Question

A $280$ days old radioactive substance shows an activity of $6000\;{\text{dps}}$ , $140$ days later its activity becomes $3000\;{\text{dps}}$. What is its initial activity?
A. $20000\;{\text{dps}}$
B. $24000\;{\text{dps}}$
C. $12000\;{\text{dps}}$
D. $6000\;{\text{dps}}$

Answer 1

We can use the formula for the radioactivity to find the initial activity. The decay constant is a constant value which can be used for solving both cases. And the half- life of a radioactive sample is the time taken for half of the sample to degenerate.

Complete step-by-step solution:
Given a $280$ days old radioactive substance shows an activity of $6000\;{\text{dps}}$. And $280 + 140 = 420$ days shows activity of $3000\;{\text{dps}}$.
The decay constant is the inverse of the average lifetime of a radioactive sample before decaying.
The expression for decay constant is given as,
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{A_0}}}{A}$
Where, ${A_0}$ is the initial activity, $A$ is the activity at time $t$ .
Here we are taking the number of days as time.
Substituting the values in the above expression, we get
$\lambda = \dfrac{{2.303}}{{280}}\log \dfrac{{{A_0}}}{{6000}}........\left( 1 \right)$
And $\lambda = \dfrac{{2.303}}{{420}}\log \dfrac{{{A_0}}}{{3000}}.............\left( 2 \right)$
Since both equations are foe a constant we can equate both equations.
Equating the two equations,
$\dfrac{{2.303}}{{280}}\log \dfrac{{{A_0}}}{{6000}} = \dfrac{{2.303}}{{420}}\log \dfrac{{{A_0}}}{{3000}} \\\ \log \dfrac{{{A_0}}}{{6000}} = \dfrac{1}{{1.5}}\log \dfrac{{{A_0}}}{{3000}} \\\$
Taking the exponential on both sides and solving the above equation we get, ${A_0} = 24000\;{\text{dps}}$
The initial activity of the radioactive sample is $24000\;{\text{dps}}$.
The answer is option B.

Note:-
Alternative method:-
Activity $6000\;{\text{dps}}$ reduces to its half $3000\;{\text{dps}}$with later $140$ days. Hence the half -life of the radioactive sample is $140$ days.
Therefore in $280$ days, that is double the half- life. Thus the activity will remain $\left( {\dfrac{1}{2} \times \dfrac{1}{2}} \right) = \dfrac{1}{4}$ of the initial activity.
Thus initial activity is four times the activity of $280$ days.
Initial activity$= 4 \times 6000\;{\text{dps = 24000}}\;{\text{dps}}$
Thus the initial activity is ${\text{24000}}\;{\text{dps}}$.