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# A solution containing active cobalt ${}_{27}^{60}Co$ having activity of $0.8\mu Ci$ and decay constant $\lambda$ is injected in an animal’s body. If $1$$c{m^3} of blood is drawn from the animal’s body after 10$$hrs$ of injection, the activity found was $300$ decays per minute. What is the volume of blood that is flowing in the body? ($1Ci = 3.7 \times {10^{10}}$ decays per second and at $t = 10$$hrs {e^{ - \lambda t}} = 0.84)(A) 6litres(B) 7litres(C) 4litres(D) 5litres Last updated date: 16th Jun 2024 Total views: 385.2k Views today: 8.85k Answer Verified 385.2k+ views Hint:Here, we have a condition in which the animal is injected with cobalt {}_{27}^{60}Co, this cobalt will mix in the blood and will start to decay. Consider there are N active nuclei at an instant of time t. Find the number nuclei decayed in a time dt and from there you can find the activity. Then find the decay rate after 10$$hrs$. Keep in mind that you have taken a sample of
$1$$c{m^3}, so accordingly use the method of fraction to get the total volume of blood. Complete step by step solution: The number nuclei decayed in a time dt is proportional to the active numbers of nuclei and the time interval dt, so, we have dN \propto Ndt \to dN = - \lambda Ndt (negative sign indicates the decay). Let us Integrate this equation, \int\limits_{{N_0}}^N {\dfrac{{dN}}{N}} = - \lambda \int\limits_0^t {dt} \\ \ln \dfrac{N}{{{N_0}}} = - \lambda t \\ N = {N_0}{e^{ - \lambda t}} \\ Now, to get the activity, differentiate it, we will get - \dfrac{{dN}}{{dt}} = {N_0}\lambda {e^{ \lambda t}}. This is the activity. Initially at t = 0, activity is 0.8\mu Ci, so, we have {N_0}\;\lambda {e^{ - \left( {\lambda \left( 0 \right)} \right)}} = 0.8 \times {10^{ - 6}} \times 3.7 \times {10^{10}} {N_0}\;\lambda = 0.8 \times {10^{ - 6}} \times 3.7 \times {10^{10}} \\ {N_0}\;\lambda = 2.96 \times {10^4} \\ Now, let us take the total volume of blood flowing in the animal’s body as V. As we have taken only1$$c{m^3}$ of blood, the number of nuclei taken in the sample will be $\left( {\dfrac{1}{V}} \right) \times N = \dfrac{{{N_0}}}{V}{e^{ - \lambda t}}$ .

Now, the activity after $10$$hrs is 300decays per minute. Therefore, we will have activity given as \left( {\dfrac{{{N_0}}}{V}{e^{ - \lambda t}}} \right)\lambda = \dfrac{{300}}{{60}} decays per second. Substituting the value {N_0}\;\lambda = 2.96 \times {10^4} and {e^{ - \lambda t}} = 0.84 at t = 10$$hrs$ in the above equation, we will get,
$\dfrac{{\left( {{N_0}\lambda } \right)\left( {{e^{ - \lambda t}}} \right)}}{V} = 5 \\ \dfrac{{\left( {2.96 \times {{10}^4}} \right)\left( {0.84} \right)}}{5} = V \\$
$\to V = 4972.8$$c{m^3}$ which is approximately $5$litres.

Hence, the volume of blood that is flowing in the body is $5$litres.

Hence, Option (D) is correct.

Note: Remember to change the unit of activity from curie to decay per second. Note the method we used to find the number of nuclei present at any instant of time and also how we calculated the activity of the sample from the equation by differentiating it. Also notice that we considered the fraction of nuclei that will be available in the taken sample, otherwise, you will get the wrong answer.