Question

The half-life of a radioactive substance is 3.6 days. How much of 20 mg of this radioactive substance will remain after 36 days?
A. 0.0019 mg
B. 1.019 mg
C. 1.109 mg
D. 0.019 mg

Answer 1

Hint:It is possible to estimate how many nuclei in a radioactive sample decay over time using the law of radioactive decay. The half-life is the length of time needed for the original quantity of atoms to fall to half of what it was initially,

Formula used :
The law of radioactive decay,
\[N = {N_0}{e^{ - \lambda t}}\]
Where, \[N\] – number of atoms present at t
\[{N_0}\] - number of atoms present at \[t = 0\]
\[\lambda \] - decay or disintegration constant
The half-life of the radioactive substance is,
\[{T_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.6931}}{\lambda }\]
Where, \[{T_{\dfrac{1}{2}}}\] - half-life

Complete step by step solution:
The law of radioactive decay helps us to estimate the quantity of atoms decaying over time. For a given radioactive sample, it takes infinite time to completely decay, but the radioactive decay law will give us the approximate time for it to decay during a period of time. It is given by the following equation,
\[N = {N_0}{e^{ - \lambda t}}\]

The half-life is used to evaluate the time needed for the radioactive sample to decrease to one half of its whole content at the start. It is useful to know the half-life of a radioactive substance to know the activity of it. The half-life is given as,
\[{T_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.6931}}{\lambda }\]

According to question, \[{T_{\dfrac{1}{2}}} = 3.6\,days\], \[{N_0} = 20mg\] and \[t = 36\,days\]
To find, \[N = ?\]
To find the remaining radioactive substance of 20 mg after 36 days, we need to rearrange the above two formulae as,
\[N = {N_0}{e^{ - \dfrac{{\ln 2}}{{{T_{\dfrac{1}{2}}}}}t}}\]
\[\Rightarrow N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{T_{\dfrac{1}{2}}}}}}}\]
Substituting the above values in this equation, we get,
\[N = 20 \times {\left( {\dfrac{1}{2}} \right)^{\dfrac{{36}}{{3.6}}}}\]
\[\Rightarrow N = 20\times {\left( {\dfrac{1}{2}} \right)^{\dfrac{{36}}{{36{\rm{x1}}{{\rm{0}}^{ - 1}}}}}}\]
\[\Rightarrow N = 20 \times {\left( {\dfrac{1}{2}} \right)^{10}}\]
\[\Rightarrow N = 20\times \dfrac{1}{{1024}}\]
\[\therefore N = 0.019\,mg\]

Hence, the correct answer is option D.

Note: The half-life given is 3.6 days, which can be written as \[36 \times {\rm{1}}{{\rm{0}}^{ - 1}}\]days and hence the calculations are made easier. While rearranging the formulae, the exponential and natural log are inverse functions and hence cancel each other and the remaining 2, which has a negative sign, becomes a denominator.