A and B are two substances undergoing radioac | vedclass

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(A) The radioactive decay follows first-order kinetics: $[N]_t = [N]_0 e^{-kt}$,where $k = \frac{\ln 2}{t_{1/2}}$.For substance $A$: $[A]_t = [A]_0 e^

Vedclass · Accepted Answer

$15$

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(A) The radioactive decay follows first-order kinetics: $[N]_t = [N]_0 e^{-kt}$,where $k = \frac{\ln 2}{t_{1/2}}$.For substance $A$: $[A]_t = [A]_0 e^{-k_A t}$,where $k_A = \frac{\ln 2}{15}$.For substance $B$: $[B]_t = [B]_0 e^{-k_B t}$,where $k_B = \frac{\ln 2}{5}$ and $[B]_0 = 4[A]_0$.We want to find time $t$ such that $[A]_t = [B]_t$.$[A]_0 e^{-(\frac{\ln 2}{15})t} = 4[A]_0 e^{-(\frac{\ln 2}{5})t}$.Dividing both sides by $[A]_0 e^{-(\frac{\ln 2}{5})t}$:$e^{(\frac{\ln 2}{5} - \frac{\ln 2}{15})t} = 4$.Taking the natural logarithm on both sides:$t(\frac{\ln 2}{5} - \frac{\ln 2}{15}) = \ln 4$.$t(\ln 2)(\frac{1}{5} - \frac{1}{15}) = 2 \ln 2$.$t(\frac{3-1}{15}) = 2$.$t(\frac{2}{15}) = 2$.$t = 15 \ min$.

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