In an old rock,the ratio of uranium to lead n | vedclass

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(A) Let $N_U$ be the number of uranium nuclei remaining and $N_{Pb}$ be the number of lead nuclei formed.Given that the ratio $N_U : N_{Pb} = 1:1$,we

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$4.5 	imes 10^9$ years

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(A) Let $N_U$ be the number of uranium nuclei remaining and $N_{Pb}$ be the number of lead nuclei formed.Given that the ratio $N_U : N_{Pb} = 1:1$,we have $N_U = N_{Pb}$.The initial number of uranium nuclei $N_0 = N_U + N_{Pb} = N_U + N_U = 2N_U$.The law of radioactive decay is given by $N_U = N_0 \cdot (1/2)^{t/T_{1/2}}$.Substituting the values: $N_U = (2N_U) \cdot (1/2)^{t/T_{1/2}}$.Dividing by $N_U$: $1 = 2 \cdot (1/2)^{t/T_{1/2}}$,which simplifies to $1/2 = (1/2)^{t/T_{1/2}}$.Comparing the exponents: $t/T_{1/2} = 1$.Therefore,$t = T_{1/2} = 4.5 	imes 10^9$ years.

In an old rock,the ratio of uranium to lead nuclei is $1:1$. The half-life of uranium is $4.5 \times 10^9$ years. If the rock initially contained only uranium nuclei,how old is the rock?

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