Get Answers to all your Questions

header-bg qa

A solution containing active cobalt \frac{60}{27}  Co having activity of 0.8 μCi and decay constant λ is injected in an animal’s body. If 1 cm^{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? (1 Ci=3.7×^{10^{10}}  decays per second and at t=10 hrs e^{-\lambda t}=0.84)

  • Option 1)

    6 liters

  • Option 2)

    7 liters

  • Option 3)

    4 liters

  • Option 4)

    5 liters

 

Answers (2)

best_answer

As we learned

 

Number of nuclei after disintegration -

N=N_{0}e^{-\lambda t} or A=A_{0}e^{-\lambda t}

- wherein

Number of nucleor activity at a time is exponentional function

 

 let total volume of blood is V

Initial activity A_{o}=0.8\mu ci

Activity at t=t_{1}, A=A_{o}\cdot e^{-\lambda t}

Activity of 1cm^{3}=\left ( \frac{A}{V} \right )\cdot 1cm^{3}

                    (A')

v=(1cm^{3})\cdot \frac{A_{0}e^{-\lambda t}}{A_{1}}=(1cm^{3})\times \frac{8\times 10^{-7}\times 3.7\times 10^{10}\times 0.84}{\left ( \frac{300}{60} \right )}

V = 4.97 li

 


Option 1)

6 liters

Option 2)

7 liters

Option 3)

4 liters

Option 4)

5 liters

Posted by

Plabita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE