Two radioactive materials A and B have decay | vedclass

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Question

$N = N _{0} e ^{-\lambda t }$

$\frac{ N _{ B }}{ N _{ A }}=\frac{ e ^{-\lambda_{2} t }}{ e ^{-\lambda_{1} t }}= e ^{-\lambda_{2} t } \cdot e ^{\lam

Vedclass · Accepted Answer

$9$

Vedclass · Answer

$N = N _{0} e ^{-\lambda t }$

$\frac{ N _{ B }}{ N _{ A }}=\frac{ e ^{-\lambda_{2} t }}{ e ^{-\lambda_{1} t }}= e ^{-\lambda_{2} t } \cdot e ^{\lambda_{1} t }$

$e ^{1}= e ^{\left(\lambda_{1}-\lambda_{2}\right) t }$

$\left(\lambda_{1}-\lambda_{2}\right) t =1$

$t =\frac{1}{\lambda_{1}-\lambda_{2}}=\frac{1}{25 \lambda-16 \lambda}=\frac{1}{9 \lambda}$

Two radioactive materials $A$ and $B$ have decay constants $25 \lambda$ and $16 \lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $B$ to that of $A$ will be "$e$" after a time $\frac{1}{a \lambda}$. The value of $a$ is $......$

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