Question

Half-life of a substance is \[20\] minutes, then the time between \[33\% \]decay and \[67\% \] decay will be:
\[(i){\text{ 20}}\] Minutes
\[(ii){\text{ 40}}\] Minutes
\[(iii){\text{ 50}}\] Minutes
\[(iv){\text{ 10 }}\]Minutes

Answer 1

Hint: Half-life of substance is defined as when the \[50\% \] decay of the radioactive substance takes place. With the help of half-life we will find the decay constant. Then by using decay constant we can find the time required for \[33\% \]decay and \[67\% \]decay. Thus we can find the difference of time between them.
Formula Used:
\[(i){\text{ }}\lambda {\text{ = }}\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}\] , where \[\lambda \] is decay constant and \[{t_{\dfrac{1}{2}}}\] is the halftime for the completion of reaction.
\[(ii){\text{ t = }}\dfrac{{2.303}}{\lambda }{\log _{10}}\dfrac{{{N_ \circ }}}{N}\] , where \[\lambda \] is decay constant , \[t\] is the time taken for completion of reaction, \[{N_ \circ }\] is the initial amount of substance and \[N\] is the amount of substance left after decay.

Complete answer:
The half-life of a radioactive substance is given which is equal to \[20\] minutes. With the help of half-time we will find the value of decay constant by using the relation as,
\[{\text{ }}\lambda {\text{ = }}\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}\]
It is given that, \[{t_{\dfrac{1}{2}}}{\text{ = 20}}\], on substituting the value we get,
\[ \Rightarrow \] \[{\text{ }}\lambda {\text{ = }}\dfrac{{0.693}}{{20}}\]
\[ \Rightarrow \] \[{\text{ }}\lambda {\text{ = 0}}{\text{.03465}}\] Per minute
Thus we get the decay constant for the radioactive substance. Now we will find the time for \[33\% \]decay of the substance of its initial value. This can be find by using the relation,
\[{\text{ t = }}\dfrac{{2.303}}{\lambda }{\log _{10}}\dfrac{{{N_ \circ }}}{N}\]
Where, \[{N_ \circ }\] is the initial concentration of substance. For \[33\% \] decay the value of \[N\]will be equal to,\[N{\text{ = 100 - 33 = 67}}\]
On substituting the values in the equation we get the time for \[33\% \] decay as,
\[{\text{ }}{{\text{t}}_{33\% }}{\text{ = }}\dfrac{{2.303}}{\lambda }{\log _{10}}\dfrac{{{N_ \circ }}}{N}\]
\[ \Rightarrow \] \[{\text{ }}{{\text{t}}_{33\% }}{\text{ = }}\dfrac{{2.303}}{{0.03465}}{\log _{10}}\dfrac{{100}}{{67}}\]
\[ \Rightarrow \] \[{\text{ }}{{\text{t}}_{33\% }}{\text{ = 11}}{\text{.6}}\] Minutes
Similarly for \[67\% \]decay the value of \[N\]will be equal to \[N{\text{ = 100 - 67 = 33}}\] , on substituting the values we get time for \[67\% \] decay as,
\[{\text{ }}{{\text{t}}_{67\% }}{\text{ = }}\dfrac{{2.303}}{\lambda }{\log _{10}}\dfrac{{{N_ \circ }}}{N}\]
\[ \Rightarrow \] \[{\text{ }}{{\text{t}}_{67\% }}{\text{ = }}\dfrac{{2.303}}{{0.03465}}{\log _{10}}\dfrac{{100}}{{33}}\]
\[ \Rightarrow \] \[{\text{ }}{{\text{t}}_{67\% }}{\text{ = 32}}\] Minutes
Hence we get the time for each decay respectively. Now the time taken between \[33\% \]decay and \[67\% \] decay will be.
\[\Delta t{\text{ = }}{{\text{t}}_{33\% }}{\text{ - }}{{\text{t}}_{67\% }}\]
\[ \Rightarrow \] \[\Delta t{\text{ = 11}}{\text{.6 - 32}}\] Minutes
\[ \Rightarrow \] \[\Delta t{\text{ = 20}}{\text{.4 }} \approx {\text{ 20}}\] Minutes
Therefore the correct option is \[(i){\text{ 20}}\] Minutes.

Note:
Since the value of time comes to be in negative numbers, but time cannot be a negative number. We will ignore the negative sign and thus time is always a non-negative number. We can also convert minutes into seconds. \[N\] is the amount of substance left after the decay percentage.