Question

The isotope Francium $224$ has a half life of $20$ minutes. A sample of the isotope has an initial activity of $9$ disintegrations per second. The approximate activity of the sample an hour later will be what? Multiple choice question. Please explain why.

Answer 1

Some unstable nucleus undergoes decomposition to form a stable nucleus. This process is known as radioactive decay or radioactive decomposition. The given half life is $20$ minutes. Given time is one hour, by substituting this time in the below formula gives the approximate activity of the sample an hour later.
$N\left( t \right) = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{t_{1/2}}}}}}$
$N\left( t \right)$ is the quantity of the substance remaining
${N_0}$ is the initial amount
$t$ is the time given
${t_{1/2}}$ is half -life.

Complete Step By Step Answer:
Francium is a radioactive element that can undergo radioactive decay or decomposition to form a stable nucleus and has a half-life of $20$ minutes.
Given that the isotope Francium $224$ has a half-life of $20$ minutes.
The given time is an hour which means one hour means $60$ minutes. Which means the sample is passed through three half-lives.
As the initial quantity is same, substitute the half lime in ${t_{1/2}}$ and $60$ minutes in $t$
$N\left( t \right) = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{{60}}{{20}}}}$
By simplifying the above formula,
$N\left( t \right) = {N_0}{\left( {\dfrac{1}{2}} \right)^3}$
Thus, the approximate activity of the sample an hour later which is known as disintegration is $\dfrac{{{N_0}}}{8}$ .

Note:
The initial we have taken is the same, but the half-life time and given time is $60$ minutes, from these both the number of half-lives that sample passes can be known. From the half-lives passed the amount of activity of the sample can be determined. For three one half-life the value is $\dfrac{{{A_0}}}{2}$ , and for the two half-lives the value is $\dfrac{{{A_0}}}{4}$ . These constant values are given the activity of the sample.