બે રેડિઓએક્ટિવ પદાર્થો A અને B ના ક્ષય-નિયતાં | vedclass

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Question

Given : $\lambda_{A}=5 \lambda$, $\lambda_{B}=\lambda$

At $t=0,\left(N_{0}\right)_{A}=\left(N_{0}\right)_{B}$

$\frac{N_{A}}{N_{B}}=\left(\frac{1

Vedclass · Accepted Answer

$\frac{1}{{2\lambda }}$

Vedclass · Answer

Given : $\lambda_{A}=5 \lambda$, $\lambda_{B}=\lambda$

At $t=0,\left(N_{0}ight)_{A}=\left(N_{0}ight)_{B}$

$\frac{N_{A}}{N_{B}}=\left(\frac{1}{e}ight)^{2}$

According to radioactive decay, $\frac{N}{N_{0}}=e^{-\lambda t}$

$	herefore \,\frac{N_{A}}{\left(N_{0}ight)_{A}} =e^{-\lambda_{A} t} $  ..... $(i)$

$\frac{N_{B}}{\left(N_{0}ight)_{B}} =e^{-\lambda_{B} t}$  ..... $(ii)$

Divide $(i)$ by $(ii)$, we get

$\frac{N_{A}}{N_{B}}=e^{-\left(\lambda_{A}-\lambda_{B}ight) t}$  or, $\frac{N_{A}}{N_{B}}=e^{-(5 \lambda-\lambda) t}$

or, $\left(\frac{1}{e}ight)^{2}=e^{-4 \lambda t}$ or,  $\left(\frac{1}{e}ight)^{2}=\left(\frac{1}{e}ight)^{4 \lambda t}$

or, $4 \lambda t=2$ $ \Rightarrow \quad t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}$

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