Question

The half-life of radioactive radon is $3.8$ days. The time at the end of which $\frac{1}{{20}}th$ of the radon sample will remain undecayed is (given ${\log _{10}}e = 0.4343$ )
A) $3.8\,\,days$
B) $16.5\,\,days$
C) $33\,\,days$
D) $76\,\,days$

Answer 1

‌To solve this question, we must first understand the concept of half-life of Radioactive substance. Then we need to assess a formula for half-life which includes initial and final content of the radioactive substance, for calculating the half-life and then only we can conclude the correct answer.

Complete step by step solution:
Before we move forward with the solution of this given question, let us first understand some basic concepts:
Half-life of a radioactive substance ${t_{1/2}}$ measures the time it takes for a given amount of the substance to become reduced by half as a consequence of decay, and therefore, the emission of radiation.
It is related to the radioactive decay constant $k$ as ${t_{1/2}} = \,\,\frac{{\ln 2}}{k}$ .
Also its relation with mean life ${\lambda _m} = {k_1}$ ​ is
${t_{1/2}} = \,\,{\lambda _m}\ln 2$
Step 1: In this step we will enlist all the given properties:
Half-life ${t_{1/2}}$ $= 3.8$ days
Final amount remaining $= \frac{1}{{20}}$ of initial content
Step 2: In this step we will calculate the required time:
As we know that, $N = \,\,{N_ \circ }{e^{ - \lambda t}}$
And, ${t_{1/2}} = \,\,\frac{{\ln 2}}{k}$
Now, substituting the value of $\lambda$in the first formula:
$\frac{N}{{{N_ \circ }}} = \,\,{e^{ - \frac{{\ln 2}}{{3.8}}t}}$ ; where $t$ is the required time
$\Rightarrow \frac{1}{{20}} = \,\,{e^{ - \frac{{\ln 2}}{{3.8}}t}}$
$\Rightarrow t = \,\,16.5\,\,days$
So, clearly we can conclude that the correct answer is Option B.

Note: A half-life usually describes the decay of discrete entities, such as radioactive atoms. In that case, it does not work to use the definition that states "half-life is the time required for exactly half of the entities to decay". For example, if there is just one radioactive atom, and its half-life is one second, there will not be "half of an atom" left after one second.