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The activity of a radioactive sample is measured as ${N_o}$ counts per minute at $t = 0$ and ${N_o}/e$ counts per minute $t = 5{\kern 1pt} $ minutes. The time, (in minute) at which the activity reduces to half its value, is:
A) ${\log _e}\dfrac{2}{5}$
B) $\dfrac{5}{{{{\log }_e}2}}$
C) $5{\log _{10}}2$
D) $5{\log _e}2$

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Last updated date: 13th Jun 2024
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Answer
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Hint: To solve this question, you need to consider the decay of reactivity as a first-order reaction or process and thus use equations for first order reaction to find the answer to the asked question. The half-life of a first-order reaction is: ${t_{1/2}} = \dfrac{{{{\log }_e}2}}{\lambda }$

Complete step by step answer:
As explained in the hint section of the solution to the asked question, we need to consider the decay of reactivity as a first-order reaction or process and use its equation to first, find the value of $\lambda $ , which is the decay constant of the decaying quantity.
The equation of a first-order reaction is given as:
$N = {N_o}{e^{ - \lambda t}}$
Where, ${N_o}$ is the initial decay rate
$N$ is the decay rate at time $t$
And, $\lambda $ is the decay constant
The question has already told us that the initial decay rate is ${N_o}$
The decay rate after time $t = 5$ min is given as ${N_o}/e$
Substituting in the values, we get:
$\Rightarrow {N_o}/e = {N_o}{e^{ - 5\lambda }}$
After solving, we get:
$
\Rightarrow - 5\lambda = - 1 \\
\Rightarrow \lambda = \dfrac{1}{5} \\
 $
Now, we have found the value of the decay constant as: $\lambda = \dfrac{1}{5}$
We can find the half-life for the reaction, which is basically what the question is asking since half-life is the time taken at which the activity of the quantity gets reduced to half.
We already know that for a first-order half-life can be found out using the equation:
${t_{1/2}} = \dfrac{{{{\log }_e}2}}{\lambda }$
Substituting in the value of the decay constant in the equation, we get:
$\Rightarrow {t_{1/2}} = \dfrac{{{{\log }_e}2}}{{\dfrac{1}{5}}}$
$\Rightarrow {t_{1/2}} = 5{\log _e}2$

Hence, We can see that the option (D) is the correct option as the value matches what we found out by solving the question.

Note: Many students do not take such reactions as first-order and stay confused about what and which formulae to use to solve the question. Another way of solving the question would have been to put $N = \dfrac{{{N_o}}}{2}$ and find the value of $t$ using the value of $\lambda = \dfrac{1}{5}$ .