Question

Half-life of a radioactive substance is 1 hour. The time taken for the disintegration of 87.5% substance is
A. 2 hrs
B. 3 hrs
C. 3.5 hrs
D. 4 hrs

Answer 1

Hint: Use the formula for the decay constant. Using this formula determines the half-life of the radioactive substance. Then use the formula for the population of the radioactive substance at any time t and determine the time required for the disintegration of the substance.

Formula used:
The decay constant \[\lambda \] is given by
\[\lambda = \dfrac{{0.693}}{T}\] …… (1)
Here, \[T\] is the half-life of a radioactive element.
The equation for the population of the radioactive element present at any time is
\[N = {N_0}{e^{ - \lambda t}}\] …… (2)
Here, \[{N_0}\] is the initial population of the radioactive element, \[N\] is the radioactive element at time \[t\] and \[\lambda \] is the decay constant of the decay.

Complete step by step solution:
We have given that the half-life of the radioactive substance is one hour.
\[T = 1\,{\text{h}}\]
We can determine the decay constant using equation (1).
Substitute for in equation (1).
\[\lambda = \dfrac{{0.693}}{{1\,{\text{h}}}}\]
\[ \Rightarrow \lambda = 0.693\,{{\text{h}}^{{\text{ - 1}}}}\]
Hence, the decay constant is \[0.693\,{{\text{h}}^{{\text{ - 1}}}}\].
The radioactive substance disintegrated 87.5%. Hence, the radioactive substance available at time \[t\] is 12.5%.
Hence, the fraction of the radioactive substance as that of the initial population of the radioactive substance at time \[t\] is 0.125.
\[\dfrac{N}{{{N_0}}} = 0.125\]
Rearrange equation (2) for \[\dfrac{N}{{{N_0}}}\].
\[\dfrac{N}{{{N_0}}} = {e^{ - \lambda t}}\]
Substitute \[0.125\] for \[\dfrac{N}{{{N_0}}}\] and \[0.693\,{{\text{h}}^{{\text{ - 1}}}}\] for \[\lambda \] in the above equation.
\[0.125 = {e^{ - \left( {0.693\,{{\text{h}}^{{\text{ - 1}}}}} \right)t}}\]
Take natural log on both sides of the above equation.
\[{\log _e}\left( {0.125} \right) = {\log _e}{e^{ - \left( {0.693\,{{\text{h}}^{{\text{ - 1}}}}} \right)t}}\]
\[ \Rightarrow {\log _e}\left( {0.125} \right) = - \left( {0.693\,{{\text{h}}^{{\text{ - 1}}}}} \right)t\]
\[ \Rightarrow - 2.0794 = - \left( {0.693\,{{\text{h}}^{{\text{ - 1}}}}} \right)t\]
\[ \Rightarrow t = \dfrac{{2.0794}}{{0.693\,{{\text{h}}^{{\text{ - 1}}}}}}\]
\[ \Rightarrow t = 3\,{\text{h}}\]
Therefore, the time required for the disintegration is \[3\,{\text{h}}\].

So, the correct answer is “Option B”.

Note:
One can also solve the same question by another method. One can directly use the formula for the population of the radioactive substance at any time t in terms of the half-life period of the radioactive substance instead of determining the decay constant of the decay of the radioactive substance.