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

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.