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# After a time interval equal to 3 half-lives how many times would the activity of a radioactive element be of its initial activity?A) ${2^3}$B) ${3^2}$C) $\dfrac{1}{{{3^2}}}$D) $\dfrac{1}{{{2^3}}}$

Last updated date: 24th Jun 2024
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Hint: The law of radioactive decay states that the number of nuclei undergoing the decay in a sample per unit time is proportional to the total number of nuclei in the sample. Thus the radioactive decay of a radioactive element has an exponential nature i.e., the activity of the element will be exponentially decreasing.

Formulas used:
i) The activity of a radioactive element for a time $t$ is given by, $R = {R_0}{e^{ - \lambda t}}$ where ${R_0}$ is the activity at the time $t = 0$ and $\lambda$ is the radioactive decay constant.
ii) The radioactive decay constant is given by $\lambda = \dfrac{{\ln 2}}{{{T_{1/2}}}}$ where ${T_{1/2}}$ is the half-life period of the radioactive element.

Complete step by step answer:
Step 1: Describe the problem at hand.
The given problem demands the factor by which the activity of the radioactive element will be after a time of 3 half-lives to be found out. If the period of the half-life of the element is ${T_{1/2}}$, then this means that we have to determine the activity of the radioactive element when the time is $t = 3{T_{1/2}}$.

Step 2: Express the activity of the element for the time $t = 3{T_{1/2}}$.
Generally, we express the activity of a radioactive element for a time $t$ as
$R = {R_0}{e^{ - \lambda t}}$ --------(1) where ${R_0}$ is the activity at the time $t = 0$ and $\lambda$ is the radioactive decay constant.
The relation for the radioactive decay constant is given as $\lambda = \dfrac{{\ln 2}}{{{T_{1/2}}}}$ -------- (2)
Then substituting equation (2) in (1) we get, $R = {R_0}{e^{\left[ {\left( {\dfrac{{ - \ln 2}}{{{T_{1/2}}}}} \right) \times t} \right]}}$ and the activity for the time $t = 3{T_{1/2}}$ will be $R = {R_0}{e^{\left[ {\left( {\dfrac{{ - \ln 2}}{{{T_{1/2}}}}} \right) \times \left( {3{T_{1/2}}} \right)} \right]}}$ ----------- (3)

Simplifying the power of $e$ in equation (3) we get, $R = {R_0}{e^{\left( { - 3\ln 2} \right)}} = {R_0}{e^{ - \ln {2^3}}}$
$\Rightarrow R = \dfrac{{{R_0}}}{{{2^3}}}$
Thus the activity becomes $\dfrac{1}{{{2^3}}}$ of its initial activity after the given time.

So the correct option is (D).

Note: The power of the exponential $e$ in equation (3) is simplified using the relation $a\ln b = \ln {b^a}$. In equation (3), $a = 3$ and $b = 2$ and so we get the term ${e^{\ln {2^3}}}$. The exponential of the natural logarithm will be the argument of the natural logarithm i.e., ${e^{\ln a}} = a$. The negative sign in the power in equation (3) indicates that the reciprocal has to be taken i.e., ${e^{ - \ln a}} = \dfrac{1}{a}$ .