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The half-life of the isotope ${ }_{11} \mathrm{Na}^{24}$ is $15 \mathrm{hrs}$. How much time does it take for $\dfrac{7}{8}$ th of a sample of this isotope to decay
A. $75\, \mathrm{hrs}$
B. $65\, \mathrm{hrs}$
C. $55\, \mathrm{hrs}$
D. $45\, \mathrm{hrs}$

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Last updated date: 18th May 2024
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Answer
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Hint:The Half life of the radioactive sample is the time taken by the sample to reach half of the original sample. In this problem half-life of the element is given and we have to find the time taken for $\dfrac{7}{8}$ th of a sample to decay. Therefore, we can use the direct equation connecting all these factors to find the solution.

Formula used:
We can use the following formula to calculate the time taken for $\dfrac{7}{8} t h$ of a sample to decay:
$\dfrac{N}{N_{0}}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}}$
Where $\mathrm{N}$ is the amount of sample remaining after $\mathrm{t}$ time.
$N_{0}$ is the original amount of sample.
$\mathrm{T}$ is the half-life of the isotope.

Complete step by step solution:
Here in this question half-life of the isotope ${ }_{11} N a^{24}$ is given as $15 \mathrm{hrs}$. After time the $\dfrac{7}{8}$ th of the sample is decayed. We have to find the time taken for this decay. To use the given we have to find how much sample is remaining now.

If we consider the amount of original sample present as 1,then
Undecayed sample, $N=1-\dfrac{7}{8}=\dfrac{1}{8}$
And $\mathrm{T}=15 \mathrm{hrs}$
Therefore,
$\dfrac{N}{N_{0}}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{T}} \Rightarrow \dfrac{1}{8}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{15}}$
Taking natural logarithm, we get:
$\ln \dfrac{1}{8}=\dfrac{t}{15} \ln \dfrac{1}{2}$

On further solving we get:
$\dfrac{t}{15}=\dfrac{\ln \dfrac{1}{8}}{\ln \dfrac{1}{2}} \\
\Rightarrow \dfrac{t}{15}=3 \\
\therefore t=15 \times 3=45 \text { hours }$
That is, if half-life of a sample is $15 \mathrm{hrs}$, then time taken for $\dfrac{7}{8}$ of the sample to decay is $45 \mathrm{hrs}$.

Therefore, the answer is option D.

Notes: One of the possible mistakes most people make is they consider the amount of sample as $\dfrac{7}{8}$. But actually, $\mathrm{N}$ in the equation stands for the remaining amount of sample. We approximate the original amount of sample to 1 since here fraction is considered. If the amount was given in percentage, we take the original amount as 100.