Question

A patient undergoing treatment for thyroid cancer receives a dose of radioactive iodine $\left( {^{131}I} \right)$ , which has a half-life of $8.05\;$ days. If the original dose contained $12mg$ of $\left( {^{131}I} \right)$ , calculate the mass of $^{131}I$ remains after $16.1\;$ days?
A.$6mg$
B.$9mg$
C.$2mg$
D.$3mg$

Answer 1

Hint: The given question is based on the concept of half-life of a substance that is the time it takes for the substance to reach half its initial value. The time after which we have to find the concentration of the substance is exactly two times the half life of the radioactive iodine. This means that by using the relation that connects the initial concentration, half-life and final concentration of iodine we can find the answer to this question.
FORMULA USED:
${N_t}$ = ${N_0}$$ \times {2^{ - \dfrac{t}{T}}}$
Where ${N_t}$ is the concentration of the radioactive iodine after the time given in the question.
${N_0}$ is the initial concentration of radioactive iodine
$t$ is the time after which the concentration is required to be obtained
$T$ is the half life of the radioactive iodine.

Complete step by step solution:
It is required to find the concentration of iodine after $16.1\;$ days have passed. This means that a certain amount of iodine has decayed and the concentration of iodine has reduced. To find this answer we must plug in the values given in the question into the formula given.
The initial concentration of the radioactive iodine will be, ${N_0}$= $12mg$
The half life of the iodine will be, $t$= $8.05\;$ days.
The time after which the concentration is required to be found, $T$= $16.1\;$ days.
Therefore, substituting the values given in the formula,
${N_t}$ = ${N_0} \times {2^{ - \dfrac{t}{T}}}$
we get,
${N_t}$ = $12 \times {2^{ - \dfrac{{16.1}}{{8.05}}}}$
$\dfrac{{16.1\;}}{{8.05}}$ = $2$
Therefore, putting this value in the above step we will get,
${N_t}$=$12 \times {2^{ - 2}}$
The exponent on $2$ is negative meaning it will divide $12$ as shown below:
${N_t}$=$\dfrac{{12}}{{{2^2}}}$
This will further give,
${N_t}$= $\dfrac{{12}}{4}$
Therefore, ${N_t}$ = $3$
Thus, the concentration of iodine after $16.1\;$ days in the question will be $3mg$ that is the mass of radioactive iodine remaining will be $3mg$.
Therefore, the correct option to the answer will be option D that is, $3mg$.

Note:
-It is important to keep in mind that the formula ${N_t}$ = ${N_0} \times {2^{ - \dfrac{t}{T}}}$ helps to derive ${N_t}$ which is the concentration that is remaining.
-Half life can be defined as the time required for half of the mass of a reactant to undergo radioactive decay. Therefore, with every half life that passes, the concentration of the substance decreases by half of the previous concentration.