Question

Calculate the time required for 60$\%$ of a sample of radon to undergo decay. (Given ${T_{\dfrac{1}{2}}}$ of radon $= 3.8$ days).

Answer 1

In this type of questions, we make use of the decay equation. We can find the decay constant from the half-life of radon and the amount of radon left in the sample is also known. By using these values in the decay equation, we can obtain the required answer.

Formula used:
The decay equation for a radioactive material is given as
$N = {N_0}{e^{ - \lambda t}}$
Here N represents the number of nuclei of the radioactive material in a given sample at some time t while ${N_0}$ represents the number of nuclei of the radioactive material in a given sample initially at t = 0.
$\lambda$ is known as the decay constant. It is related to the half- life ${T_{\dfrac{1}{2}}}$ of the radioactive material by the following relation.
$\lambda = \dfrac{{0.693}}{{{T_{\dfrac{1}{2}}}}}$

Complete step by step answer:
We are given a sample of radon which is a radioactive material and is undergoing decay. We need to find out how much time is required for the 60$\%$ of samples to undergo decay.
Let initially we have ${N_0}$ number of radon nuclei in the sample. After time t, the number of radioactive nuclei reduces to N. Now if 60$\%$ of sample has decayed then only 40$\%$ of nuclei are left. Therefore, we have
$\dfrac{N}{{{N_0}}} = \dfrac{{40}}{{100}} = \dfrac{2}{5}$
We are also given the half-life of radon from which we can calculate the decay constant for radon.
${T_{\dfrac{1}{2}}}$ of radon $= 3.8{\text{ days}} = 3.8 \times 24 \times 60 \times 60s$
This is done in the following way.
$\lambda = \dfrac{{0.693}}{{{T_{\dfrac{1}{2}}}}} = \dfrac{{0.693}}{{3.8 \times 24 \times 60 \times 60}} = 2.11 \times {10^{ - 6}}{s^{ - 1}}$
Now we can use all these known values in the decay equation to calculate the value of t. This can be done in the following way.
$N = {N_0}{e^{ - \lambda t}} \\\ \dfrac{N}{{{N_0}}} = {e^{ - \lambda t}} \\\$
Taking natural logarithm on the both sides of this equation, we get
$\ln \dfrac{N}{{{N_0}}} = - \lambda t \\\ \Rightarrow t = \dfrac{1}{\lambda }\ln \dfrac{{{N_0}}}{N} \\\$
Now inserting all the known parameters, we get
$t = \dfrac{1}{{2.11 \times {{10}^{ - 6}}}}\ln \dfrac{5}{2} \\\ = \dfrac{{0.916}}{{2.11 \times {{10}^{ - 6}}}} = 0.434 \times {10^6}s = 4.34 \times {10^5}s = 5.02{\text{ days}} \\\$
This is the required solution.

Note:
It should be noted that in the question we are given the percentage of the radon sample that has decayed. But in the decay equation, we always talk about the amount of nuclei which are left after decay.