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(C) Let $N_0$ be the initial number of atoms of $X$. At any time $T$,the number of atoms of $X$ remaining is $N_X(T) = N_0 e^{-\lambda T}$.Since $X$ d

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$t$

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(C) Let $N_0$ be the initial number of atoms of $X$. At any time $T$,the number of atoms of $X$ remaining is $N_X(T) = N_0 e^{-\lambda T}$.Since $X$ decays to $Y$,the number of atoms of $Y$ formed at time $T$ is $N_Y(T) = N_0(1 - e^{-\lambda T})$.At the intersection point,$N_X(T) = N_Y(T)$.Therefore,$N_0 e^{-\lambda T} = N_0(1 - e^{-\lambda T})$.$e^{-\lambda T} = 1 - e^{-\lambda T} \implies 2e^{-\lambda T} = 1 \implies e^{-\lambda T} = \frac{1}{2}$.Taking the natural logarithm on both sides,$-\lambda T = \ln(1/2) = -\ln(2)$.So,$T = \frac{\ln(2)}{\lambda}$.Since the half-life $t = \frac{\ln(2)}{\lambda}$,we get $T = t$.

The graph represents the decay of a newly prepared sample of radioactive nuclide $X$ to a stable nuclide $Y$. The half-life of $X$ is $t$. The growth curve for $Y$ intersects the decay curve for $X$ after time $T$. What is the time $T$?

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