Question

A radioactive material of half-life time of 69.3 days kept in a container \[\dfrac{2}{3}rd\] of the substance remains undecayed after (given, \[\ln \dfrac{3}{2}=0.4\])
(A) 20 days
(B) 25 days
(C) 40 days
(D) 50 days

Answer 1

Hint: Radioactivity refers to the phenomenon in which the substance decays by emission of radiation. Half-life is defined as the time taken by the material in which the number of undecayed atoms becomes half. A material containing unstable nuclei is considered radioactive

Complete step by step answer: We know there exists a relationship between the decay constant, \[\lambda \]and half-life \[{{T}_{1/2}}\]. It states \[{{T}_{1/2}}\lambda =0.693\]
Given, \[{{T}_{1/2}}\]= 69.3 days
Thus \[\begin{align}
  & 69.3\times \lambda =0.693 \\
 & \lambda =0.01/days \\
\end{align}\]
Now we have calculated the decay constant and it is given that \[\dfrac{2}{3}rd\]of the substance remains undecayed.
Let the number of initial atoms be \[{{N}_{0}}\]. Then the number of undecayed atoms be N. then according to the question,
\[N=\dfrac{2}{3}{{N}_{0}}\]
Now using the law of radioactivity, \[N={{N}_{0}}{{e}^{-\lambda t}}\]
\[\begin{align}
  & \dfrac{2}{3}{{N}_{0}}={{N}_{0}}{{e}^{-\lambda t}} \\
 & \dfrac{2}{3}={{e}^{-0.01t}} \\
\end{align}\]
\[0.66={{e}^{-0.01t}}\]
Using the natural log, we get,
\[Ln(0.66)=-0.01t\]
-0.41=-0.01t
t=40 days
Hence, the correct option is (C).

Additional Information: Half-life is the time for half the radioactive nuclei in any sample to undergo radioactive decay. For example, after 2 half-lives, there will be one fourth the original material remains, after three half-lives one eight the original material remains, and so on. Half-life is a convenient way to assess the rapidity of a decay

Note: While solving such problems we have to keep in mind that while using the formula \[N={{N}_{0}}{{e}^{-\lambda t}}\], the quantity on LHS is the number of atoms or nuclei which are undecayed after time t and \[{{N}_{0}}\]is the original number of atoms or nuclei at a time, t=0.