Answer

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**Hint:**First use the equation for the rate of radioactive decay, i.e. \[\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt\] , for the decay of half of the radioactive substance (i.e. \[N = 50\] , \[{N_0} = 100\] and \[t = 6hours = 360\min \] ) and calculate the value of \[k\] . Then substitute this value of \[k\] in \[\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt\] , where \[t = 18hours = 1080\min \] and \[{N_0} = 36grams\] , to calculate the value of \[N\] . Then calculate the value of \[{N_0} - N\] to reach the solution.

**Complete answer:**

Before moving on to the mathematical calculations, it would be beneficial for us to discuss radioactive disintegration (radioactive decay) and half-life.

Radioactive disintegration also called radioactive decay is the spontaneous decay of an atomic nucleus which results in the release of matter and energy from the nucleus. A radioactive substance has unstable nuclei that do not have the required binding energy (the energy required to separate all the constituents of the nucleus away from each other) to keep the nucleus intact.

Half-life – If we have a given amount of radioactive substance, then the time taken for half of that substance to decay is known as the half-life of that radioactive substance.

We know that the equation of the rate of radioactive decay is

\[\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt\]

Here, \[N = \] The amount of radioactive substance left after time \[t\]

\[{N_0} = \] The initial amount of radioactive substance

\[k = \] The rate of constant of radioactive decay

\[t = \] Time

So, for the amount of the substance to decrease by half, \[N = 50\]

\[{N_0} = 100\]

\[t = 6hours = 360\min \]

So, \[\ln \left[ {\dfrac{{50}}{{100}}} \right] = - k \times 360\]

\[\ln \left[ {\dfrac{1}{2}} \right] = - k \times 360\]

\[k = \dfrac{{ - \ln \left[ 2 \right]}}{{ - 360}}\]

We know that \[\ln \left[ 2 \right] = 0.693\]

\[k = \dfrac{{0.693}}{{360}} = 0.0019\min \]

When the radioactive substance has been decaying for \[18hours\] , then \[{N_0} = 36\] grams

\[t = 18hours = 1080\min \]

\[k = 0.0019\min \]

So, \[\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt\]

\[ \Rightarrow \ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - 0.0019 \times 1080\]

\[ \Rightarrow \ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - 2.052\]

\[ \Rightarrow \dfrac{N}{{{N_0}}} = {\ln ^{ - 1}}\left[ { - 2.052} \right]\]

We know that \[{\ln ^{ - 1}}\left[ { - 2.052} \right] = 0.1284777\]

\[ \Rightarrow \dfrac{N}{{36}} = {\text{0}}{\text{.1284777}}\]

\[N = 4.625gram\]

We have to calculate the amount of radioactive substance decayed, i.e. \[{N_0} - N = 36 - 4.625\]

\[{N_0} - N = 31.375gram\]

Therefore \[31.375gram\] of a radioactive substance decays in \[18\] hours, if the half-life of that radioactive substance is $6$ hours.

**So, the correct answer is “Option A”.**

**Note:**

To solve such types of problems in which we use the equation for the rate of radioactive decay, i.e. \[\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt\] , it is very important to know how to calculate the log values ( like in this question we used the value of \[\ln \left[ 2 \right]\] and \[{\ln ^{ - 1}}\left[ { - 2.052} \right]\] ). Most of the time it is already given in the problem, but it was not in the given problem.

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