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(C) Radioactive decay follows first-order kinetics. Let $N_0$ be the initial number of uranium atoms. At time $t$,the number of uranium atoms remainin

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$4.5 	imes 10^9$ years

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(C) Radioactive decay follows first-order kinetics. Let $N_0$ be the initial number of uranium atoms. At time $t$,the number of uranium atoms remaining is $N_t$ and the number of lead atoms formed is $N_0 - N_t$. Given that the number of uranium atoms equals the number of lead atoms,$N_t = N_0 - N_t$,which implies $N_t = N_0 / 2$. This means the time elapsed is equal to one half-life period $(t_{1/2})$. Therefore,the age of the rock $t = t_{1/2} = 4.5 	imes 10^9$ years.

In a radioactive decay,a uranium atom transforms into a lead atom. If a rock sample from the moon contains an equal number of uranium and lead atoms,and the half-life $(t_{1/2})$ for uranium is $4.5 \times 10^9$ years,then the age of the rock will be:

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