A radio nuclide A_1 with decay constant lambd | vedclass

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Question

Suppose $N_{1}$ and $N_{2}$ are the number of two radionuclides $A_{1}, A_{2}$ at time $t$

Then

$\frac{d N_{1}}{d t}=-\lambda_{1} N_{1}(1)$

$

Vedclass · Accepted Answer

$\frac{{\ln \left( {{\lambda _2}/{\lambda _1}} \right)}}{{{\lambda _2} - {\lambda _1}}}$

Vedclass · Answer

Suppose $N_{1}$ and $N_{2}$ are the number of two radionuclides $A_{1}, A_{2}$ at time $t$

Then

$\frac{d N_{1}}{d t}=-\lambda_{1} N_{1}(1)$

$\frac{d N_{2}}{d t}=\lambda_{1} N_{1}-\lambda_{2} N_{2}(2)$

From $(1)$

$N_{1}=N_{10} e^{-\lambda_{1} t}$

where $N_{10}$ is the initial number of nuclides $A_{1}$ at time $t=0$

From $(2)$

$\left(\frac{d N_{2}}{d t}+\lambda_{2} N_{2}\right) e^{\lambda_{2} t}=\lambda_{1} N_{10} e^{-\left(\lambda_{1}-\lambda_{2}\right) t}$

$\operatorname{or}\left(N_{2} e^{\lambda_{2} t}\right)=\operatorname{const} \frac{\lambda_{1} N_{10}}{\lambda_{1}-\lambda_{2}} e^{-\left(\lambda_{1}-\lambda_{2}\right) t}$

since $N_{2}=0$ at $t=0$

constant $N_{2}=\frac{\lambda_{1} N_{10}}{\lambda_{1}-\lambda_{2}}$

Thus $=\frac{\lambda_{1} N_{10}}{\lambda_{1}-\lambda_{2}}\left(e^{-\lambda_{2} t-e^{-\lambda_{1} t}}\right)$

(b) The activity of nuclide $A_{2}$ is $\lambda_{2} N_{2} .$ This is maximum when $N_{2}$ is maximum. That happens when

$\frac{d N_{2}}{d t}=0$

This require $\lambda_{2} e^{-\lambda_{2} t_{m}=\lambda_{1} e^{-\lambda_{2} tm}}$

$\operatorname{or} t_{m}=\frac{\ln \left(\lambda_{1} / \lambda_{2}\right)}{\lambda_{1}-\lambda_{2}}$

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