $\dfrac{18}{3}\,=\,6$
Now, we will use the formula and calculate the amount of undecayed isotope left.
\[N\,=\,{{N}_{o}}{{\left( \dfrac{1}{2} \right)}^{n}}\]
Where, N is the amount of substance left undecayed
${{N}_{o}}$is the initial amount of the substance
On putting the values given in the question, we get,
\[N\,=\,256{{\left( \dfrac{1}{2} \right)}^{6}}\]
Therefore, N = 4g
So, the mass of the isotope left undecayed after 18hours is 4g.
Hence, the answer of the given question is option (C).
Note:
This question can be solved by using the formula of the rate constant. In that case we will first find the rate constant using the half-life and then we will use the rate constant to find the amount of undecayed isotope. But this is a long process and involves antilog so avoid it.
\[N\,=\,{{N}_{o}}{{\left( \dfrac{1}{2} \right)}^{n}}\]
Where, N is the amount of substance left undecayed
${{N}_{o}}$is the initial amount of the substance
\[n\,=\,\dfrac{T}{{{t}_{{}^{1}/{}_{2}}}}\] where, T is the time taken in disintegration and ${{T}_{{}^{1}/{}_{2}}}$ is the half- life of the substance
Complete answer:
Let’s look at the solution to the given question:
The disintegration of radioactive substances is a first order reaction. This means that the rate of disintegration is dependent on the concentration of one of the reactants only.
Radioactive substances are highly unstable and keep on emitting energy continuously. Uranium, thorium, etc. are some examples of radioactive substances.
It is given in the question that,${{N}_{o}}\,=\,256g$, T= 18hours and ${{t}_{{}^{1}/{}_{2}}}$= 3hours.
First we will calculate ‘n’
\[n\,=\,\dfrac{T}{{{t}_{{}^{1}/{}_{2}}}}\]
On putting the values given in the question, we get,
n = $\dfrac{18}{3}\,=\,6$
Now, we will use the formula and calculate the amount of undecayed isotope left.
\[N\,=\,{{N}_{o}}{{\left( \dfrac{1}{2} \right)}^{n}}\]
Where, N is the amount of substance left undecayed
${{N}_{o}}$is the initial amount of the substance
On putting the values given in the question, we get,
\[N\,=\,256{{\left( \dfrac{1}{2} \right)}^{6}}\]
Therefore, N = 4g
So, the mass of the isotope left undecayed after 18hours is 4g.
Hence, the answer of the given question is option (C).
Note:
This question can be solved by using the formula of the rate constant. In that case we will first find the rate constant using the half-life and then we will use the rate constant to find the amount of undecayed isotope. But this is a long process and involves antilog so avoid it.
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