Two radioactive elements R and S disinte | vedclass

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Question

$\mathrm{N}_{1}=\mathrm{N}_{01} \mathrm{e}^{-\lambda_{2} \mathrm{t}}$

$\mathrm{N}_{2}=\mathrm{N}_{02} \mathrm{e}^{-\lambda_{2} \mathrm{z}}$

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Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

$\mathrm{N}_{1}=\mathrm{N}_{01} \mathrm{e}^{-\lambda_{2} \mathrm{t}}$

$\mathrm{N}_{2}=\mathrm{N}_{02} \mathrm{e}^{-\lambda_{2} \mathrm{z}}$

$\frac{{{{m{N}}_1}}}{{{{m{N}}_2}}} = \frac{2}{1} 	imes {{m{e}}^{ - \left( {{\lambda _1} - {\lambda _2}} ight)t}}$

${\lambda _1}t = 3\ln 2$

${\lambda _2}{m{t}} = 3\ln 2 	imes \frac{3}{{4.5}} = 2\ln 2$

$ \Rightarrow \frac{{{{m{N}}_1}}}{{{{m{N}}_2}}} = 2 	imes {{m{e}}^{ - \ln 2}} = 1$

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