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(C) Let $N_0$ be the initial number of atoms of radioactive nuclide $X$. At any time $t$,the number of atoms of $X$ remaining is given by $N_X(t) = N_

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$	au$

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(C) Let $N_0$ be the initial number of atoms of radioactive nuclide $X$. At any time $t$,the number of atoms of $X$ remaining is given by $N_X(t) = N_0 e^{-\lambda t}$,where $\lambda = \frac{\ln(2)}{	au}$.Since $X$ decays into $Y$,the number of atoms of $Y$ at time $t$ is $N_Y(t) = N_0 - N_X(t) = N_0(1 - e^{-\lambda t})$.At the intersection point $T$,$N_X(T) = N_Y(T)$.Therefore,$N_0 e^{-\lambda T} = N_0(1 - e^{-\lambda T})$.Dividing by $N_0$,we get $e^{-\lambda T} = 1 - e^{-\lambda T}$,which implies $2e^{-\lambda T} = 1$,or $e^{-\lambda T} = \frac{1}{2}$.Taking the natural logarithm on both sides,$-\lambda T = \ln(1/2) = -\ln(2)$.Thus,$\lambda T = \ln(2)$.Substituting $\lambda = \frac{\ln(2)}{	au}$,we get $\left(\frac{\ln(2)}{	au}ight) T = \ln(2)$.Solving for $T$,we find $T = 	au$.

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 $\tau$. The growth curve for $Y$ intersects the decay curve for $X$ after time $T$. What is the time $T$?

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