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

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.