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  • A radioactive element decays into another element with a decay constant <

A radioactive element \mathrm{A} decays into another element \mathrm{B} with a decay constant \mathrm{4 \lambda}, which then decays into a third element \mathrm{C} (decay constant \mathrm{=9 \lambda }).If act the atoms in the beginning consisted of atoms of type A only, then the ratio of the number of atoms of \mathrm{A} to that of type \mathrm{B} (ie.\mathrm{N_{A}/N_{B}}) when the number of atoms of \mathrm{B} is at a maximum 6 is      

Option: 1

\mathrm{\frac{3}{2}}


Option: 2

\frac{9}{4}


Option: 3

1


Option: 4

\mathrm{ln3/ln 2}


Answers (1)

best_answer

\mathrm{\frac{d N B}{d t}=\lambda_A N_A- \lambda_B N_B=4 \lambda N_A-9 \lambda N_B=0 } (when NB is max.)

\mathrm{\Rightarrow 4 N_A-9 N_B=0 }
\mathrm{\Rightarrow \frac{N_A}{N_B}=\frac{9}{4} }.

                                

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