Question

The number of atoms of a radioactive substance of half life T is \[{{N}_{0}}\]at t=0. The time necessary to decay from \[\dfrac{{{N}_{0}}}{2}\]atoms to \[\dfrac{{{N}_{0}}}{10}\]atoms will be
A. \[\dfrac{5T}{2}\]
B. \[T\log \dfrac{5}{2}\]
C. \[T\log 5\]
D. \[T\log 2\]

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, half life is T. at time t=0 the number of atoms is \[{{N}_{0}}\]
Now using the law of radioactivity, \[N={{N}_{0}}{{e}^{-\lambda t}}\]
For the given condition: \[\dfrac{{{N}_{0}}}{2}={{N}_{0}}{{e}^{-\lambda {{t}_{1}}}}\]and \[\dfrac{{{N}_{0}}}{10}={{N}_{0}}{{e}^{-\lambda {{t}_{2}}}}\]
Solving them, \[\dfrac{1}{2}={{e}^{-\lambda {{t}_{1}}}}\]and \[\dfrac{1}{10}={{e}^{-\lambda {{t}_{2}}}}\]
Solving them further,
\[\lambda {{t}_{1}}=\ln 2\]-----(1)
\[\lambda {{t}_{2}}=\ln 10\]----(2)
Subtracting (2) from (1) we get, \[\lambda ({{t}_{2}}-{{t}_{1}})=\ln 10-\ln 2\]
We know there exists a relationship between the decay constant, \[\lambda \]and half-life \[{{T}_{1/2}}\]. It states \[{{T}_{1/2}}\lambda =0.693\]
So, taking ln on both sides, we get
$
({{t}_{2}}-{{t}_{1}})=\dfrac{T(\log 10-\log 2)}{0.693} \\
\implies ({{t}_{2}}-{{t}_{1}})=\dfrac{T(\log 10-\log 2)}{\log 2} \\
\implies ({{t}_{2}}-{{t}_{1}})=T[\dfrac{\log 5}{\log 2}] \\
\therefore ({{t}_{2}}-{{t}_{1}})=T\log \dfrac{5}{2} \\
$

So, the correct answer is “Option B”.

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.