Question

The half-life period of radon is 3.8 days. After how many days will only one-twentieth of a radon sample be left over?

Answer 1

Hint: First determine the value of decay constant for the given value of half-life of radon. Now we can substitute the value in the equation given below:
$t\,=\,\dfrac{2.303}{\lambda }\log \dfrac{a}{a-x}$
- Where, ‘t’ is time taken for the specified amount of decay, ‘$\text{ }\!\!\lambda\!\!\text{ }$’is the decay constant, ‘a’ is initial concentration of reactant or radioisotope and ‘x’ is concentration of reactant or radioisotope that has decayed or disintegrated.

Complete step by step answer:
- To solve the question first, we will find the value of decay constant $\left( \text{ }\!\!\lambda\!\!\text{ } \right)$
$\lambda =\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}}\,=\,\dfrac{0.693}{3.8}=0.182/day$
${{t}_{{}^{1}/{}_{2}}}$ in the equation is the half-life value for radon
- Let us now find the time taken such that only one-twentieth of a radon sample is left over.
Here the value $\left( \text{a-x} \right)$ will be equal to the one –twentieth value of initial concentration.
$\left( a-x \right)\,=\,\dfrac{1}{20}a$
$\lambda \,=\,0.182/day$
Hence let’s substitute these values in the equation,
$t\,=\,\dfrac{2.303}{\lambda }\log \dfrac{a}{a-x}$
By substituting we get the value,
$t\,=\,\dfrac{2.303}{\lambda }\log \dfrac{a}{{}^{a}/{}_{20}}$
$t\,=\,\dfrac{2.303}{0.182}\log \,20$
$t\,=\,\dfrac{2.303}{0.182}\times 1.3010$
$t\,=\,16.46\,days$
From the above calculation we can conclude that the time taken by a radon sample such that only one-twentieth of the original sample is left is 16.46 days.

Additional Information :
- Most first order reactions are radioactive decay reactions. The first order reaction finds its application to convert unstable nuclei to a stable nucleus. And the energy released during this process is harnessed and is used as an energy source in many processes.
- Radioactive decay also known as radioactive disintegration or nuclear disintegration is the process by which an unstable atomic nucleus loses energy in the form of radiation to gain stability. Any element which consists of unstable nuclei is considered a radioactive element.

Note: In case, you are not able to determine the order of the reaction, take a look at the unit of decay constant. The unit of decay constant is different for every order of reaction. The unit of decay constant for first order reaction is ${{\min }^{-1}}$.