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A radioactive material of half life \mathrm{\top } has produced in a nuclear reactor at different instants. the quantity produced second time was twice of that produced first time .If now their present activities are \mathrm{A_1 } and \mathrm{A_2 } respectively then their age difference equals

Option: 1

\mathrm{\frac{\top }{\ln 2}\left|\ln \frac{A_1}{A_2}\right|}


Option: 2

\mathrm{\top \ln \left(\frac{A_1}{A_2}\right)}


Option: 3

\mathrm{\frac{\top}{\ln 2}\left|\ln \frac{A_2}{2 A_1}\right|}


Option: 4

\mathrm{\top\left|\ln \frac{A_2}{2 A_1}\right|}


Answers (1)

best_answer

Half life of both are equal.

\mathrm{A_1=A_{01} \cdot e^{-\lambda t_1}=\lambda \cdot N_1 \cdot e^{-\lambda t_1}}\quad -(1)
\mathrm{A_2=A_{02} \cdot e^{-\lambda t_2}=\lambda N_2 \cdot e^{-\lambda t_2}}\quad -(2)

\mathrm{\text { Since } N_2 =2 N_1}

\mathrm{\Rightarrow A_2 =2 \lambda N_1 \cdot e^{-\lambda t_2}-\text { (2) } }
\mathrm{\Rightarrow \frac{A_1}{A_2} =\frac{1}{2} \cdot e^{\lambda\left(t_2-t_1\right) } }.

\mathrm{\Rightarrow \ln \left(\frac{2 A_1}{A_2}\right) =\lambda\left(t_2-t_1\right) }
\mathrm{t_2-t_1 =A g e \text { Difference }=\frac{1}{\lambda} \ln \left(\frac{2 A_1}{A_2}\right)}
\mathrm{=\frac{\top}{\ln 2}\left|\ln \frac{A_2}{2 A_1}\right| }.


 

Posted by

Shailly goel

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